获取列表中每行最常用的单词

Question

我有一个数据框 df 有 50,000 多行：

>>> df
                 message                          words  wordCount  uniqueWordCount
0             my name is                 [my, name, is]          3                3
1  happy birthday to you     [happy, birthday, to, you]          4                4
2         la la la la la           [la, la, la, la, la]          5                1
3 you are you that is it  [you, are, you, that, is, it]          6                5
...

我想创建一个新列，其中包含message 中最常用的 3 个单词。 到目前为止我所做的工作，但需要相当长的时间。

>>> df["mostFrequent"] = df["message"].apply(
      lambda x: sorted(
        textblob.TextBlob(x).word_counts, key=textblob.TextBlob(x).word_counts.get, reverse=True)[:3])

>>> df["mostFrequent"]
0 [my, name, is]
1 [happy, birthday, to]
2 [la]
3 [you, are, that]
...

有没有更有效的方法？

Answer 1

将自定义 lambda 函数与 collections.Counter 一起使用：

from collections import Counter
f = lambda x: [word for word, word_count in Counter(x).most_common(3)]
df["mostFrequent"] = df["words"].apply(f)
print (df)
                  message                          words  wordCount  \
0              my name is                 [my, name, is]          3   
1   happy birthday to you     [happy, birthday, to, you]          4   
2          la la la la la           [la, la, la, la, la]          5   
3  you are you that is it  [you, are, you, that, is, it]          6   

   uniqueWordCount           mostFrequent  
0                3         [my, name, is]  
1                4  [happy, birthday, to]  
2                1                   [la]  
3                5       [you, are, that]