Dou*_* AA 73 python floating-point split
是否存在将数字1234.5678分成两部分的pythonic方法,(1234, 0.5678)即整数部分和小数部分？
mhy*_*itz 121
用途math.modf:
import math
x = 1234.5678
math.modf(x) # (0.5678000000000338, 1234.0)
utd*_*mir 55
我们可以使用一个不着名的内置功能; divmod:
>>> s = 1234.5678
>>> i, d = divmod(s, 1)
>>> i
1234.0
>>> d
0.5678000000000338
mac*_*mac 36
>>> 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,所以如果你想像人类那样表示它,你需要使用小数库
Mar*_*som 13
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)
我提出了两个语句，可以将正数和负数分为整数和分数，而不影响精度（位溢出）和速度。
代码
# Divide a number (x) into integer and fraction
i = int(x) # Get integer
f = (x*1e17 - i*1e17) / 1e17 # Get fraction
该值的正值和负值100.1323将按如下方式除法 ( 100, 0.1323) 和 ( -100, -0.1323) ，其中math.modf结果将为 ( 0.13230000000000075, 100.0) 和 ( -0.13230000000000075,-100.0 )。
速度测试
性能测试表明这两条语句的速度比math.modf只要不将这两个语句放入自己的函数或方法中（C/C++ 扩展对此进行了改进），这两个语句就会比 更快。
test.py:
#!/usr/bin/env python
import math
import cProfile
""" Get the performance of both statements and math.modf """
X = -100.1323 # The number to be divided into integer and fraction
LOOPS = range(5 * 10 ** 6) # Number of loops
def scenario_a():
""" Get the performance of the statements """
for _ in LOOPS:
i = int(X) # -100
f = (X*1e17-i*1e17)/1e17 # -0.1323
def scenario_b():
""" Tests the speed of the statements when integer need to be 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 scenario_c():
""" Tests the speed of the statements in a function """
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():
""" Tests the speed of math.modf """
for _ in LOOPS:
f, i = math.modf(X) # (-0.13230000000000075, -100.0)
def scenario_e():
""" Tests the speed of math.modf when the integer part should be integer """
for _ in LOOPS:
f, i = math.modf(X) # (-0.13230000000000075, -100.0)
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()')
结果：
4 function calls in 1.357 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 1.357 1.357 <string>:1(<module>)
1 1.357 1.357 1.357 1.357 test.py:11(scenario_a)
1 0.000 0.000 1.357 1.357 {built-in method builtins.exec}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
4 function calls in 1.858 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 1.858 1.858 <string>:1(<module>)
1 1.858 1.858 1.858 1.858 test.py:18(scenario_b)
1 0.000 0.000 1.858 1.858 {built-in method builtins.exec}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
5000004 function calls in 2.744 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 2.744 2.744 <string>:1(<module>)
1 1.245 1.245 2.744 2.744 test.py:26(scenario_c)
5000000 1.499 0.000 1.499 0.000 test.py:29(modf)
1 0.000 0.000 2.744 2.744 {built-in method builtins.exec}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
5000004 function calls in 1.904 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 1.904 1.904 <string>:1(<module>)
1 1.073 1.073 1.904 1.904 test.py:37(scenario_d)
1 0.000 0.000 1.904 1.904 {built-in method builtins.exec}
5000000 0.831 0.000 0.831 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 2.547 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 2.547 2.547 <string>:1(<module>)
1 1.696 1.696 2.547 2.547 test.py:43(scenario_e)
1 0.000 0.000 2.547 2.547 {built-in method builtins.exec}
5000000 0.851 0.000 0.851 0.000 {built-in method math.modf}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
C/C++ 扩展
我尝试用 C/C++ 支持来编译这两个语句，结果更好。使用Python扩展模块，可以获得比Python更快、更准确的方法math.modf.
math2.pyx:
def modf(number):
cdef float num = <float> number
cdef int i = <int> num
return i, (num*1e17 - i*1e17) / 1e17
请参阅Cython 基础知识
test.py:
#!/usr/bin/env python
import math
import cProfile
import math2
""" Get the performance of both statements and math.modf """
X = -100.1323 # The number to be divided into integers and fractions
LOOPS = range(5 * 10 ** 6) # Number of loops
def scenario_a():
""" Tests the speed of the statements in a function using C/C++ support """
for _ in LOOPS:
i, f = math2.modf(X) # (-100, -0.1323)
def scenario_b():
""" Tests the speed of math.modf """
for _ in LOOPS:
f, i = math.modf(X) # (-0.13230000000000075, -100.0)
if __name__ == '__main__':
cProfile.run('scenario_a()')
cProfile.run('scenario_b()')
结果：
5000004 function calls in 1.629 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 1.629 1.629 <string>:1(<module>)
1 1.100 1.100 1.629 1.629 test.py:10(scenario_a)
1 0.000 0.000 1.629 1.629 {built-in method builtins.exec}
5000000 0.529 0.000 0.529 0.000 {math2.modf}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
5000004 function calls in 1.802 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 1.802 1.802 <string>:1(<module>)
1 1.010 1.010 1.802 1.802 test.py:16(scenario_b)
1 0.000 0.000 1.802 1.802 {built-in method builtins.exec}
5000000 0.791 0.000 0.791 0.000 {built-in method math.modf}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
笔记
模除正数的速度更快，但如果没有一些额外的工作就无法处理负数，这将使负数的除法速度变慢。然而，这是除正数最快的方法。
i, f = int(x), x*1e17%1e17/1e17 # Divide a number (x) into integer and fraction
正值的除法100.1323如下（100, 0.1323），负值的除法错误（-100, 0.8677）。