展开 pandas 数据框列 [英] exploding a pandas dataframe column

问题描述

我有一个看起来像这样的Pandas Dataframe:

I have a Pandas Dataframe that looks something like this:

text = ["abcd", "efgh", "ijkl", "mnop", "qrst", "uvwx", "yz"]

labels = ["label_1, label_2", 
          "label_1, label_3, label_2", 
          "label_2, label_4", 
          "label_1, label_2, label_5", 
          "label_2, label_3", 
          "label_3, label_5, label_1, label_2", 
          "label_1, label_3"]

df = pd.DataFrame(dict(text=text, labels=labels))
df



   text                              labels
0  abcd                    label_1, label_2
1  efgh           label_1, label_3, label_2
2  ijkl                    label_2, label_4
3  mnop           label_1, label_2, label_5
4  qrst                    label_2, label_3
5  uvwx  label_3, label_5, label_1, label_2
6    yz                    label_1, label_3

我想将数据框格式化为以下格式:

I would like to format the dataframe into something like this:

text  label_1  label_2  label_3  label_4  label_5

abcd        1.0      1.0      0.0      0.0      0.0
efgh        1.0      1.0      1.0      0.0      0.0
ijkl        0.0      1.0      0.0      1.0      0.0
mnop        1.0      1.0      0.0      0.0      1.0
qrst        0.0      1.0      1.0      0.0      0.0
uvwx        1.0      1.0      1.0      0.0      1.0
yz          1.0      0.0      1.0      0.0      0.0

我该怎么做? (我知道我可以通过执行df.labels.str.split(",")之类的操作来分割标签中的字符串并将其转换为列表，但是不确定如何从那里开始.

How can I accomplish this? (I know I can split the strings in the labels and convert them into lists by doing something like df.labels.str.split(",") but not sure as to how to proceed from there.

(因此，基本上，我想将标签列中的那些关键字转换成其自己的列，并在它们出现在预期输出中时填充为1)

(so basically I'd like to convert those keywords in the labels columns into its own columns and fill in 1 whenever they appear as shown in expected output)

推荐答案

您可以使用

You can use pd.Series.str.get_dummies and combine with the text series:

dummies = df['labels'].str.replace(' ', '').str.get_dummies(',')
res = df['text'].to_frame().join(dummies)

print(res)

   text  label_1  label_2  label_3  label_4  label_5
0  abcd        1        1        0        0        0
1  efgh        1        1        1        0        0
2  ijkl        0        1        0        1        0
3  mnop        1        1        0        0        1
4  qrst        0        1        1        0        0
5  uvwx        1        1        1        0        1
6    yz        1        0        1        0        0

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