[python] How to get numbers after decimal point?

How do I get the numbers after a decimal point?

For example, if I have 5.55, how do i get .55?

5.55 % 1

Keep in mind this won't help you with floating point rounding problems. I.e., you may get:

0.550000000001

Or otherwise a little off the 0.55 you are expecting.

Another example using modf

from math import modf
number = 1.0124584

# [0] decimal, [1] integer
result = modf(number)
print(result[0])
# output = 0124584
print(result[1])
# output = 1

What about:

a = 1.3927278749291
b = a - int(a)

b
>> 0.39272787492910011

Or, using numpy:

import numpy
a = 1.3927278749291
b = a - numpy.fix(a)

Float numbers are not stored in decimal (base10) format. Have a read through the python documentation on this to satisfy yourself why. Therefore, to get a base10 representation from a float is not advisable.

Now there are tools which allow storage of numeric data in decimal format. Below is an example using the Decimal library.

from decimal import *

x = Decimal('0.341343214124443151466')
str(x)[-2:] == '66'  # True

y = 0.341343214124443151466
str(y)[-2:] == '66'  # False

Try Modulo:

5.55%1 = 0.54999999999999982

Another crazy solution is (without converting in a string):

number = 123.456
temp = 1

while (number*temp)%10 != 0:
    temp = temp *10
    print temp
    print number

temp = temp /10
number = number*temp
number_final = number%temp
print number_final

import math

x = 1245342664.6
print( (math.floor(x*1000)%1000) //100 )

It definitely worked

>>> n=5.55
>>> if "." in str(n):
...     print "."+str(n).split(".")[-1]
...
.55

Just using simple operator division '/' and floor division '//' you can easily get the fraction part of any given float.

number = 5.55

result = (number/1) - (number//1)

print(result)

This is only if you want toget the first decimal

print(int(float(input()) * 10) % 10)

Or you can try this

num = float(input())
b = num - int(num) 
c = b * 10
print(int(c))

similar to the accepted answer, even easier approach using strings would be

def number_after_decimal(number1):
    number = str(number1)
    if 'e-' in number: # scientific notation
        number_dec = format(float(number), '.%df'%(len(number.split(".")[1].split("e-")[0])+int(number.split('e-')[1])))
    elif "." in number: # quick check if it is decimal
        number_dec = number.split(".")[1]
    return number_dec

import math
orig = 5.55
whole = math.floor(orig)    # whole = 5.0
frac = orig - whole         # frac = 0.55

Another option would be to use the re module with re.findall or re.search:

import re


def get_decimcal(n: float) -> float:
    return float(re.search(r'\.\d+', str(n)).group(0))


def get_decimcal_2(n: float) -> float:
    return float(re.findall(r'\.\d+', str(n))[0])


def get_int(n: float) -> int:
    return int(n)


print(get_decimcal(5.55))
print(get_decimcal_2(5.55))
print(get_int(5.55))

Output

0.55
0.55
5

If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.

Source

How to get rid of additional floating numbers in python subtraction?

Use floor and subtract the result from the original number:

>> import math #gives you floor.
>> t = 5.55 #Give a variable 5.55
>> x = math.floor(t) #floor returns t rounded down to 5..
>> z = t - x #z = 5.55 - 5 = 0.55

You can use this:

number = 5.55
int(str(number).split('.')[1])

Example:

import math
x = 5.55
print((math.floor(x*100)%100))

This is will give you two numbers after the decimal point, 55 from that example. If you need one number you reduce by 10 the above calculations or increase depending on how many numbers you want after the decimal.

