将单数替换为“空格和数字”Python : Replace _2/28

Question

DF_Column:         ASSESSMENT
            ASSESSMENT 21/28, MISSED
            ASSESSMENT 1/28, MISSED
            ASSESSMENT 2/28, MISSED
            ASSESSMENT 23/28, MISSED
            ASSESSMENT 23/28, MISSED
            ASSESSMENT 5/28, MISSED

我在python想要的是：

               ASSESSMENT
        ASSESSMENT 21/28, MISSED
        ASSESSMENT  1/28, MISSED
        ASSESSMENT  2/28, MISSED
        ASSESSMENT 23/28, MISSED
        ASSESSMENT 23/28, MISSED
        ASSESSMENT  5/28, MISSED

如果是单数/28，评估后应该有双倍的空间。

Answer 1

您可以使用Python找到日期模式并用一个新字符串替换，使用Python雷尼在长度为2的字符串中重新对齐日期。

import pandas as pd
import re

df = pd.read_csv('sample.csv', sep='\n')
print(df)

def fix_alignment(m):
    g1 = m.group(1).rjust(2, ' ')
    g2 = m.group(2)
    return f'{g1}/{g2}'

df['NewAlignment'] = df['ASSESSMENT'].apply(
        lambda row: re.sub(r'\b(\d{1})/(\d{2})', fix_alignment, row)
)
print(df)

df输出

                 ASSESSMENT              NewAlignment
0  ASSESSMENT 21/28, MISSED  ASSESSMENT 21/28, MISSED
1   ASSESSMENT 1/28, MISSED  ASSESSMENT  1/28, MISSED
2   ASSESSMENT 2/28, MISSED  ASSESSMENT  2/28, MISSED
3  ASSESSMENT 23/28, MISSED  ASSESSMENT 23/28, MISSED
4  ASSESSMENT 23/28, MISSED  ASSESSMENT 23/28, MISSED
5   ASSESSMENT 5/28, MISSED  ASSESSMENT  5/28, MISSED

Answer 2

这使用字段中的两个空格作为拆分字符。然后确保中间的子字符串长6个字符(包括空格)。我在一个由您的列组成的数组上测试了这一点。

['{} {:>6} {}'.format(*i.split(' ')) for i in col]

下面是我刚在dataframe上使用的代码

df['ASSESSMENT'] = ['{} {:>6} {}'.format(*i.split(' ')) for i in df['ASSESSMENT']]