How to use a decimal range step value

Question

Is there a way to step between 0 and 1 by 0 1    I thought I could do it like the following  but it failed   for i in range 0  1  0 1       print i   Instead  it says that the step argument cannot be zero  which I did not expect

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Increase the magnitude of i for the loop and then reduce it when you need it   for i   100 in range 0  100  10       print i   100 0   EDIT  I honestly cannot remember why I thought that would work syntactically  for i in range 0  11  1       print i   10 0   That should have the desired output

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more itertools is a third-party library that implements a numeric range tool   import more itertools as mit   for x in mit numeric range 0  1  0 1       print     1f   format x     Output  0 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9   This tool also works for Decimal and Fraction

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You can use this function   def frange start end step       return map lambda x  x step  range int start 1  step  int end 1  step

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Rather than using a decimal step directly  it s much safer to express this in terms of how many points you want  Otherwise  floating-point rounding error is likely to give you a wrong result   You can use the linspace function from the NumPy library  which isn t part of the standard library but is relatively easy to obtain   linspace takes a number of points to return  and also lets you specify whether or not to include the right endpoint    gt  gt  gt  np linspace 0 1 11  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9   1      gt  gt  gt  np linspace 0 1 10 endpoint False  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9     If you really want to use a floating-point step value  you can  with numpy arange    gt  gt  gt  import numpy as np  gt  gt  gt  np arange 0 0  1 0  0 1  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9     Floating-point rounding error will cause problems  though  Here s a simple case where rounding error causes arange to produce a length-4 array when it should only produce 3 numbers    gt  gt  gt  numpy arange 1  1 3  0 1  array  1    1 1  1 2  1 3

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And if you do this often  you might want to save the generated list r  r map lambda x  x 10 0 range 0 10   for i in r      print i

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Building on  xrange  start   stop   step     you can define a generator that accepts and produces any type you choose  stick to types supporting   and  lt      gt  gt  gt  def drange start  stop  step           r   start         while r  lt  stop              yield r             r    step               gt  gt  gt  i0 drange 0 0  1 0  0 1   gt  gt  gt     g    x for x in i0    0    0 1    0 2    0 3    0 4    0 5    0 6    0 7    0 8    0 9    1    gt  gt  gt

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To counter the float precision issues  you could use the Decimal module  This demands an extra effort of converting to Decimal from int or float while writing the code  but you can instead pass str and modify the function if that sort of convenience is indeed necessary  from decimal import Decimal   def decimal range  args        zero  one   Decimal  0    Decimal  1        if len args     1          start  stop  step   zero  args 0   one     elif len args     2          start  stop  step   args    one       elif len args     3          start  stop  step   args     else          raise ValueError  Expected 1 or 2 arguments  got  s    len args        if not all  type arg     Decimal for arg in  start  stop  step             raise ValueError  Arguments must be passed as  lt type  Decimal gt           neglect bad cases     if  start    stop  or  start  gt  stop and step  gt   zero  or                              start  lt  stop and step  lt   zero           return         current   start     while abs current   lt  abs stop           yield current         current    step  Sample outputs - from decimal import Decimal as D  list decimal range D  2        Decimal  0    Decimal  1    list decimal range D  2    D  4 5        Decimal  2    Decimal  3    Decimal  4    list decimal range D  2    D  4 5    D  0 5        Decimal  2    Decimal  2 5    Decimal  3 0    Decimal  3 5    Decimal  4 0    list decimal range D  2    D  4 5    D  -0 5          list decimal range D  2    D  -4 5    D  -0 5        Decimal  2       Decimal  1 5       Decimal  1 0       Decimal  0 5       Decimal  0 0       Decimal  -0 5       Decimal  -1 0       Decimal  -1 5       Decimal  -2 0       Decimal  -2 5       Decimal  -3 0       Decimal  -3 5       Decimal  -4 0

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My solution   def seq start  stop  step 1  digit 0       x   float start      v          while x  lt   stop          v append round x digit           x    step     return v

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It can be done using Numpy library  arange   function allows steps in float  But  it returns a numpy array which can be converted to list using tolist   for our convenience   for i in np arange 0  1  0 1  tolist       print i

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Add auto-correction for the possibility of an incorrect sign on step   def frange start step stop       step    2   stop gt start   step lt 0  -1     return  start i step for i in range int  stop-start  step

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sign   lambda x   1  -1  x  lt  0  def frange start  stop  step       i   0     r len str step  split      -1       args  start stop step      if not step  return        if all int i   float i  for i in args           start stop step map int args      if sign step   1          while start   i   step  lt  stop              yield round start   i   step r              i    1     else          while start   i   step  gt  stop              yield round start   i   step r              i    1

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And if you do this often  you might want to save the generated list r  r map lambda x  x 10 0 range 0 10   for i in r      print i

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It can be done using Numpy library  arange   function allows steps in float  But  it returns a numpy array which can be converted to list using tolist   for our convenience   for i in np arange 0  1  0 1  tolist       print i

