How to check if a float value is a whole number

Question

I am trying to find the largest cube root that is a whole number  that is less than 12 000    processing   True n   12000 while processing      n -  1     if n     1 3      checks to see if this has decimals or not   I am not sure how to check if it is a whole number or not though  I could convert it to a string then use indexing to check the end values and see whether they are zero or not  that seems rather cumbersome though  Is there a simpler way

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You can use a modulo operation for that   if  n     1 0 3     1    0      print  We have a decimal number here

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The above answers work for many cases but they miss some  Consider the following   fl   sum  0 1  10     this is 0 9999999999999999  but we want to say it IS an int   Using this as a benchmark  some of the other suggestions don t get the behavior we might want   fl is integer     False  fl   1    0       False   Instead try   def isclose a  b  rel tol 1e-09  abs tol 0 0       return abs a-b   lt   max rel tol   max abs a   abs b    abs tol   def is integer fl       return isclose fl  round fl     now we get   is integer fl      True   isclose comes with Python 3 5   and for other Python s you can use this mostly equivalent definition  as mentioned in the corresponding PEP

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You can use the round function to compute the value    Yes in python as many have pointed when we compute the value of a cube root  it will give you an output with a little bit of error  To check if the value is a whole number you can use the following function   def cube integer n       if round n   1 0 3 0    3    n          return True     return False   But remember that int n  is equivalent to math floor and because of this if you find the int 41063625   1 0 3 0   you will get 344 instead of 345   So please be careful when using int withe cube roots

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You don t need to loop or to check anything  Just take a cube root of 12 000 and round it down    r   int 12000   1 3 0   print r r r   10648

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gt  gt  gt  def is near integer n  precision 8  get integer False           if get integer              return int round n  precision           else              return round n     round n  precision       gt  gt  gt  print is near integer 10648     1 0 3    True  gt  gt  gt  print is near integer 10648     1 0 3   get integer True   22  gt  gt  gt  for i in  4 9  5 1  4 99  5 01  4 999  5 001  4 9999  5 0001  4 99999  5 000 01  4 999999  5 000001           print i  is near integer i  4       4 9 False 5 1 False 4 99 False 5 01 False 4 999 False 5 001 False 4 9999 False 5 0001 False 4 99999 True 5 00001 True 4 999999 True 5 000001 True  gt  gt  gt

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You could use this   if k    int k       print str k      is a whole number

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All the answers are good but a sure fire method would be   def whole  n        return  n 10  10  0    The function returns True if it s a whole number else False    I know I m a bit late but here s one of the interesting methods which I made     Edit  as stated by the comment below  a cheaper equivalent test would be   def whole n       return n 1  0

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Try using   int val     val   It will give lot more precision than any other methods

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Wouldn t it be easier to test the cube roots  Start with 20  20  3   8000  and go up to 30  30  3   27000   Then you have to test fewer than 10 integers   for i in range 20  30       print  Trying  0   format i       if i    3  gt  12000          print  Maximum integral cube root less than 12000   0   format i - 1           break

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We can use the modulo     operator  This tells us how many remainders we have when we divide x by y - expresses as x   y  Every whole number must divide by 1  so if there is a remainder  it must not be a whole number    This function will return a boolean  True or False  depending on whether n is a whole number   def is whole n       return n   1    0

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How about   if x 1  0      print  is integer

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To check if a float value is a whole number  use the float is integer   method    gt  gt  gt   1 0  is integer   True  gt  gt  gt   1 555  is integer   False   The method was added to the float type in Python 2 6   Take into account that in Python 2  1 3 is 0  floor division for integer operands    and that floating point arithmetic can be imprecise  a float is an approximation using binary fractions  not a precise real number   But adjusting your loop a little this gives    gt  gt  gt  for n in range 12000  -1  -1           if  n     1 0 3   is integer                print n      27 8 1 0   which means that anything over 3 cubed   including 10648  was missed out due to the aforementioned imprecision    gt  gt  gt   4  3      1 0 3  3 9999999999999996  gt  gt  gt  10648     1 0 3  21 999999999999996   You d have to check for numbers close to the whole number instead  or not use float   to find your number  Like rounding down the cube root of 12000    gt  gt  gt  int 12000     1 0 3   22  gt  gt  gt  22    3 10648   If you are using Python 3 5 or newer  you can use the math isclose   function to see if a floating point value is within a configurable margin    gt  gt  gt  from math import isclose  gt  gt  gt  isclose  4  3      1 0 3   4  True  gt  gt  gt  isclose 10648     1 0 3   22  True   For older versions  the naive implementation of that function  skipping error checking and ignoring infinity and NaN  as mentioned in PEP485   def isclose a  b  rel tol 1e-9  abs tol 0 0       return abs a - b   lt   max rel tol   max abs a   abs b    abs tol

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Just a side info  is integer is doing internally    import math isInteger    math floor x     x       Not exactly in python  but the cpython implementation is implemented as mentioned above

[python] How to check if a float value is a whole number

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