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?
This question is related to
split
floating-point
python
intpart,decimalpart = int(value),value-int(value)
Works for positive numbers.
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])
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)
This also works for me
>>> val_int = int(a)
>>> val_fract = a - val_int
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)
We can use a not famous built-in function; divmod:
>>> s = 1234.5678
>>> i, d = divmod(s, 1)
>>> i
1234.0
>>> d
0.5678000000000338
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 -> 100, 0.1323 or -100.1323 -> -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
""" 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():
""" The integers (i) and the fractions (f)
come out as integer and float. """
for _ in LOOPS:
i = int(X) # -100
f = (X*1e17-i*1e17)/1e17 # -0.1323
def scenario_b():
""" The integers (i) and the fractions (f)
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 scenario_c():
""" Performance test 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():
for _ in LOOPS:
f, i = math.modf(X) # (-100.0, -0.13230000000000075)
def scenario_e():
""" Convert the integer part to real integer. """
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 <string>:1(<module>)
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 <string>:1(<module>)
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 <string>:1(<module>)
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 <string>:1(<module>)
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 <string>:1(<module>)
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
>>> 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
Source: Stackoverflow.com