How to get numbers after decimal point?

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Method 1

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.

Method 2

Use modf:

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

Method 3

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)

Method 4

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.

Method 5

An easy approach for you:

number_dec = str(number-int(number))[1:]

Method 6

Try Modulo:

5.55%1 = 0.54999999999999982

Method 7

To make it work with both positive and negative numbers:
try abs(x)%1. For negative numbers, without with abs, it will go wrong.

5.55 % 1

output 0.5499999999999998

-5.55 % 1

output 0.4500000000000002

Method 8

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

Method 9

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

Method 10

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

Method 11

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)

Method 12

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)

Method 13

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

Method 14

This is a solution I tried:

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

Method 15

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

Method 16

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

Method 17

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.

Method 18

import math

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

It definitely worked

Method 19

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')

Method 20

You can use this:

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

Method 21

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))

Method 22

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

Method 23

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

Method 24

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

Method 25

a = 12.587

b = float(‘0.’ + str(a).split(‘.’)[-1])

Method 26

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.

Method 27

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.

Method 28

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

output: 0.55

Method 29

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])

Method 30

If you are using pandas:

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

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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