Range with step of type float

Question

The documentation basically says that range must behave exactly as this implementation  for positive step    def range start  stop  step     x   start   while True      if x  gt   stop  return     yield x     x    step   It also says that its arguments must be integers  Why is that  Isn t that definition also perfectly valid if step is a float   In my case  I am esp  needing a range function which accepts a float type as its step argument  Is there any in Python or do I need to implement my own     More specific  How would I translate this C code directly to Python in a nice way  i e  not just doing it via a while-loop manually    for float x   0  x  lt  10  x    0 5f

User · Answer

Probably because you can t have part of an iterable  Also  floats are imprecise

User · Answer

In order to be able to use decimal numbers in a range expression a cool way for doing it is the following   x   0 1 for x in range 0  10

User · Answer

One explanation might be floating point rounding issues  For example  if you could call  range 0  0 4  0 1    you might expect an output of   0  0 1  0 2  0 3    but you in fact get something like   0  0 1  0 2000000001  0 3000000001    due to rounding issues  And since range is often used to generate indices of some sort  it s integers only   Still  if you want a range generator for floats  you can just roll your own   def xfrange start  stop  step       i   0     while start   i   step  lt  stop          yield start   i   step         i    1

User · Answer

Here is a special case that might be good enough       1 0 divStep  x for x in range start divStep  stop divStep     In your case this would be    for float x   0  x  lt  10  x    0 5f                   gt  start   0 stop    10 divstep   1  5   2  This needs to be int  thats why I said  special case    and so    gt  gt  gt     5 x for x in range 0 2  10 2    0 0  0 5  1 0  1 5  2 0  2 5  3 0  3 5  4 0  4 5  5 0  5 5  6 0  6 5  7 0  7 5  8 0  8 5  9 0  9 5

User · Answer

This is what I would use   numbers    float x  10 for x in range 10     rather than   numbers    x 0 1 for x in range 10   that would return    0 0  0 1  0 2  0 30000000000000004  0 4  0 5  0 6000000000000001  0 7000000000000001  0 8  0 9    hope it helps

User · Answer

The problem with floating point is that you may not get the same number of items as you expected  due to inaccuracy  This can be a real problem if you are playing with polynomials where the exact number of items is quite important   What you really want is an arithmetic progression  the following code will work quite happily for int  float and complex     and strings  and lists      def arithmetic progression start  step  length       for i in xrange length           yield start   i   step   Note that this code stands a better chance of your last value being within a bull s roar of the expected value than any alternative which maintains a running total    gt  gt  gt  10000   0 0001  sum 0 0001 for i in xrange 10000    1 0  0 9999999999999062   gt  gt  gt  10000    1 3    sum 1 3  for i in xrange 10000    3333 333333333333  3333 3333333337314    Correction  here s a competetive running-total gadget   def kahan range start  stop  step       assert step  gt  0 0     total   start     compo   0 0     while total  lt  stop          yield total         y   step - compo         temp   total   y         compo    temp - total  - y         total   temp   gt  gt  gt  list kahan range 0  1  0 0001   -1  0 9999  gt  gt  gt  list kahan range 0  3333 3334  1 3    -1  3333 333333333333  gt  gt  gt

User · Answer

You could use numpy arange   EDIT  The docs prefer numpy linspace  Thanks  Droogans for noticing

User · Answer

When you add floating point numbers together  there s often a little bit of error  Would a range 0 0  2 2  1 1  return  0 0  1 1  or  0 0  1 1  2 199999999   There s no way to be certain without rigorous analysis   The code you posted is an OK work-around if you really need this  Just be aware of the possible shortcomings

[python] Range with step of type float

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