How to truncate float values

Question

I want to remove digits from a float to have a fixed number of digits after the dot  like  1 923328437452   1 923  I need to output as a string to another function  not print  Also I want to ignore the lost digits  not round them

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A general and simple function to use   def truncate float number  length          Truncate float numbers  up to the number specified     in length that must be an integer         number   number   pow 10  length      number   int number      number   float number      number    pow 10  length      return number

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gt  gt  gt  floor  1 23658945    10  4    10  4 1 2365     divide and multiply by 10  number of desired digits

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First  the function  for those who just want some copy-and-paste code   def truncate f  n          Truncates pads a float f to n decimal places without rounding        s        format f      if  e  in s or  E  in s          return   0   1 f   format f  n      i  p  d   s partition          return     join  i   d  0  n   n      This is valid in Python 2 7 and 3 1   For older versions  it s not possible to get the same  intelligent rounding  effect  at least  not without a lot of complicated code   but rounding to 12 decimal places before truncation will work much of the time   def truncate f  n          Truncates pads a float f to n decimal places without rounding        s      12f    f     i  p  d   s partition          return     join  i   d  0  n   n      Explanation  The core of the underlying method is to convert the value to a string at full precision and then just chop off everything beyond the desired number of characters  The latter step is easy  it can be done either with string manipulation  i  p  d   s partition          join  i   d  0  n   n      or the decimal module  str Decimal s  quantize Decimal  0   1    -n    rounding ROUND DOWN     The first step  converting to a string  is quite difficult because there are some pairs of floating point literals  i e  what you write in the source code  which both produce the same binary representation and yet should be truncated differently  For example  consider 0 3 and 0 29999999999999998  If you write 0 3 in a Python program  the compiler encodes it using the IEEE floating-point format into the sequence of bits  assuming a 64-bit float   0011111111010011001100110011001100110011001100110011001100110011   This is the closest value to 0 3 that can accurately be represented as an IEEE float  But if you write 0 29999999999999998 in a Python program  the compiler translates it into exactly the same value  In one case  you meant it to be truncated  to one digit  as 0 3  whereas in the other case you meant it to be truncated as 0 2  but Python can only give one answer  This is a fundamental limitation of Python  or indeed any programming language without lazy evaluation  The truncation function only has access to the binary value stored in the computer s memory  not the string you actually typed into the source code 1  If you decode the sequence of bits back into a decimal number  again using the IEEE 64-bit floating-point format  you get  0 2999999999999999888977697537484345957637      so a naive implementation would come up with 0 2 even though that s probably not what you want  For more on floating-point representation error  see the Python tutorial   It s very rare to be working with a floating-point value that is so close to a round number and yet is intentionally not equal to that round number  So when truncating  it probably makes sense to choose the  nicest  decimal representation out of all that could correspond to the value in memory  Python 2 7 and up  but not 3 0  includes a sophisticated algorithm to do just that  which we can access through the default string formatting operation        format f    The only caveat is that this acts like a g format specification  in the sense that it uses exponential notation  1 23e 4  if the number is large or small enough  So the method has to catch this case and handle it differently  There are a few cases where using an f format specification instead causes a problem  such as trying to truncate 3e-10 to 28 digits of precision  it produces 0 0000000002999999999999999980   and I m not yet sure how best to handle those   If you actually are working with floats that are very close to round numbers but intentionally not equal to them  like 0 29999999999999998 or 99 959999999999994   this will produce some false positives  i e  it ll round numbers that you didn t want rounded  In that case the solution is to specify a fixed precision     0   1 f   format f  sys float info dig   n   2    The number of digits of precision to use here doesn t really matter  it only needs to be large enough to ensure that any rounding performed in the string conversion doesn t  bump up  the value to its nice decimal representation  I think sys float info dig   n   2 may be enough in all cases  but if not that 2 might have to be increased  and it doesn t hurt to do so   In earlier versions of Python  up to 2 6  or 3 0   the floating point number formatting was a lot more crude  and would regularly produce things like   gt  gt  gt  1 1 1 1000000000000001   If this is your situation  if you do want to use  nice  decimal representations for truncation  all you can do  as far as I know  is pick some number of digits  less than the full precision representable by a float  and round the number to that many digits before truncating it  A typical choice is 12      12f    f   but you can adjust this to suit the numbers you re using     1Well    I lied  Technically  you can instruct Python to re-parse its own source code and extract the part corresponding to the first argument you pass to the truncation function  If that argument is a floating-point literal  you can just cut it off a certain number of places after the decimal point and return that  However this strategy doesn t work if the argument is a variable  which makes it fairly useless  The following is presented for entertainment value only   def trunc introspect f  n          Truncates pads the float f to n decimal places by looking at the caller s source code        current frame   None     caller frame   None     s   inspect stack       try          current frame   s 0          caller frame   s 1          gen   tokenize tokenize io BytesIO caller frame 4  caller frame 5   encode  utf-8    readline          for token type  token string          in gen              if token type    tokenize NAME and token string    current frame 3                   next gen    left parenthesis                 token type  token string            next gen    float literal                 if token type    tokenize NUMBER                      try                          cut point   token string index        n   1                     except ValueError    no decimal in string                         return token string          0    n                     else                          if len token string   lt  cut point                              token string     0     cut point - len token string                           return token string  cut point                  else                      raise ValueError  Unable to find floating-point literal  this probably means you called    with a variable   format current frame 3                    break     finally          del s  current frame  caller frame   Generalizing this to handle the case where you pass in a variable seems like a lost cause  since you d have to trace backwards through the program s execution until you find the floating-point literal which gave the variable its value  If there even is one  Most variables will be initialized from user input or mathematical expressions  in which case the binary representation is all there is

