round doesn t seem to be rounding properly

Question

The documentation for the round   function states that you pass it a number  and the positions past the decimal to round  Thus it should do this   n   5 59 round n  1    5 6   But  in actuality  good old floating point weirdness creeps in and you get   5 5999999999999996   For the purposes of UI  I need to display 5 6  I poked around the Internet and found some documentation that this is dependent on my implementation of Python  Unfortunately  this occurs on both my Windows dev machine and each Linux server I ve tried  See here also   Short of creating my own round library  is there any way around this

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Code   x1   5 63 x2   5 65 print float    2f    round x1 1       gives you  5 6  print float    2f    round x2 1       gives you  5 7    Output   5 6 5 7

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I am doing   int round  x   0     In this case  we first round properly at the unit level  then we convert to integer to avoid printing a float   so    gt  gt  gt  int round 5 59 0   6   I think this answer works better than formating the string  and it also makes more sens to me to use the round function

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Here is an easy way to round a float number to any number of decimal places  and it still works in 2021  float number   12 234325335563 rounded   round float number  3    3 is the number of decimal places to be returned You can pass any number in place of 3 depending on how many decimal places you want to return  print rounded   And this will print  12 234

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You can use the string format operator    similar to sprintf   mystring      2f    5 5999

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Floating point math is vulnerable to slight  but annoying  precision inaccuracies   If you can work with integer or fixed point  you will be guaranteed precision

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Formatting works correctly even without having to round      1f    n

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round 5 59  1  is working fine  The problem is that 5 6 cannot be represented exactly in binary floating point    gt  gt  gt  5 6 5 5999999999999996  gt  gt  gt     As Vinko says  you can use string formatting to do rounding for display   Python has a module for decimal arithmetic if you need that

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If you use the Decimal module you can approximate without the use of the  round  function   Here is what I ve been using for rounding especially when writing monetary applications    Decimal str 16 2   quantize Decimal   01    rounding ROUND UP    This will return a Decimal Number which is 16 20

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Take a look at the Decimal module     Decimal    is based on a floating-point   model which was designed with people   in mind  and necessarily has a   paramount guiding principle       computers must provide an arithmetic   that works in the same way as the   arithmetic that people learn at   school         excerpt from the decimal   arithmetic specification    and      Decimal numbers can be represented   exactly  In contrast  numbers like 1 1   and 2 2 do not have an exact   representations in binary floating   point  End users typically would not   expect 1 1    2 2 to display as   3 3000000000000003  as it does with binary floating point    Decimal provides the kind of operations that make it easy to write apps that require floating point operations and also need to present those results in a human readable format  e g   accounting

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I would avoid relying on round   at all in this case  Consider  print round 61 295  2   print round 1 295  2     will output  61 3 1 29   which is not a desired output if you need solid rounding to the nearest integer  To bypass this behavior go with math ceil    or math floor   if you want to round down    from math import ceil decimal count   2 print ceil 61 295   10    decimal count    10    decimal count  print ceil 1 295   10    decimal count    10    decimal count    outputs  61 3 1 3   Hope that helps

