How to get numbers after decimal point?

问题

How do I get the numbers after a decimal point?

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

回答1:

An easy approach for you:

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

回答2:

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.

回答3:

Use modf:

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

回答4:

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)

回答5:

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.

回答6:

Try Modulo:

5.55%1 = 0.54999999999999982

回答7:

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

回答8:

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

回答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

回答10:

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)

回答11:

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

回答12:

This is a solution I tried:

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

回答13:

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

回答14:

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.

回答15:

import math

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

It definitely worked

回答16:

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.

回答17:

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.

回答18:

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

output: 0.55

回答19:

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])

回答20:

If you are using pandas:

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

回答21:

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?

回答22:

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

来源：https://stackoverflow.com/questions/3886402/how-to-get-numbers-after-decimal-point

标签

python

floating-point

decimal