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This is my solution to get ranges with float steps  Using this function it s not necessary to import numpy  nor install it  I m pretty sure that it could be improved and optimized  Feel free to do it and post it here   from   future   import division from math import log  def xfrange start  stop  step        old start   start  backup this value      digits   int round log 10000  10    1  get number of digits     magnitude   10  digits     stop   int magnitude   stop   convert from      step   int magnitude   step   0 1 to 10  e g        if start    0          start   10   digits-1      else          start   10   digits  start      data         create array       calc number of iterations     end loop   int  stop-start   step      if old start    0          end loop    1      acc   start      for i in xrange 0  end loop           data append acc magnitude          acc    step      return data  print xfrange 1  2 1  0 1  print xfrange 0  1 1  0 1  print xfrange -1  0 1  0 1    The output is    1 0  1 1  1 2  1 3  1 4  1 5  1 6  1 7  1 8  1 9  2 0   0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9  1 0  1 1   -1 0  -0 9  -0 8  -0 7  -0 6  -0 5  -0 4  -0 3  -0 2  -0 1  0 0

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The range   built-in function returns a sequence of integer values  I m afraid  so you can t use it to do a decimal step     I d say just use a while loop   i   0 0 while i  lt   1 0      print i     i    0 1   If you re curious  Python is converting your 0 1 to 0  which is why it s telling you the argument can t be zero

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Lots of the solutions here still had floating point errors in Python 3 6 and didnt do exactly what I personally needed    Function below takes integers or floats  doesnt require imports and doesnt return floating point errors   def frange x  y  step       if int x   y   step      x   y   step           r   list range int x   int y   int step        else          f   10     len str step   - str step  find      - 1          rf   list range int x   f   int y   f   int step   f            r    i   f for i in rf       return r

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And if you do this often  you might want to save the generated list r  r map lambda x  x 10 0 range 0 10   for i in r      print i

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The trick to avoid round-off problem is to use a separate number to move through the range  that starts and half the step ahead of start     floating point range def frange a  b  stp 1 0     i   a stp 2 0   while i lt b      yield a     a    stp     i    stp   Alternatively  numpy arange can be used

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You can use this function   def frange start end step       return map lambda x  x step  range int start 1  step  int end 1  step

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This one liner will not clutter your code  The sign of the step parameter is important   def frange start  stop  step       return  x step start for x in range 0 round abs  stop-start  step  0 5001           int  stop-start  step lt 0  -2 1

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Here s a solution using itertools   import itertools  def seq start  end  step       if step    0          raise ValueError  step must not be 0       sample count   int abs end - start    step      return itertools islice itertools count start  step   sample count    Usage Example   for i in seq 0  1  0 1       print i

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Increase the magnitude of i for the loop and then reduce it when you need it   for i   100 in range 0  100  10       print i   100 0   EDIT  I honestly cannot remember why I thought that would work syntactically  for i in range 0  11  1       print i   10 0   That should have the desired output

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Python s range   can only do integers  not floating point  In your specific case  you can use a list comprehension instead    x   0 1 for x in range 0  10      Replace the call to range with that expression     For the more general case  you may want to write a custom function or generator

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frange start  stop  precision   def frange a b i       p   10  i     sr   a p     er    b p    1     p   float p      return map lambda x  x p  xrange sr er    In  gt frange -1 1 1   Out gt  -1 0  -0 9  -0 8  -0 7  -0 6  -0 5  -0 4  -0 3  -0 2  -0 1  0 0  0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9  1 0

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For completeness of boutique  a functional solution   def frange a b s     return    if s  gt  0 and a  gt  b or s  lt  0 and a  lt  b or s  0 else  a  frange a s b s

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start and stop are inclusive rather than one or the other  usually stop is excluded  and without imports  and using generators  def rangef start  stop  step  fround 5               Yields sequence of numbers from start  inclusive  to stop  inclusive      by step  increment  with rounding set to n digits        param start  start of sequence      param stop  end of sequence      param step  int or float increment  e g  1 or 0 001       param fround  float rounding  n decimal places      return              try          i   0         while stop  gt   start and step  gt  0              if i  0                  yield start             elif start  gt   stop                  yield stop             elif start  lt  stop                  if start    0                      yield 0                 if start    0                      yield start             i    1             start    step             start   round start  fround          else              pass     except TypeError as e          yield  type-error      format e      else          pass     passing print list rangef -100 0 10 0 1    print list rangef -100 0 0 5    print list rangef -1 1 0 2    print list rangef -1 1 0 1    print list rangef -1 1 0 05    print list rangef -1 1 0 02    print list rangef -1 1 0 01    print list rangef -1 1 0 005      failing  type-error  print list rangef  1   10   1     print list rangef 1 10  1          Python 3 6 2  v3 6 2 5fd33b5  Jul  8 2017  04 57 36   MSC v 1900 64   bit  AMD64

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NumPy is a bit overkill  I think    p 10 for p in range 0  10    0 0  0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9    Generally speaking  to do a step-by-1 x up to y you would do  x 100 y 2  p x for p in range 0  int x y     0 0  0 01  0 02  0 03       1 97  1 98  1 99     1 x produced less rounding noise when I tested

