根据原始DF(Python 3，Pandas)对列表中的值进行分类

Question

名为KW的虚构df如下所示：

 Group    Subgroup    Word
 orange    zebra      keys
 green     lion       mouse
 blue      horse      captain

我目前拥有的代码采用"单词"列下的每个单词，并将某些字母替换为字典中的其他字母，每次一个。之后，创建所有这些拼写错误的列表。因此，使用KW df:

kw = df[['Word',"Group","Subgroup"]]
words = kw.to_dict()["Word"].values()
md = {"m":"w","o":"z"}
md = {k: v.split(',') for k, v in md.items()}
newwords = []
for word in words:
    newwords.append(word)
    for c in md:
        occ = word.count(c)
        pos = 0
        for _ in range(occ):
            pos = word.find(c, pos)
            for r in md[c]:
                tmp = word[:pos] + r + word[pos+1:]
                newwords.append(tmp)
            pos += 1

返回

Word
keys
mouse
wouse
mzuse
captain

我想做的基本上是根据被操纵的原始单词将这些拼写错误重新分类到相应的组/子群中。因此，理想情况下，与其吐出一个单独的拼写错误列表，不如像这样：

Group    Subgroup    Word
orange   zebra       keys
green    lion        mouse
green    lion        wouse
green    lion        mzuse
blue     horse       captain

Answer 1

不知何故，我们需要将新单词与原始单词关联起来。您可以通过在newwords中存储2个元组（如('mouse', 'wouse')）来实现这一点。然后，您可以将newwords转换为DataFrame，并使用pd.merge通过连接原始单词将newwords与kw合并

import pandas as pd
df = pd.read_table('data', sep='s+')
kw = df[['Word',"Group","Subgroup"]]
words = df['Word']
md = {"m":"w","o":"z"}
md = {k: v.split(',') for k, v in md.items()}
newwords = []
for word in words:
    # Save both the original word and the new word
    newwords.append((word, word))
    for c in md:
        occ = word.count(c)
        pos = 0
        for _ in range(occ):
            pos = word.find(c, pos)
            for r in md[c]:
                tmp = word[:pos] + r + word[pos+1:]
                newwords.append((word, tmp))
newwords = pd.DataFrame(newwords, columns=['Word', 'New'])
# Merge on the original Word
result = pd.merge(newwords, kw, left_on='Word', right_on='Word', how='left')
result = result[['Group', 'Subgroup', 'New']]
result.columns =  ['Group', 'Subgroup', 'Word']
print(result)

产生

    Group Subgroup     Word
0  orange    zebra     keys
1   green     lion    mouse
2   green     lion    wouse
3   green     lion    mzuse
4    blue    horse  captain