如何根据列表中的单词创建新的 Pandas 列

Question

所以我有一个列表和一个数据框。我想从列表中取出单词并将其作为列的标题。如果这个词是它添加到新创建的列的行。如果它不在行中，请留空或不适用。 我应该使用 iloc 吗？

import pandas as pd
wordlist = [['this is sentence 1'],['this is sentence 2'],['this is not a sentence'],['ok who is this']]
query=['is','not']
df = pd.DataFrame(wordlist, columns = ['Name'])

for word in query:
    if word in df['Name']:
        df[word] = word
df


Output

Name                       is     not  <<column titles
0   this is sentence 1     is     NA
1   this is sentence 2     is     NA
2   this is not a sentence is     not
3   ok who is this         is     NA

Answer 1

创建一个搜索模式，然后使用Series.str.extractall 来获取单词。然后将每个唯一的单词变成一个 dummy 并聚合回原始行索引，并连接回原始 DataFrame。

import pandas as pd

pat = f'({"|".join(query)})'
#(is|not)

df_dummies = pd.get_dummies(df['Name'].str.extractall(pat)[0]).max(level=0)

df = pd.concat([df, df_dummies], axis=1)

#                     Name  is  not
#0      this is sentence 1   1    0
#1      this is sentence 2   1    0
#2  this is not a sentence   1    1
#3          ok who is this   1    0

---

如果您真的希望重复单词而不是 dummies，那么我们可以将虚拟 DataFrame 乘以列。

df_dummies = pd.get_dummies(df['Name'].str.extractall(pat)[0]).max(level=0)
df_dummies = df_dummies.mul(df_dummies.columns).replace('', np.NaN)
df = pd.concat([df, df_dummies], axis=1)

#                     Name  is  not
#0      this is sentence 1  is  NaN
#1      this is sentence 2  is  NaN
#2  this is not a sentence  is  not
#3          ok who is this  is  NaN

---

最后提醒一下，单词'this' 本身包含匹配'is'，因此上面的基本模式匹配单独的单词'is' 和'this' 的最后两个字符。如果要排除属于较长单词的匹配项，请修改搜索模式以包含查询中每个元素周围的单词边界：

pat = '(\\b' + '\\b|\\b'.join(query) + '\\b)'
#'(\\bis\\b|\\bnot\\b)'