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Python s range   can only do integers  not floating point  In your specific case  you can use a list comprehension instead    x   0 1 for x in range 0  10      Replace the call to range with that expression     For the more general case  you may want to write a custom function or generator

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import numpy as np for i in np arange 0  1  0 1        print i

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I know I m late to the party here  but here s a trivial generator solution that s working in 3 6   def floatRange  args       start  step   0  1     if len args     1          stop   args 0      elif len args     2          start  stop   args 0   args 1      elif len args     3          start  stop  step   args 0   args 1   args 2      else          raise TypeError  floatRange accepts 1  2  or 3 arguments    0  given   format len args        for num in start  step  stop          if not isinstance num   int  float                raise TypeError  floatRange only accepts float and integer arguments    0     1  given   format type num   str num        for x in range int  stop-start  step            yield start    x   step      return   then you can call it just like the original range      there s no error handling  but let me know if there is an error that can be reasonably caught  and I ll update  or you can update it  this is StackOverflow

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more itertools is a third-party library that implements a numeric range tool   import more itertools as mit   for x in mit numeric range 0  1  0 1       print     1f   format x     Output  0 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9   This tool also works for Decimal and Fraction

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Building on  xrange  start   stop   step     you can define a generator that accepts and produces any type you choose  stick to types supporting   and  lt      gt  gt  gt  def drange start  stop  step           r   start         while r  lt  stop              yield r             r    step               gt  gt  gt  i0 drange 0 0  1 0  0 1   gt  gt  gt     g    x for x in i0    0    0 1    0 2    0 3    0 4    0 5    0 6    0 7    0 8    0 9    1    gt  gt  gt

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Rather than using a decimal step directly  it s much safer to express this in terms of how many points you want  Otherwise  floating-point rounding error is likely to give you a wrong result   You can use the linspace function from the NumPy library  which isn t part of the standard library but is relatively easy to obtain   linspace takes a number of points to return  and also lets you specify whether or not to include the right endpoint    gt  gt  gt  np linspace 0 1 11  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9   1      gt  gt  gt  np linspace 0 1 10 endpoint False  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9     If you really want to use a floating-point step value  you can  with numpy arange    gt  gt  gt  import numpy as np  gt  gt  gt  np arange 0 0  1 0  0 1  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9     Floating-point rounding error will cause problems  though  Here s a simple case where rounding error causes arange to produce a length-4 array when it should only produce 3 numbers    gt  gt  gt  numpy arange 1  1 3  0 1  array  1    1 1  1 2  1 3

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Rather than using a decimal step directly  it s much safer to express this in terms of how many points you want  Otherwise  floating-point rounding error is likely to give you a wrong result   You can use the linspace function from the NumPy library  which isn t part of the standard library but is relatively easy to obtain   linspace takes a number of points to return  and also lets you specify whether or not to include the right endpoint    gt  gt  gt  np linspace 0 1 11  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9   1      gt  gt  gt  np linspace 0 1 10 endpoint False  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9     If you really want to use a floating-point step value  you can  with numpy arange    gt  gt  gt  import numpy as np  gt  gt  gt  np arange 0 0  1 0  0 1  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9     Floating-point rounding error will cause problems  though  Here s a simple case where rounding error causes arange to produce a length-4 array when it should only produce 3 numbers    gt  gt  gt  numpy arange 1  1 3  0 1  array  1    1 1  1 2  1 3

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frange start  stop  precision   def frange a b i       p   10  i     sr   a p     er    b p    1     p   float p      return map lambda x  x p  xrange sr er    In  gt frange -1 1 1   Out gt  -1 0  -0 9  -0 8  -0 7  -0 6  -0 5  -0 4  -0 3  -0 2  -0 1  0 0  0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9  1 0

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Here is my solution which works fine with float range -1  0  0 01  and works without floating point representation errors  It is not very fast  but works fine       from decimal import Decimal  def get multiplier  from   to  step       digits          for number in   from   to  step           pre   Decimal str number     1         digit   len str pre   - 2         digits append digit      max digits   max digits      return float 10     max digits     def float range  from   to  step  include False          Generates a range list of floating point values over the Range  start  stop         with step size step        include True - allows to include right value to if possible           Works fine with floating point representation                mult   get multiplier  from   to  step        print mult     int from   int round  from   mult       int to   int round  to   mult       int step   int round step   mult         print int from int to int step     if include          result   range int from  int to   int step  int step          result    r for r in result if r  lt   int to      else          result   range int from  int to  int step        print result     float result    r   mult for r in result      return float result   print float range -1  0  0 01 include False   assert float range 1 01  2 06  5 05   1  True       1 01  1 06  1 11  1 16  1 21  1 26  1 31  1 36  1 41  1 46  1 51  1 56  1 61  1 66  1 71  1 76  1 81  1 86  1 91  1 96  2 01  2 06   assert float range 1 01  2 06  5 05   1  False      1 01  1 06  1 11  1 16  1 21  1 26  1 31  1 36  1 41  1 46  1 51  1 56  1 61  1 66  1 71  1 76  1 81  1 86  1 91  1 96  2 01

