splitting a number into the integer and decimal parts

Question

Is there a pythonic way of splitting a number such as 1234 5678 into two parts  1234  0 5678  i e  the integer part and the decimal part

User · Answer

We can use a not famous built-in function  divmod    gt  gt  gt  s   1234 5678  gt  gt  gt  i  d   divmod s  1   gt  gt  gt  i 1234 0  gt  gt  gt  d 0 5678000000000338

User · Answer

If you don t mind using NumPy  then   In  319   real   np array  1234 5678    In  327   integ  deci   int np floor real    np asscalar real   1   In  328   integ  deci Out 328    1234  0 5678000000000338

User · Answer

Use math modf   import math x   1234 5678 math modf x     0 5678000000000338  1234 0

User · Answer

I have come up with two statements that can divide positive and negative numbers into integers and fractions without compromising accuracy  bit overflow  and speed  x   100 1323   A number to be divided into integers and fractions    The two statement to divided a number into integers and fractions i   int x    A positive or negative integer f    x 1e17-i 1e17  1e17   A positive or negative fraction  E g  100 1323 - gt  100  0 1323 or -100 1323 - gt  -100  -0 1323 Speedtest The performance test shows that the two statements are faster than math modf  as long as they are not put into their own function or method  test py     usr bin env python import math import cProfile   quot  quot  quot  Get the performance of both statements and math modf   quot  quot  quot   X   -100 1323   The number to be divided into integers and fractions LOOPS   range 5 10  6    Number of loops  def scenario a         quot  quot  quot  The integers  i  and the fractions  f          come out as integer and float   quot  quot  quot      for   in LOOPS          i   int X    -100         f    X 1e17-i 1e17  1e17   -0 1323  def scenario b         quot  quot  quot  The integers  i  and the fractions  f          come out as float          NOTE  The only difference between this               and math modf is the accuracy   quot  quot  quot      for   in LOOPS          i   int X    -100         i  f   float i    X 1e17-i 1e17  1e17    -100 0  -0 1323   def scenario c         quot  quot  quot  Performance test of the statements in a function   quot  quot  quot      def modf x           i   int x          return i   x 1e17-i 1e17  1e17      for   in LOOPS          i  f   modf X     -100  -0 1323   def scenario d        for   in LOOPS          f  i   math modf X     -100 0  -0 13230000000000075   def scenario e         quot  quot  quot  Convert the integer part to real integer   quot  quot  quot      for   in LOOPS          f  i   math modf X     -100 0  -0 13230000000000075          i   int i    -100  if   name         main         cProfile run  scenario a         cProfile run  scenario b         cProfile run  scenario c         cProfile run  scenario d         cProfile run  scenario e      Output           4 function calls in 1 312 seconds     Ordered by  standard name     ncalls  tottime  percall  cumtime  percall filename lineno function          1    0 000    0 000    1 312    1 312  lt string gt  1  lt module gt           1    1 312    1 312    1 312    1 312 test py 10 scenario a          1    0 000    0 000    1 312    1 312  built-in method builtins exec          1    0 000    0 000    0 000    0 000  method  disable  of   lsprof Profiler  objects             4 function calls in 1 887 seconds     Ordered by  standard name     ncalls  tottime  percall  cumtime  percall filename lineno function          1    0 000    0 000    1 887    1 887  lt string gt  1  lt module gt           1    1 887    1 887    1 887    1 887 test py 18 scenario b          1    0 000    0 000    1 887    1 887  built-in method builtins exec          1    0 000    0 000    0 000    0 000  method  disable  of   lsprof Profiler  objects             5000004 function calls in 2 797 seconds     Ordered by  standard name     ncalls  tottime  percall  cumtime  percall filename lineno function          1    0 000    0 000    2 797    2 797  lt string gt  1  lt module gt           1    1 261    1 261    2 797    2 797 test py 27 scenario c    5000000    1 536    0 000    1 536    0 000 test py 31 modf          1    0 000    0 000    2 797    2 797  built-in method builtins exec          1    0 000    0 000    0 000    0 000  method  disable  of   lsprof Profiler  objects             5000004 function calls in 1 852 seconds     Ordered by  standard name     ncalls  tottime  percall  cumtime  percall filename lineno function          1    0 000    0 000    1 852    1 852  lt string gt  1  lt module gt           1    1 050    1 050    1 852    1 852 test py 38 scenario d          1    0 000    0 000    1 852    1 852  built-in method builtins exec    5000000    0 802    0 000    0 802    0 000  built-in method math modf          1    0 000    0 000    0 000    0 000  method  disable  of   lsprof Profiler  objects             5000004 function calls in 2 467 seconds     Ordered by  standard name     ncalls  tottime  percall  cumtime  percall filename lineno function          1    0 000    0 000    2 467    2 467  lt string gt  1  lt module gt           1    1 652    1 652    2 467    2 467 test py 42 scenario e          1    0 000    0 000    2 467    2 467  built-in method builtins exec    5000000    0 815    0 000    0 815    0 000  built-in method math modf          1    0 000    0 000    0 000    0 000  method  disable  of   lsprof Profiler  objects   NOTE  The statement can be faster with modulo  but modulo can not be used to split negative numbers into integer and fraction parts  i  f   int x   x 1e17 1e17 1e17   x can not be negative

User · Answer

This also works for me   gt  gt  gt  val int   int a   gt  gt  gt  val fract   a - val int

User · Answer

gt  gt  gt  a   147 234  gt  gt  gt  a   1 0 23400000000000887  gt  gt  gt  a    1 147 0  gt  gt  gt    If you want the integer part as an integer and not a float  use int a  1  instead  To obtain the tuple in a single passage   int a  1   a 1   EDIT  Remember that the decimal part of a float number is approximate  so if you want to represent it as a human would do  you need to use the decimal library

User · Answer

intpart decimalpart   int value  value-int value    Works for positive numbers

User · Answer

This is the way I do it   num   123 456 split num   str num  split      int part   int split num 0   decimal part   int split num 1

User · Answer

This variant allows getting desired precision    gt  gt  gt  a   1234 5678  gt  gt  gt   lambda x  y   int x   int x y    y y   a  1e0   1234  0 0   gt  gt  gt   lambda x  y   int x   int x y    y y   a  1e1   1234  0 5   gt  gt  gt   lambda x  y   int x   int x y    y y   a  1e15   1234  0 5678

[split] splitting a number into the integer and decimal parts

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