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The problem is only when last digit is 5  Eg  0 045 is internally stored as 0 044999999999999    You could simply increment last digit to 6 and round off  This will give you the desired results   import re   def custom round num  precision 0         Get the type of given number     type num   type num        If the given type is not a valid number type  raise TypeError     if type num not in  int  float  Decimal           raise TypeError  type    doesn t define   round   method  format type num   name           If passed number is int  there is no rounding off      if type num    int          return num       Convert number to string      str num   str num  lower         We will remove negative context from the number and add it back in the end     negative number   False     if num  lt  0          negative number   True         str num   str num 1         If number is in format 1e-12 or 2e 13  we have to convert it to       to a string in standard decimal notation      if  e-  in str num            For 1 23e-7  e power   7         e power   int re findall  e- 0-9     str num  0  2              For 1 23e-7  number   123         number      join str num split  e-   0  split               zeros                Number of zeros   e power - 1   6         for i in range e power - 1               zeros   zeros    0            Scientific notation 1 23e-7 in regular decimal   0 000000123         str num    0     zeros   number     if  e   in str num            For 1 23e 7  e power   7         e power   int re findall  e   0-9     str num  0  2              For 1 23e 7  number characteristic   1           characteristic is number left of decimal point          number characteristic   str num split  e    0  split      0            For 1 23e 7  number mantissa   23           mantissa is number right of decimal point          number mantissa   str num split  e    0  split      1            For 1 23e 7  number   123         number   number characteristic   number mantissa         zeros                Eg  for this condition   1 23e 7         if e power  gt   len number mantissa                 Number of zeros   e power - mantissa length   5             for i in range e power - len number mantissa                    zeros   zeros    0                Scientific notation 1 23e 7 in regular decimal   12300000 0             str num   number   zeros     0            Eg  for this condition   1 23e 1         if e power  lt  len number mantissa                 In this case  we only need to shift the decimal e power digits to the right               So we just copy the digits from mantissa to characteristic and then remove               them from mantissa              for i in range e power                   number characteristic   number characteristic   number mantissa i              number mantissa   number mantissa i                 Scientific notation 1 23e 1 in regular decimal   12 3             str num   number characteristic         number mantissa       characteristic is number left of decimal point      characteristic part   str num split      0        mantissa is number right of decimal point      mantissa part   str num split      1        If number is supposed to be rounded to whole number        check first decimal digit  If more than 5  return       characteristic   1 else return characteristic     if precision    0          if mantissa part and int mantissa part 0    gt   5              return type num int characteristic part    1          return type num characteristic part        Get the precision of the given number      num precision   len mantissa part        Rounding off is done only if number precision is       greater than requested precision     if num precision  lt   precision          return num       Replace the last  5  with 6 so that rounding off returns desired results     if str num -1      5           str num   re sub  5     6   str num      result   round type num str num   precision        If the number was negative  add negative context back     if negative number          result   result   -1     return result

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You can switch the data type to an integer    gt  gt  gt  n   5 59  gt  gt  gt  int n   10    10 0 5 5  gt  gt  gt  int n   10   0 5  56   And then display the number by inserting the locale s decimal separator   However  Jimmy s answer is better

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Works Perfect   format 5 59    1f     to display float format 5 59    1f     to round

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What about   round n 1  epsilon

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You get  5 6  if you do str round n  1   instead of just round n  1

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Another potential option is   def hard round number  decimal places 0               Function      - Rounds a float value to a specified number of decimal places     - Fixes issues with floating point binary approximation rounding in python     Requires      -  number           - Type  int float         - What  The number to round     Optional      -  decimal places           - Type  int          - What  The number of decimal places to round to         - Default  0     Example              hard round 5 6 1                      return int number  10  decimal places  0 5   10  decimal places

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It s a big problem indeed  Try out this code   print    2f     round  2 4 4 3 5 6 3 4 4  8 2      It displays 4 85  Then you do    print  Media     1f     round  2 4 4 3 5 6 3 4 4  8 1      and it shows 4 8  Do you calculations by hand the exact answer is 4 85  but if you try    print  Media     20f     round  2 4 4 3 5 6 3 4 4  8 20      you can see the truth  the float point is stored as the nearest finite sum of fractions whose denominators are powers of two

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I can t help the way it s stored  but at least formatting works correctly       1f    round n  1    Gives you  5 6

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Here s where I see round failing  What if you wanted to round these 2 numbers to one decimal place  23 45 23 55 My education was that from rounding these you should get  23 4 23 6 the  rule  being that you should round up if the preceding number was odd  not round up if the preceding number were even  The round function in python simply truncates the 5

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printf the sucker   print    1f    5 59    returns 5 6

[python] round() doesn't seem to be rounding properly

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