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start and stop are inclusive rather than one or the other  usually stop is excluded  and without imports  and using generators  def rangef start  stop  step  fround 5               Yields sequence of numbers from start  inclusive  to stop  inclusive      by step  increment  with rounding set to n digits        param start  start of sequence      param stop  end of sequence      param step  int or float increment  e g  1 or 0 001       param fround  float rounding  n decimal places      return              try          i   0         while stop  gt   start and step  gt  0              if i  0                  yield start             elif start  gt   stop                  yield stop             elif start  lt  stop                  if start    0                      yield 0                 if start    0                      yield start             i    1             start    step             start   round start  fround          else              pass     except TypeError as e          yield  type-error      format e      else          pass     passing print list rangef -100 0 10 0 1    print list rangef -100 0 0 5    print list rangef -1 1 0 2    print list rangef -1 1 0 1    print list rangef -1 1 0 05    print list rangef -1 1 0 02    print list rangef -1 1 0 01    print list rangef -1 1 0 005      failing  type-error  print list rangef  1   10   1     print list rangef 1 10  1          Python 3 6 2  v3 6 2 5fd33b5  Jul  8 2017  04 57 36   MSC v 1900 64   bit  AMD64

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Building on  xrange  start   stop   step     you can define a generator that accepts and produces any type you choose  stick to types supporting   and  lt      gt  gt  gt  def drange start  stop  step           r   start         while r  lt  stop              yield r             r    step               gt  gt  gt  i0 drange 0 0  1 0  0 1   gt  gt  gt     g    x for x in i0    0    0 1    0 2    0 3    0 4    0 5    0 6    0 7    0 8    0 9    1    gt  gt  gt

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For completeness of boutique  a functional solution   def frange a b s     return    if s  gt  0 and a  gt  b or s  lt  0 and a  lt  b or s  0 else  a  frange a s b s

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Python s range   can only do integers  not floating point  In your specific case  you can use a list comprehension instead    x   0 1 for x in range 0  10      Replace the call to range with that expression     For the more general case  you may want to write a custom function or generator

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I know I m late to the party here  but here s a trivial generator solution that s working in 3 6   def floatRange  args       start  step   0  1     if len args     1          stop   args 0      elif len args     2          start  stop   args 0   args 1      elif len args     3          start  stop  step   args 0   args 1   args 2      else          raise TypeError  floatRange accepts 1  2  or 3 arguments    0  given   format len args        for num in start  step  stop          if not isinstance num   int  float                raise TypeError  floatRange only accepts float and integer arguments    0     1  given   format type num   str num        for x in range int  stop-start  step            yield start    x   step      return   then you can call it just like the original range      there s no error handling  but let me know if there is an error that can be reasonably caught  and I ll update  or you can update it  this is StackOverflow

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The range   built-in function returns a sequence of integer values  I m afraid  so you can t use it to do a decimal step     I d say just use a while loop   i   0 0 while i  lt   1 0      print i     i    0 1   If you re curious  Python is converting your 0 1 to 0  which is why it s telling you the argument can t be zero

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My versions use the original range function to create multiplicative indices for the shift  This allows same syntax to the original range function  I have made two versions  one using float  and one using Decimal  because I found that in some cases I wanted to avoid the roundoff drift introduced by the floating point arithmetic   It is consistent with empty set results as in range xrange   Passing only a single numeric value to either function will return the standard range output to the integer ceiling value of the input parameter  so if you gave it 5 5  it would return range 6     Edit  the code below is now available as package on pypi  Franges     frange py from math import ceil   find best range function available to version  2 7 x   3 x x  try       xrange   xrange except NameError       xrange   range  def frange start  stop   None  step   1          frange generates a set of floating point values over the      range  start  stop  with step size step      frange  start   stop    step            if stop is None          for x in  xrange int ceil start                 yield x     else            create a generator expression for the index values         indices    i for i in  xrange 0  int  stop-start  step                yield results         for i in indices              yield start   step i     drange py import decimal from math import ceil   find best range function available to version  2 7 x   3 x x  try       xrange   xrange except NameError       xrange   range  def drange start  stop   None  step   1  precision   None          drange generates a set of Decimal values over the     range  start  stop  with step size step      drange  start   stop   step   precision            if stop is None          for x in  xrange int ceil start                 yield x     else            find precision         if precision is not None              decimal getcontext   prec   precision           convert values to decimals         start   decimal Decimal start          stop   decimal Decimal stop          step   decimal Decimal step            create a generator expression for the index values         indices                 i for i in  xrange                  0                     stop-start  step  to integral value                                       yield results         for i in indices              yield float start   step i      testranges py import frange import drange list frange frange 0  2  0 5      0 0  0 5  1 0  1 5  list drange drange 0  2  0 5  precision   6      0 0  0 5  1 0  1 5  list frange frange 3      0  1  2  list frange frange 3 5      0  1  2  3  list frange frange 0 10  -1