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If you fancy some mathemagic  this works for  ve numbers    gt  gt  gt  v   1 923328437452  gt  gt  gt  v - v   1e-3 1 923

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Here is an easy way   def truncate num  res 3       return  floor num pow 10  res  0 5   pow 10  res    for num   1 923328437452  this outputs 1 923

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So many of the answers given for this question are just completely wrong  They either round up floats  rather than truncate  or do not work for all cases   This is the top Google result when I search for  Python truncate float   a concept which is really straightforward  and which deserves better answers  I agree with Hatchkins that using the decimal module is the pythonic way of doing this  so I give here a function which I think answers the question correctly  and which works as expected for all cases   As a side-note  fractional values  in general  cannot be represented exactly by binary floating point variables  see here for a discussion of this   which is why my function returns a string    from decimal import Decimal  localcontext  ROUND DOWN  def truncate number  places       if not isinstance places  int           raise ValueError  Decimal places must be an integer        if places  lt  1          raise ValueError  Decimal places must be at least 1          If you want to truncate to 0 decimal places  just do int number        with localcontext   as context          context rounding   ROUND DOWN         exponent   Decimal str 10    - places           return Decimal str number   quantize exponent  to eng string

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value  value to be truncated   n  number of values after decimal  value   0 999782 n   3 float int value 1en   1e-n

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Am also a python newbie and after making use of some bits and pieces here  I offer my two cents  print str int time time     str datetime now   microsecond   3    str int time time     will take the time epoch as int and convert it to string and join with    str datetime now   microsecond   3  which returns the microseconds only  convert to string and truncate to first 3 chars

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Just wanted to mention that the old  make round   with floor    trick of  round f    floor f 0 5    can be turned around to make floor   from round    floor f    round f-0 5    Although both these rules break around negative numbers  so using it is less than ideal   def trunc f  n       if f  gt  0          return     f     n   f - 0 5 10  -n       elif f    0          return     f     n  f      elif f  lt  0          return     f     n   f   0 5 10  -n

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There is an easy workaround in python 3  Where to cut I defined with an help variable decPlace to make it easy to adapt   f   1 12345 decPlace  4 f cut   int f   10  decPlace   10  decPlace   Output   f   1 1234   Hope it helps

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round 1 923328437452  3    See Python s documentation on the standard types  You ll need to scroll down a bit to get to the round function  Essentially the second number says how many decimal places to round it to

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Simple python script -  n   1 923328437452 n   float int n   1000   n   1000

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def precision value  precision               param  value  takes a float     param  precision  int  number of decimal places     returns a float             x   10 0  precision     num   int value   x   x     return num precision 1 923328437452  3            1 923

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n   1 923328437452 str n   4

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def trunc num  digits      sp   str num  split         return     join  sp 0   sp 1   digits      This should work   It should give you the truncation you are looking for

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When using a pandas df this worked for me  import math def truncate number  digits  - gt  float      stepper   10 0    digits     return math trunc stepper   number    stepper  df  trunc     df  float val   apply lambda x  truncate x 1   df  trunc   df  trunc   map     1f   format

