使用Levenshtein距离替换另一列中的单词wrt单词

Question

假设我有一个数据帧df1:

Sr       A              B                            C
1      rains         It rain there.             It rains there
2      plane         This is a vertical planes  This is a vertical plane
3      tree          Plant a trees              Plant a tree

列C是我的预期输出.我需要比较B列的字符串中的每个单词和A中的单词,如果Levenshtein距离为1,则将其替换.

我的方法:

import jellyfish as jf
def word_replace(str1):
    comp = #don't know how to store value of column A in this variable.
    for word in str1.split():
        if jf.levenshtein_distance(word,comp) == 1:
           word = comp
        else:
            pass
    return str1

df1['C'] = df1['B'].apply(word_replace)

第二件事,如果列A有双字"near miss"怎么样？我如何修改上述代码？例如:

 Sr       A              B                            C
  1     near miss        that was a ner mis          that was a near miss

Answer 1

您已经在一个问题中提出了两个问题,这对Stack Overflow来说绝不是一个好主意.我只想回答你的第一个问题,如果你想让别人看看你的第二个问题,那么我建议你专门为它写一个新问题.

pd.DataFrame.apply可以跨行或跨列工作,您希望单独处理每一行,因此您必须传递axis=1关键字参数.

下面是一些解决您问题的代码,它使用列表理解,利用三元运算符来选择需要替换的单词.然后使用这个列表连接在一起str.join().

最初你的代码在迭代字符串上进行迭代,但是这不起作用,因为你在迭代列表时无法修改它们.它还假设函数的输入是一个字符串,这是不正确的,因为它将是一个pandas.Series对象.

这是一段简化的代码,并没有考虑标点符号之类的内容,我将其作为练习留给读者.

import pandas as pd
import jellyfish as jf

data1 =  {'A':['rains','plane','tree'],'B':['It rain there','This is a vertical planes','Plant a trees']}
df1 = pd.DataFrame(data1)

def word_replace(row):
    comp = row['A']
    str1 = row['B']

    out = ' '.join([comp if jf.levenshtein_distance(word, comp) == 1
                    else word for word in str1.split()])
    return out

df1['C'] = df1.apply(word_replace, axis=1)