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Increase the magnitude of i for the loop and then reduce it when you need it   for i   100 in range 0  100  10       print i   100 0   EDIT  I honestly cannot remember why I thought that would work syntactically  for i in range 0  11  1       print i   10 0   That should have the desired output

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Rather than using a decimal step directly  it s much safer to express this in terms of how many points you want  Otherwise  floating-point rounding error is likely to give you a wrong result   You can use the linspace function from the NumPy library  which isn t part of the standard library but is relatively easy to obtain   linspace takes a number of points to return  and also lets you specify whether or not to include the right endpoint    gt  gt  gt  np linspace 0 1 11  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9   1      gt  gt  gt  np linspace 0 1 10 endpoint False  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9     If you really want to use a floating-point step value  you can  with numpy arange    gt  gt  gt  import numpy as np  gt  gt  gt  np arange 0 0  1 0  0 1  array   0     0 1   0 2   0 3   0 4   0 5   0 6   0 7   0 8   0 9     Floating-point rounding error will cause problems  though  Here s a simple case where rounding error causes arange to produce a length-4 array when it should only produce 3 numbers    gt  gt  gt  numpy arange 1  1 3  0 1  array  1    1 1  1 2  1 3

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To counter the float precision issues  you could use the Decimal module  This demands an extra effort of converting to Decimal from int or float while writing the code  but you can instead pass str and modify the function if that sort of convenience is indeed necessary  from decimal import Decimal   def decimal range  args        zero  one   Decimal  0    Decimal  1        if len args     1          start  stop  step   zero  args 0   one     elif len args     2          start  stop  step   args    one       elif len args     3          start  stop  step   args     else          raise ValueError  Expected 1 or 2 arguments  got  s    len args        if not all  type arg     Decimal for arg in  start  stop  step             raise ValueError  Arguments must be passed as  lt type  Decimal gt           neglect bad cases     if  start    stop  or  start  gt  stop and step  gt   zero  or                              start  lt  stop and step  lt   zero           return         current   start     while abs current   lt  abs stop           yield current         current    step  Sample outputs - from decimal import Decimal as D  list decimal range D  2        Decimal  0    Decimal  1    list decimal range D  2    D  4 5        Decimal  2    Decimal  3    Decimal  4    list decimal range D  2    D  4 5    D  0 5        Decimal  2    Decimal  2 5    Decimal  3 0    Decimal  3 5    Decimal  4 0    list decimal range D  2    D  4 5    D  -0 5          list decimal range D  2    D  -4 5    D  -0 5        Decimal  2       Decimal  1 5       Decimal  1 0       Decimal  0 5       Decimal  0 0       Decimal  -0 5       Decimal  -1 0       Decimal  -1 5       Decimal  -2 0       Decimal  -2 5       Decimal  -3 0       Decimal  -3 5       Decimal  -4 0

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This is my solution to get ranges with float steps  Using this function it s not necessary to import numpy  nor install it  I m pretty sure that it could be improved and optimized  Feel free to do it and post it here   from   future   import division from math import log  def xfrange start  stop  step        old start   start  backup this value      digits   int round log 10000  10    1  get number of digits     magnitude   10  digits     stop   int magnitude   stop   convert from      step   int magnitude   step   0 1 to 10  e g        if start    0          start   10   digits-1      else          start   10   digits  start      data         create array       calc number of iterations     end loop   int  stop-start   step      if old start    0          end loop    1      acc   start      for i in xrange 0  end loop           data append acc magnitude          acc    step      return data  print xfrange 1  2 1  0 1  print xfrange 0  1 1  0 1  print xfrange -1  0 1  0 1    The output is    1 0  1 1  1 2  1 3  1 4  1 5  1 6  1 7  1 8  1 9  2 0   0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9  1 0  1 1   -1 0  -0 9  -0 8  -0 7  -0 6  -0 5  -0 4  -0 3  -0 2  -0 1  0 0

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Suprised no-one has yet mentioned the recommended solution in the Python 3 docs      See also          The linspace recipe shows how to implement a lazy version of range    that suitable for floating point applications       Once defined  the recipe is easy to use and does not require numpy or any other external libraries  but functions like numpy linspace    Note that rather than a step argument  the third num argument specifies the number of desired values  for example   print linspace 0  10  5     linspace 0  10  5  print list linspace 0  10  5       0 0  2 5  5 0  7 5  10    I quote a modified version of the full Python 3 recipe from  Andrew Barnert below   import collections abc import numbers  class linspace collections abc Sequence          linspace start  stop  num  - gt  linspace object      Return a virtual sequence of num numbers from start to stop  inclusive        If you need a half-open range  use linspace start  stop  num 1   -1               def   init   self  start  stop  num           if not isinstance num  numbers Integral  or num  lt   1              raise ValueError  num must be an integer  gt  1           self start  self stop  self num   start  stop  num         self step    stop-start   num-1      def   len   self           return self num     def   getitem   self  i           if isinstance i  slice               return  self x  for x in range  i indices len self             if i  lt  0              i   self num   i         if i  gt   self num              raise IndexError  linspace object index out of range           if i    self num-1              return self stop         return self start   i self step     def   repr   self           return                  format type self    name                                           self start  self stop  self num      def   eq   self  other           if not isinstance other  linspace               return False         return   self start  self stop  self num                      other start  other stop  other num       def   ne   self  other           return not self  other     def   hash   self           return hash  type self   self start  self stop  self num