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Most answers are way too complicated in my opinion  how about this   digits   2    Specify how many digits you want  fnum    122 485221  truncated float   float fnum  fnum find        digits   1     gt  gt  gt  122 48   Simply scanning for the index of     and truncate as desired  no rounding   Convert string to float as final step   Or in your case if you get a float as input and want a string as output   fnum   str 122 485221     convert float to string first truncated float   fnum  fnum find        digits   1     string output

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Something simple enough to fit in a list-comprehension  with no libraries or other external dependencies  For Python   3 6  it s very simple to write with f-strings   The idea is to let the string-conversion do the rounding to one more place than you need and then chop off the last digit    gt  gt  gt  nout   3    desired number of digits in output  gt  gt  gt   f  x   nout 1 f    -1  for x in  2 3  4 5  8 9  9 8  5 4  3 2     0 666    0 800    0 888    1 125    1 250    1 500     Of course  there is rounding happening here  namely for the fourth digit   but rounding at some point is unvoidable  In case the transition between truncation and rounding is relevant  here s a slightly better example    gt  gt  gt  nacc   6    desired accuracy  maximum 15    gt  gt  gt  nout   3    desired number of digits in output  gt  gt  gt   f  x   nacc f    - nacc-nout   for x in  2 9999  2 99999  2 999999  2 9999999    gt  gt  gt    2 999    2 999    2 999    3 000     Bonus  removing zeros on the right   gt  gt  gt  nout   3    desired number of digits in output  gt  gt  gt   f  x   nout 1 f    -1  rstrip  0   for x in  2 3  4 5  8 9  9 8  5 4  3 2     0 666    0 8    0 888    1 125    1 25    1 5

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The truely pythonic way of doing it is  from decimal import    with localcontext   as ctx      ctx rounding   ROUND DOWN     print Decimal  1 923328437452   quantize Decimal  0 001      or shorter   from decimal import Decimal as D  ROUND DOWN  D  1 923328437452   quantize D  0 001    rounding ROUND DOWN    Update  Usually the problem is not in truncating floats itself  but in the improper usage of float numbers before rounding    For example  int 0 7 3 100  100    2 09    If you are forced to use floats  say  you re accelerating your code with numba   it s better to use cents as  internal representation  of prices   70 3    210  and multiply divide the inputs outputs

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The result of round is a float  so watch out  example is from Python 2 6     gt  gt  gt  round 1 923328437452  3  1 923  gt  gt  gt  round 1 23456  3  1 2350000000000001   You will be better off when using a formatted string    gt  gt  gt     3f    1 923328437452  1 923   gt  gt  gt     3f    1 23456  1 235

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use numpy round  import numpy as np precision   3 floats    1 123123123  2 321321321321  new float   np round floats  precision

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At my Python 2 7 prompt    gt  gt  gt  int 1 923328437452   1000  1000 0 1 923

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int 16 5   this will give an integer value of 16  i e  trunc  won t be able to specify decimals  but guess you can do that by   import math   def trunc invalue  digits       return int invalue math pow 10 digits   math pow 10 digits

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I did something like this   from math import trunc   def truncate number  decimals 0       if decimals  lt  0          raise ValueError  truncate received an invalid value of decimals       format decimals       elif decimals    0          return trunc number      else          factor   float 10  decimals          return trunc number factor  factor

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Short and easy variant  def truncate float value  digits after point 2       pow 10   10    digits after point     return  float int value   pow 10      pow 10   gt  gt  gt  truncate float 1 14333  2   gt  gt  gt  1 14   gt  gt  gt  truncate float 1 14777  2   gt  gt  gt  1 14    gt  gt  gt  truncate float 1 14777  4   gt  gt  gt  1 1477

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You can do   def truncate f  n       return math floor f   10    n    10    n   testing    gt  gt  gt  f 1 923328437452  gt  gt  gt   truncate f  n  for n in range 5    1 0  1 9  1 92  1 923  1 9233

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If you mean when printing  then the following should work   print    3f    number

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The core idea given here seems to me to be the best approach for this problem   Unfortunately  it has received less votes while the later answer that has more votes is not complete  as observed in the comments   Hopefully  the implementation below provides a short and complete solution for truncation    x000D   x000D  def trunc num  digits   x000D      l   str float num   split      x000D      digits   min len l 1    digits  x000D      return  l 0      l 1   digits   x000D   x000D   x000D    which should take care of all corner cases found here and here

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def trunc f n     return     16f    f    n-16

[python] How to truncate float values?

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