Safest way to convert float to integer in python

Question

Python s math module contain handy functions like floor  amp  ceil  These functions take a floating point number and return the nearest integer below or above it  However these functions return the answer as a floating point number  For example   import math f math floor 2 3    Now f returns   2 0   What is the safest way to get an integer out of this float  without running the risk of rounding errors  for example if the float is the equivalent of 1 99999  or perhaps I should use another function altogether

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df  Column Name   df  Column Name   astype int

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You could use the round function   If you use no second parameter    of significant digits  then I think you will get the behavior you want   IDLE output    gt  gt  gt  round 2 99999999999  3  gt  gt  gt  round 2 6  3  gt  gt  gt  round 2 5  3  gt  gt  gt  round 2 4  2

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All integers that can be represented by floating point numbers have an exact representation  So you can safely use int on the result  Inexact representations occur only if you are trying to represent a rational number with a denominator that is not a power of two   That this works is not trivial at all  It s a property of the IEEE floating point representation that int  floor        if the magnitude of the numbers in question is small enough  but different representations are possible where int floor 2 3   might be 1   To quote from Wikipedia      Any integer with absolute value less than or equal to 224 can be exactly represented in the single precision format  and any integer with absolute value less than or equal to 253 can be exactly represented in the double precision format

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If you need to convert a string float to an int you can use this method   Example   38 0  to 38  In order to convert this to an int you can cast it as a float then an int  This will also work for float strings or integer strings    gt  gt  gt  int float  38 0    38  gt  gt  gt  int float  38    38   Note  This will strip any numbers after the decimal    gt  gt  gt  int float  38 2    38

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Another code sample to convert a real float to an integer using variables   vel  is a real float number and converted to the next highest INTEGER   newvel    import arcpy math  os  sys  arcpy da     with arcpy da SearchCursor densifybkp  floseg vel Length   as cursor   for row in cursor      curvel   float row 1       newvel   int math ceil curvel

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Use int your non integer number  will nail it   print int 2 3     2  print int math sqrt 5      2

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math floor will always return an integer number and thus int math floor some float   will never introduce rounding errors   The rounding error might already be introduced in math floor some large float   though  or even when storing a large number in a float in the first place   Large numbers may lose precision when stored in floats

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Since you re asking for the  safest  way  I ll provide another answer other than the top answer   An easy way to make sure you don t lose any precision is to check if the values would be equal after you convert them    if int some value     some value       some value   int some value    If the float is 1 0 for example   1 0 is equal to 1  So the conversion to int will execute  And if the float is 1 1  int 1 1  equates to 1  and 1 1    1  So the value will remain a float and you won t lose any precision

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That this works is not trivial at all  It s a property of the IEEE floating point representation that int  floor        if the magnitude of the numbers in question is small enough  but different representations are possible where int floor 2 3   might be 1    This post explains why it works in that range   In a double  you can represent 32bit integers without any problems  There cannot be any rounding issues  More precisely  doubles can represent all integers between and including 253 and -253   Short explanation  A double can store up to 53 binary digits  When you require more  the number is padded with zeroes on the right   It follows that 53 ones is the largest number that can be stored without padding  Naturally  all  integer  numbers requiring less digits can be stored accurately   Adding one to 111 omitted 111  53 ones  yields 100   000   53 zeroes   As we know  we can store 53 digits  that makes the rightmost zero padding   This is where 253 comes from     More detail  We need to consider how IEEE-754 floating point works     1 bit    11   8     52   23        bits double single precision   sign    exponent   mantissa     The number is then calculated as follows  excluding special cases that are irrelevant here       -1sign  times  1 mantissa  times 2exponent - bias   where bias   2exponent - 1 - 1  i e  1023 and 127 for double single precision respectively   Knowing that multiplying by 2X simply shifts all bits X places to the left  it s easy to see that any integer must have all bits in the mantissa that end up right of the decimal point to zero   Any integer except zero has the following form in binary      1x   x where the x-es represent the bits to the right of the MSB  most significant bit     Because we excluded zero  there will always be a MSB that is one mdash which is why it s not stored  To store the integer  we must bring it into the aforementioned form  -1sign  times  1 mantissa  times 2exponent - bias   That s saying the same as shifting the bits over the decimal point until there s only the MSB towards the left of the MSB  All the bits right of the decimal point are then stored in the mantissa   From this  we can see that we can store at most 52 binary digits apart from the MSB   It follows that the highest number where all bits are explicitly stored is  111 omitted 111    that s 53 ones  52   implicit 1  in the case of doubles    For this  we need to set the exponent  such that the decimal point will be shifted 52 places  If we were to increase the exponent by one  we cannot know the digit right to the left after the decimal point   111 omitted 111x    By convention  it s 0  Setting the entire mantissa to zero  we receive the following number   100 omitted 00x    100 omitted 000    That s a 1 followed by 53 zeroes  52 stored and 1 added due to the exponent   It represents 253  which marks the boundary  both negative and positive  between which we can accurately represent all integers  If we wanted to add one to 253  we would have to set the implicit zero  denoted by the x  to one  but that s impossible

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Combining two of the previous results  we have   int round some float     This converts a float to an integer fairly dependably

[python] Safest way to convert float to integer in python?

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