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This one liner will not clutter your code  The sign of the step parameter is important   def frange start  stop  step       return  x step start for x in range 0 round abs  stop-start  step  0 5001           int  stop-start  step lt 0  -2 1

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Best Solution  no rounding error  gt  gt  gt  step    1  gt  gt  gt  N   10       number of data points  gt  gt  gt    x   pow step  -1  for x in range 0  N   1      0 0  0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9  1 0   Or  for a set range instead of set data points  e g  continuous function   use   gt  gt  gt  step    1  gt  gt  gt  rnge   1       NOTE range   1  i e  span of data points  gt  gt  gt  N   int rnge   step  gt  gt  gt    x   pow step -1  for x in range 0  N   1      0 0  0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9  1 0   To implement a function  replace x   pow step  -1  with f  x   pow step  -1     and define f  For example   gt  gt  gt  import math  gt  gt  gt  def f x           return math sin x    gt  gt  gt  step    1  gt  gt  gt  rnge   1       NOTE range   1  i e  span of data points  gt  gt  gt  N   int rnge   step   gt  gt  gt    f  x   pow step -1    for x in range 0  N   1      0 0  0 09983341664682815  0 19866933079506122  0 29552020666133955  0 3894183423086505    0 479425538604203  0 5646424733950354  0 644217687237691  0 7173560908995228   0 7833269096274834  0 8414709848078965

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sign   lambda x   1  -1  x  lt  0  def frange start  stop  step       i   0     r len str step  split      -1       args  start stop step      if not step  return        if all int i   float i  for i in args           start stop step map int args      if sign step   1          while start   i   step  lt  stop              yield round start   i   step r              i    1     else          while start   i   step  gt  stop              yield round start   i   step r              i    1

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x   0 1 for x in range 0  10      in Python 2 7x gives you the result of       0 0  0 1  0 2  0 30000000000000004  0 4  0 5  0 6000000000000001  0 7000000000000001  0 8  0 9    but if you use     round x   0 1  1  for x in range 0  10     gives you the desired       0 0  0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9

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Python s range   can only do integers  not floating point  In your specific case  you can use a list comprehension instead    x   0 1 for x in range 0  10      Replace the call to range with that expression     For the more general case  you may want to write a custom function or generator

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The range   built-in function returns a sequence of integer values  I m afraid  so you can t use it to do a decimal step     I d say just use a while loop   i   0 0 while i  lt   1 0      print i     i    0 1   If you re curious  Python is converting your 0 1 to 0  which is why it s telling you the argument can t be zero

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scipy has a built in function arange which generalizes Python s range   constructor to satisfy your requirement of float handling    from scipy import arange

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My answer is similar to others using map    without need of NumPy  and without using lambda  though you could      To get a list of float values from 0 0 to t max in steps of dt   def xdt n       return dt float n  tlist    map xdt  range int t max dt  1

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NumPy is a bit overkill  I think    p 10 for p in range 0  10    0 0  0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9    Generally speaking  to do a step-by-1 x up to y you would do  x 100 y 2  p x for p in range 0  int x y     0 0  0 01  0 02  0 03       1 97  1 98  1 99     1 x produced less rounding noise when I tested

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Suprised no-one has yet mentioned the recommended solution in the Python 3 docs      See also          The linspace recipe shows how to implement a lazy version of range    that suitable for floating point applications       Once defined  the recipe is easy to use and does not require numpy or any other external libraries  but functions like numpy linspace    Note that rather than a step argument  the third num argument specifies the number of desired values  for example   print linspace 0  10  5     linspace 0  10  5  print list linspace 0  10  5       0 0  2 5  5 0  7 5  10    I quote a modified version of the full Python 3 recipe from  Andrew Barnert below   import collections abc import numbers  class linspace collections abc Sequence          linspace start  stop  num  - gt  linspace object      Return a virtual sequence of num numbers from start to stop  inclusive        If you need a half-open range  use linspace start  stop  num 1   -1               def   init   self  start  stop  num           if not isinstance num  numbers Integral  or num  lt   1              raise ValueError  num must be an integer  gt  1           self start  self stop  self num   start  stop  num         self step    stop-start   num-1      def   len   self           return self num     def   getitem   self  i           if isinstance i  slice               return  self x  for x in range  i indices len self             if i  lt  0              i   self num   i         if i  gt   self num              raise IndexError  linspace object index out of range           if i    self num-1              return self stop         return self start   i self step     def   repr   self           return                  format type self    name                                           self start  self stop  self num      def   eq   self  other           if not isinstance other  linspace               return False         return   self start  self stop  self num                      other start  other stop  other num       def   ne   self  other           return not self  other     def   hash   self           return hash  type self   self start  self stop  self num