See what I often do to obtain numbers after the decimal point in python 3:

a=1.22
dec=str(a).split('.')
dec= int(dec[1])

This is a solution I tried:

num = 45.7234
(whole, frac) = (int(num), int(str(num)[(len(str(int(num)))+1):]))

def fractional_part(numerator, denominator):
    # Operate with numerator and denominator to 
# keep just the fractional part of the quotient
if  denominator == 0:
      return 0
  else:
       return (numerator/ denominator)-(numerator // denominator)  
 

print(fractional_part(5, 5)) # Should be 0
print(fractional_part(5, 4)) # Should be 0.25
print(fractional_part(5, 3)) # Should be 0.66...
print(fractional_part(5, 2)) # Should be 0.5
print(fractional_part(5, 0)) # Should be 0
print(fractional_part(0, 5)) # Should be 0

What about:

a = 1.234
b = a - int(a)
length = len(str(a))

round(b, length-2)

Output:
print(b)
0.23399999999999999
round(b, length-2)
0.234

Since the round is sent to a the length of the string of decimals ('0.234'), we can just minus 2 to not count the '0.', and figure out the desired number of decimal points. This should work most times, unless you have lots of decimal places and the rounding error when calculating b interferes with the second parameter of round.

Easier if the input is a string, we can use split()

decimal = input("Input decimal number: ") #123.456

# split 123.456 by dot = ['123', '456']
after_coma = decimal.split('.')[1] 

# because only index 1 is taken then '456'
print(after_coma) # '456'

if you want to make a number type print(int(after_coma)) # 456

I've found that really large numbers with really large fractional parts can cause problems when using modulo 1 to get the fraction.

import decimal

>>> d = decimal.Context(decimal.MAX_PREC).create_decimal(
... '143000000000000000000000000000000000000000000000000000000000000000000000000000.1231200000000000000002013210000000'
... )
...
>>> d % 1
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
decimal.InvalidOperation: [<class 'decimal.DivisionImpossible'>]

I instead grabbed the integral part and subtracted it first to help simplify the rest of it.

>>> d - d.to_integral()
Decimal('0.1231200000000000000002013210')

If you are using pandas:

df['decimals'] = df['original_number'].mod(1)

def fractional_part(numerator, denominator):
    if denominator == 0:
        return 0
    else:
        return numerator / denominator - numerator // denominator

print(fractional_part(5, 5)) # Should be 0
print(fractional_part(5, 4)) # Should be 0.25
print(fractional_part(5, 3)) # Should be 0.66...
print(fractional_part(5, 2)) # Should be 0.5
print(fractional_part(5, 0)) # Should be 0
print(fractional_part(0, 5)) # Should be 0

Use modf:

>>> import math
>>> frac, whole = math.modf(2.5)
>>> frac
0.5
>>> whole
2.0

Using the decimal module from the standard library, you can retain the original precision and avoid floating point rounding issues:

>>> from decimal import Decimal
>>> Decimal('4.20') % 1
Decimal('0.20')

As kindall notes in the comments, you'll have to convert native floats to strings first.

Sometimes trailing zeros matter

In [4]: def split_float(x):
   ...:     '''split float into parts before and after the decimal'''
   ...:     before, after = str(x).split('.')
   ...:     return int(before), (int(after)*10 if len(after)==1 else int(after))
   ...: 
   ...: 

In [5]: split_float(105.10)
Out[5]: (105, 10)

In [6]: split_float(105.01)
Out[6]: (105, 1)

In [7]: split_float(105.12)
Out[7]: (105, 12)

You may want to try this:

your_num = 5.55
n = len(str(int(your_num)))
float('0' + str(your_num)[n:])

It will return 0.55.

A solution is using modulo and rounding.

import math

num = math.fabs(float(5.55))
rem = num % 1

rnd_by =   len(str(num)) - len(str(int(num))) - 1

print(str(round(rem,rnd_by)))

Your output will be 0.55

number=5.55
decimal=(number-int(number))
decimal_1=round(decimal,2)
print(decimal)
print(decimal_1)

output: 0.55

Using math module

speed of this has to be tested

from math import floor

def get_decimal(number):
    '''returns number - floor of number'''
    return number-floor(number)

Example:

n = 765.126357123

get_decimal(n)

0.12635712300004798