将数字分为整数和小数部分

是否存在将数字分割1234.5678成两个部分（(1234, 0.5678)即整数部分和小数部分）的有效方法？

用途math.modf：

import math
x = 1234.5678
math.modf(x) # (0.5678000000000338, 1234.0)

我们可以使用一个不著名的内置函数；divmod：

>>> s = 1234.5678
>>> i, d = divmod(s, 1)
>>> i
1234.0
>>> d
0.5678000000000338

>>> a = 147.234
>>> a % 1
0.23400000000000887
>>> a // 1
147.0
>>>

如果您希望整数部分为整数而不是浮点数，请int(a//1)改用。要在单个段落中获取元组：(int(a//1), a%1)

编辑：请记住，浮点数的小数部分是近似值，因此，如果您要像人类那样表示它，则需要使用十进制库

intpart,decimalpart = int(value),value-int(value)

适用于正数。

此变体允许获得所需的精度：

>>> a = 1234.5678
>>> (lambda x, y: (int(x), int(x*y) % y/y))(a, 1e0)
(1234, 0.0)
>>> (lambda x, y: (int(x), int(x*y) % y/y))(a, 1e1)
(1234, 0.5)
>>> (lambda x, y: (int(x), int(x*y) % y/y))(a, 1e15)
(1234, 0.5678)

这对我也有用

>>> val_int = int(a)
>>> val_fract = a - val_int

这是我的方法：

num = 123.456
split_num = str(num).split('.')
int_part = int(split_num[0])
decimal_part = int(split_num[1])

如果您不介意使用NumPy，则：

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)

看了几个答案之后。我提出了这两个语句，它们可以将正数和负数分为整数部分和小数部分，而不会影响精度。性能测试表明，math.modf只要不将这两个新语句放入自己的函数或方法中，它们就会比快。

i = int(x) # i contains a positive or negative integer
f = (x*1e17-i*1e17)/1e17 # f contains a positive or negative fraction

例如100.1323->100, 0.1323和-100.1323->-100, -0.1323

测试脚本：

#!/usr/bin/env python
import math
import cProfile

""" Get the performance of both statements vs math.modf. """

X = -100.1323
LOOPS = range(5*10**6)

def fun_a():
    """ The integer part (i) is an integer, and
        the fraction part (f) is a float.
        NOTE: I think this is the most correct way. """
    for _ in LOOPS:
        i = int(X) # -100
        f = (X*1e17-i*1e17)/1e17 # -0.1323

def fun_b():
    """ The integer (i) and fraction (f) part will
        come out as float.
        NOTE: The only difference between this
              and math.modf is the accuracy. """
    for _ in LOOPS:
        i = int(X) # -100
        i, f = float(i), (X*1e17-i*1e17)/1e17 # (-100.0, -0.1323)

def fun_c():
    """ Performance test of the statements in a function.
        The integer part (i) is an integer, and
        the fraction part (f) is a float. """
    def modf(x):
        i = int(x)
        return i, (x*1e17-i*1e17)/1e17

    for _ in LOOPS:
        i, f = modf(X) # (-100, -0.1323)

def fun_d():
    for _ in LOOPS:
        f, i = math.modf(X) # (-100.0, -0.13230000000000075)

def fun_e():
    """ Convert the integer part to integer. """
    for _ in LOOPS:
        f, i = math.modf(X) # (-100.0, -0.13230000000000075)
        i = int(i) # -100

if __name__ == '__main__':
    cProfile.run('fun_a()')
    cProfile.run('fun_b()')
    cProfile.run('fun_c()')
    cProfile.run('fun_d()')
    cProfile.run('fun_e()')

输出：

         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 <string>:1(<module>)
        1    1.312    1.312    1.312    1.312 new1.py:10(fun_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 <string>:1(<module>)
        1    1.887    1.887    1.887    1.887 new1.py:17(fun_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 <string>:1(<module>)
        1    1.261    1.261    2.797    2.797 new1.py:23(fun_c)
  5000000    1.536    0.000    1.536    0.000 new1.py:27(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 <string>:1(<module>)
        1    1.050    1.050    1.852    1.852 new1.py:34(fun_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 <string>:1(<module>)
        1    1.652    1.652    2.467    2.467 new1.py:38(fun_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}

注意：

使用模数语句可以更快，但是模数不能用于将负数拆分为整数和小数部分。

i, f = int(x), x*1e17%1e17/1e17 # x can not be negative