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Here s a solution using itertools   import itertools  def seq start  end  step       if step    0          raise ValueError  step must not be 0       sample count   int abs end - start    step      return itertools islice itertools count start  step   sample count    Usage Example   for i in seq 0  1  0 1       print i

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import numpy as np for i in np arange 0  1  0 1        print i

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x   0 1 for x in range 0  10      in Python 2 7x gives you the result of       0 0  0 1  0 2  0 30000000000000004  0 4  0 5  0 6000000000000001  0 7000000000000001  0 8  0 9    but if you use     round x   0 1  1  for x in range 0  10     gives you the desired       0 0  0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9

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I am only a beginner  but I had the same problem  when simulating some calculations  Here is how I attempted to work this out  which seems to be working with decimal steps   I am also quite lazy and so I found it hard to write my own range function   Basically what I did is changed my xrange 0 0  1 0  0 01  to xrange 0  100  1  and used the division by 100 0 inside the loop  I was also concerned  if there will be rounding mistakes  So I decided to test  whether there are any  Now I heard  that if for example 0 01 from a calculation isn t exactly the float 0 01 comparing them should return False  if I am wrong  please let me know    So I decided to test if my solution will work for my range by running a short test   for d100 in xrange 0  100  1       d   d100   100 0     fl   float  0 00   4 - len str d100      str d100       print d       fl   d    fl   And it printed True for each   Now  if I m getting it totally wrong  please let me know

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My solution   def seq start  stop  step 1  digit 0       x   float start      v          while x  lt   stop          v append round x digit           x    step     return v

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Similar to R s seq function  this one returns a sequence in any order given the correct step value  The last value is equal to the stop value    def seq start  stop  step 1       n   int round  stop - start  float step        if n  gt  1          return  start   step i for i in range n 1        elif n    1          return  start       else          return       Results  seq 1  5  0 5        1 0  1 5  2 0  2 5  3 0  3 5  4 0  4 5  5 0    seq 10  0  -1        10  9  8  7  6  5  4  3  2  1  0    seq 10  0  -2        10  8  6  4  2  0    seq 1  1         1

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Increase the magnitude of i for the loop and then reduce it when you need it   for i   100 in range 0  100  10       print i   100 0   EDIT  I honestly cannot remember why I thought that would work syntactically  for i in range 0  11  1       print i   10 0   That should have the desired output

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Add auto-correction for the possibility of an incorrect sign on step   def frange start step stop       step    2   stop gt start   step lt 0  -1     return  start i step for i in range int  stop-start  step

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Here is my solution which works fine with float range -1  0  0 01  and works without floating point representation errors  It is not very fast  but works fine       from decimal import Decimal  def get multiplier  from   to  step       digits          for number in   from   to  step           pre   Decimal str number     1         digit   len str pre   - 2         digits append digit      max digits   max digits      return float 10     max digits     def float range  from   to  step  include False          Generates a range list of floating point values over the Range  start  stop         with step size step        include True - allows to include right value to if possible           Works fine with floating point representation                mult   get multiplier  from   to  step        print mult     int from   int round  from   mult       int to   int round  to   mult       int step   int round step   mult         print int from int to int step     if include          result   range int from  int to   int step  int step          result    r for r in result if r  lt   int to      else          result   range int from  int to  int step        print result     float result    r   mult for r in result      return float result   print float range -1  0  0 01 include False   assert float range 1 01  2 06  5 05   1  True       1 01  1 06  1 11  1 16  1 21  1 26  1 31  1 36  1 41  1 46  1 51  1 56  1 61  1 66  1 71  1 76  1 81  1 86  1 91  1 96  2 01  2 06   assert float range 1 01  2 06  5 05   1  False      1 01  1 06  1 11  1 16  1 21  1 26  1 31  1 36  1 41  1 46  1 51  1 56  1 61  1 66  1 71  1 76  1 81  1 86  1 91  1 96  2 01

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I am only a beginner  but I had the same problem  when simulating some calculations  Here is how I attempted to work this out  which seems to be working with decimal steps   I am also quite lazy and so I found it hard to write my own range function   Basically what I did is changed my xrange 0 0  1 0  0 01  to xrange 0  100  1  and used the division by 100 0 inside the loop  I was also concerned  if there will be rounding mistakes  So I decided to test  whether there are any  Now I heard  that if for example 0 01 from a calculation isn t exactly the float 0 01 comparing them should return False  if I am wrong  please let me know    So I decided to test if my solution will work for my range by running a short test   for d100 in xrange 0  100  1       d   d100   100 0     fl   float  0 00   4 - len str d100      str d100       print d       fl   d    fl   And it printed True for each   Now  if I m getting it totally wrong  please let me know

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Similar to R s seq function  this one returns a sequence in any order given the correct step value  The last value is equal to the stop value    def seq start  stop  step 1       n   int round  stop - start  float step        if n  gt  1          return  start   step i for i in range n 1        elif n    1          return  start       else          return       Results  seq 1  5  0 5        1 0  1 5  2 0  2 5  3 0  3 5  4 0  4 5  5 0    seq 10  0  -1        10  9  8  7  6  5  4  3  2  1  0    seq 10  0  -2        10  8  6  4  2  0    seq 1  1         1

