Python 替换字符串中的分数

Question

我正在尝试替换字符串中不同格式的分数。有些格式看起来像 1/2 , 1 1/2, 1-1/2.

Input='This is the first fraction 1/2'
Input_two='This is the second fraction 3-1/8'
Input_three='This is the third fraction 20 1/4'

Output='This is the first fraction 0.5'
Output_two='This is the first fraction 3.12'
Output_three='This is the first fraction 20.25'

我尝试过的：

df['col]=df['col'].apply(lambda x: re.sub('\d\d?\d?.?\d+/\d+','1.5',str(x))
But this only works if you put the value in each time and a thousand different fractions

我也试过 from fractions import Fraction 和 from __future__ import division 但无法让它们在字符串上工作。

Answer 1

这是一个解决方案，使用 Series.str.replace 和捕获组，然后使用 lambda 将小数值转换为浮点数，然后进行替换。

df['col'].str.replace('(\s(?:\d+)?.?\d+\/\d+)', lambda x: ' '+str(eval(x.group(0).replace(' ','+').replace('-',  '+'))), regex=True)

0       This is the first fraction 0.5
1    This is the second fraction 3.125
2     This is the third fraction 20.25
Name: col, dtype: object

PS: 因为，它使用 eval 将字符串小数值转换为浮点值，它会抛出 syntaxError 如果你在问题中提到的分数部分在语法上是错误的，即实际数据与你在问题中显示的不同。

Answer 2

您可以将 fractions 包与 re.sub 一起使用。

import re
from fractions import Fraction

Inputs=[
    'This is the first fraction 1/2',
    'This is the second fraction 3-1/8',
    'This is the third fraction 20 1/4',
    'This is the first fraction 1/2 second fraction 3-1/8 third: 10 3/4'
]

def frac2string(s):
    i, f = s.groups(0)
    f = Fraction(f)
    return str(int(i) + float(f))

[re.sub(r'(?:(\d+)[-\s])?(\d+/\d+)', frac2string, s) for s in Inputs]

这给你：

['This is the first fraction 0.5',
 'This is the second fraction 3.125',
 'This is the third fraction 20.25',
 'This is the first fraction 0.5 second fraction 3.125 third: 10.75']

Python 替换字符串中的分数

Python Replace fractions in a string

python

regex

decimal

fractions