Splitting a number into the integer and decimal parts

Is there a pythonic way 开发者_运维百科of splitting a number such as 1234.5678 into two parts (1234, 0.5678) i.e. the integer part and the decimal part?

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Use math.modf:

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

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We can use a not famous built-in function; 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
>>>

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

---

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

Works for positive numbers.

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This variant allows getting desired precision:

>>> 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)

---

I have come up with two statements that can divide positive and negative numbers into integer and fraction without compromising accuracy (bit overflow) and speed.

Code

# Divide a number (x) into integer and fraction
i = int(x) # Get integer
f = (x*1e17 - i*1e17) / 1e17 # Get fraction

A positive and negative value of the value 100.1323 will be divided as follows (100, 0.1323) and (-100, -0.1323) where the math.modf result will be (0.13230000000000075, 100.0) and (-0.13230000000000075, -100.0).

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 (C/C++ extension improve this).

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()')

Result:

         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++ extension

I have tried to compile the two statements with C/C++ support and the result was even better. With Python extension module it was possible to get a method that was faster and more accurate than math.modf.

math2.pyx:

def modf(number):
    cdef float num = <float> number
    cdef int i = <int> num
    return i, (num*1e17 - i*1e17) / 1e17

See Basics of Cython

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()')

Result:

         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}

NOTE

modulo is faster to divide positive numbers, but cannot handle negative numbers without some extra work, which will make division slower for negative numbers. However, this is the fastest way to divide positive numbers.

i, f = int(x), x*1e17%1e17/1e17 # Divide a number (x) into integer and fraction

A positive value of the value 100.1323 will be divided as follows (100, 0.1323) and negative value will come out wrong (-100, 0.8677).

---

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])

---

This will accomplish the task without the issue of dropping leading zeroes (like holydrinker's answer):

Code

def extract_int_decimal():
    '''get the integer and decimal parts of a given number
    by converting it to string and using split method
    '''
    num = 1234.5678
    split_num = str(num).split('.')
    int_part = int(split_num[0])
    decimal_part = int(split_num[1]) * 10 ** -len(split_num[1])
    print("integer part:",int_part)
    print("decimal part:",decimal_part)

extract_int_decimal()

Result

integer part: 1234
decimal part: 0.5678000000000001

---

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)