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My answer is similar to others using map    without need of NumPy  and without using lambda  though you could      To get a list of float values from 0 0 to t max in steps of dt   def xdt n       return dt float n  tlist    map xdt  range int t max dt  1

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Building on  xrange  start   stop   step     you can define a generator that accepts and produces any type you choose  stick to types supporting   and  lt      gt  gt  gt  def drange start  stop  step           r   start         while r  lt  stop              yield r             r    step               gt  gt  gt  i0 drange 0 0  1 0  0 1   gt  gt  gt     g    x for x in i0    0    0 1    0 2    0 3    0 4    0 5    0 6    0 7    0 8    0 9    1    gt  gt  gt

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Best Solution  no rounding error  gt  gt  gt  step    1  gt  gt  gt  N   10       number of data points  gt  gt  gt    x   pow step  -1  for x in range 0  N   1      0 0  0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9  1 0   Or  for a set range instead of set data points  e g  continuous function   use   gt  gt  gt  step    1  gt  gt  gt  rnge   1       NOTE range   1  i e  span of data points  gt  gt  gt  N   int rnge   step  gt  gt  gt    x   pow step -1  for x in range 0  N   1      0 0  0 1  0 2  0 3  0 4  0 5  0 6  0 7  0 8  0 9  1 0   To implement a function  replace x   pow step  -1  with f  x   pow step  -1     and define f  For example   gt  gt  gt  import math  gt  gt  gt  def f x           return math sin x    gt  gt  gt  step    1  gt  gt  gt  rnge   1       NOTE range   1  i e  span of data points  gt  gt  gt  N   int rnge   step   gt  gt  gt    f  x   pow step -1    for x in range 0  N   1      0 0  0 09983341664682815  0 19866933079506122  0 29552020666133955  0 3894183423086505    0 479425538604203  0 5646424733950354  0 644217687237691  0 7173560908995228   0 7833269096274834  0 8414709848078965

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And if you do this often  you might want to save the generated list r  r map lambda x  x 10 0 range 0 10   for i in r      print i

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Lots of the solutions here still had floating point errors in Python 3 6 and didnt do exactly what I personally needed    Function below takes integers or floats  doesnt require imports and doesnt return floating point errors   def frange x  y  step       if int x   y   step      x   y   step           r   list range int x   int y   int step        else          f   10     len str step   - str step  find      - 1          rf   list range int x   f   int y   f   int step   f            r    i   f for i in rf       return r

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My versions use the original range function to create multiplicative indices for the shift  This allows same syntax to the original range function  I have made two versions  one using float  and one using Decimal  because I found that in some cases I wanted to avoid the roundoff drift introduced by the floating point arithmetic   It is consistent with empty set results as in range xrange   Passing only a single numeric value to either function will return the standard range output to the integer ceiling value of the input parameter  so if you gave it 5 5  it would return range 6     Edit  the code below is now available as package on pypi  Franges     frange py from math import ceil   find best range function available to version  2 7 x   3 x x  try       xrange   xrange except NameError       xrange   range  def frange start  stop   None  step   1          frange generates a set of floating point values over the      range  start  stop  with step size step      frange  start   stop    step            if stop is None          for x in  xrange int ceil start                 yield x     else            create a generator expression for the index values         indices    i for i in  xrange 0  int  stop-start  step                yield results         for i in indices              yield start   step i     drange py import decimal from math import ceil   find best range function available to version  2 7 x   3 x x  try       xrange   xrange except NameError       xrange   range  def drange start  stop   None  step   1  precision   None          drange generates a set of Decimal values over the     range  start  stop  with step size step      drange  start   stop   step   precision            if stop is None          for x in  xrange int ceil start                 yield x     else            find precision         if precision is not None              decimal getcontext   prec   precision           convert values to decimals         start   decimal Decimal start          stop   decimal Decimal stop          step   decimal Decimal step            create a generator expression for the index values         indices                 i for i in  xrange                  0                     stop-start  step  to integral value                                       yield results         for i in indices              yield float start   step i      testranges py import frange import drange list frange frange 0  2  0 5      0 0  0 5  1 0  1 5  list drange drange 0  2  0 5  precision   6      0 0  0 5  1 0  1 5  list frange frange 3      0  1  2  list frange frange 3 5      0  1  2  3  list frange frange 0 10  -1

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scipy has a built in function arange which generalizes Python s range   constructor to satisfy your requirement of float handling    from scipy import arange

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The range   built-in function returns a sequence of integer values  I m afraid  so you can t use it to do a decimal step     I d say just use a while loop   i   0 0 while i  lt   1 0      print i     i    0 1   If you re curious  Python is converting your 0 1 to 0  which is why it s telling you the argument can t be zero

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The trick to avoid round-off problem is to use a separate number to move through the range  that starts and half the step ahead of start     floating point range def frange a  b  stp 1 0     i   a stp 2 0   while i lt b      yield a     a    stp     i    stp   Alternatively  numpy arange can be used

[python] How to use a decimal range() step value?

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