按列表顺序替换值

Question

keywords = ['Small', 'Medium', 'Large']

df.column
0   The Small, Large, Medium
1   The fast Medium, Small XS
2   He was a Medium, Large or Small

如果一行包含关键字，我如何告诉 pandas：

1. 替换关键字，使关键字按列表顺序显示
2. 如果关键字包含后缀“XS”，请在步骤 1 中包含后缀

预期输出：

0 The Small, Medium, Large 
1 The fast Small XS, Medium
2 He was a Small, Medium or Large

Answer 1

一种方法是：

1. 将字符串拆分为与关键字匹配的单词（带或不带XS 后缀），或使用re.findall 的其他不匹配部分
2. 根据关键字列表中的索引对匹配的单词进行排序
3. 使用排序后的关键字重建单词列表
4. 将字符串重新连接在一起

你可以用这个函数做到这一点：

def sizesorter(s, keywords):
    words = re.findall(r'((?:\b(?:' + '|'.join(keywords) + r')\b)(?:\sXS)?|(?:[^\s]*(?:\s|$)))', s, re.I)
    sizes = iter(sorted([w for w in words if w.split(' ')[0] in keywords], key=lambda w:keywords.index(w.split(' ')[0])))
    words = [w if w.split(' ')[0] not in keywords else next(sizes) for w in words]
    return ''.join(words)

然后您可以将该函数应用于列。例如：

import pandas as pd
import re

df = pd.DataFrame({ 'column' : ['The Small, Large, Medium',
'The fast Medium, Small XS',
'He was a Medium, Large or Small',
'small, Large a metre'
] })

def sizesorter(s, keywords):
    words = re.findall(r'((?:\b(?:' + '|'.join(keywords) + r')\b)(?:\sXS)?|(?:[^\s]*(?:\s|$)))', s, re.I)
    sizes = iter(sorted([w for w in words if w.split(' ')[0] in keywords], key=lambda w:keywords.index(w.split(' ')[0])))
    words = [w if w.split(' ')[0] not in keywords else next(sizes) for w in words]
    return ''.join(words)
    
df.column = df.column.apply(sizesorter, args=(['Small', 'Medium', 'Large'], ))

print(df)

输出：

                            column
0         The Small, Medium, Large
1        The fast Small XS, Medium
2  He was a Small, Medium or Large

改编自this answer的单词列表的部分排序。