python - 将包含列表的列拆分为虚拟列

Question

我想将技术列拆分为仅包含一种技术的虚拟列，换句话说，获得一列Matlab，另一列Python等。

我从df['Technologies'].str.split("/").to_list()开始，但后来被阻止了。下面是我得到的数据框:

    Job                 Technologies                                       City
0   Data scientist      Matlab/Python/Pyspark/Scikit-learn/Tensorflow      Paris
1   Data architecte     Python/Java/Scala/MongoDB                          Marseille
2   Lead data scientist SPSS/SQL/Teradata/R/Python/Tensorflow/scikit-learn Nantes
3   Data scientist      C/C++/Java/Python                                  Marseille
4   Data scientist      Matlab/Python/C++/numpy/Tensorflow/scikit-learn    Bordeaux

我希望得到的结果:

    Job                 Matlab Python Pyspark Scikit-learn ... C++ Teradata City
0   Data scientist      0      1      1       1                0   0        Paris
1   Data architecte     0      0      0       0                0   1        Marseille
2   Lead data scientist 1      0      0       1                1   0        Nantes
3   Data scientist      1      0      1       0                0   1        Marseille
4   Data scientist      0      1      0       1                1   1        Bordeaux

Answer 1

这是一种方法，尽管之后您必须进行一些清洁

df = pd.get_dummies(df['Technologies'].str.split('/', Expand=True).apply(lambda x: x.str.lower(), axis=1))

重命名列

df.columns = df.columns.str.split('_').str.get(1)

对重复列求和

df = df.groupby(lambda x: x, axis=1).sum()

 c  c++  java  matlab  mongodb  numpy  pyspark  python  r  scala  scikit-learn  spss  sql  tensorflow  teradata
0  0    0     0       1        0      0        1       1  0      0             1     0    0           1         0
1  0    0     1       0        1      0        0       1  0      1             0     0    0           0         0
2  0    0     0       0        0      0        0       1  1      0             1     1    1           1         1
3  1    1     1       0        0      0        0       1  0      0             0     0    0           0         0
4  0    1     0       1        0      1        0       1  0      0             1     0    0           1         0