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Question

我有带有电影标题的数据帧和带有类型的列。例如标题为“ One”的电影是“ Action”和“ Vestern”，因为在适当的列中具有“ 1”。

   Movie  Action  Fantasy  Vestern
0    One       1        0        1
1    Two       0        0        1
2  Three       1        1        0

我的目标是创建列genres，其中将包含特定电影具有的每种流派的名称。为此，我尝试使用lambda和list comprehension，因为这样做会有所帮助。但是在运行以下代码行之后：

df['genres'] = df.apply(lambda x: [x+"|"+x for x in df.columns if x!=0])

我每行只有NaN值：

   Movie  Action  Fantasy  Vestern genres
0    One       1        0        1    NaN
1    Two       0        0        1    NaN
2  Three       1        1        0    NaN

也尝试使用groupby，但没有成功。

预期输出为：

   Movie  Action  Fantasy  Vestern          genres
0    One       1        0        1  Action|Vestern
1    Two       0        0        1         Vestern
2  Three       1        1        0  Action|Fantasy

要复制的代码：

import pandas as pd
import numpy as np

df = pd.DataFrame({"Movie":['One','Two','Three'],
                   "Action":[1,0,1],
                   "Fantasy":[0,0,1],
                   "Vestern":[1,1,0]})
print(df)

感谢您的帮助

Answer 1

import pandas as pd
import numpy as np

df = pd.DataFrame({"Movie":['One','Two','Three'],
                   "Action":[1,0,1],
                   "Fantasy":[0,0,1],
                   "Vestern":[1,1,0]})

cols = df.columns.tolist()[1:]

df['genres'] = df.apply(lambda x: "|".join(str(z) for z in [i for i in cols if x[i] !=0]) ,axis=1)
print(df)

输出

Movie  Action  Fantasy  Vestern          genres
0    One       1        0        1  Action|Vestern
1    Two       0        0        1         Vestern
2  Three       1        1        0  Action|Fantasy

Answer 2

要提高性能，可以使用dot所有列，而不是第一列，所有列都没有，最后separator，最后删除|，rstrip：

df['new'] = df.iloc[:, 1:].dot(df.columns[1:] + '|').str.rstrip('|')
print (df)
   Movie  Action  Fantasy  Vestern             new
0    One       1        0        1  Action|Vestern
1    Two       0        0        1         Vestern
2  Three       1        1        0  Action|Fantasy

或者使用列表推导来连接所有不带空字符串的值：

arr = df.iloc[:, 1:].values * df.columns[1:].values
df['new'] = ['|'.join(y for y in x if y) for x in arr]
print (df)
   Movie  Action  Fantasy  Vestern             new
0    One       1        0        1  Action|Vestern
1    Two       0        0        1         Vestern
2  Three       1        1        0  Action|Fantasy

性能：

In [54]: %timeit (jez1(df.copy()))
25.2 ms ± 2.31 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [55]: %timeit (jez2(df.copy()))
61.4 ms ± 769 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [56]: %timeit (csm(df.copy()))
1.46 s ± 35.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)



df = pd.DataFrame({"Movie":['One','Two','Three'],
                   "Action":[1,0,1],
                   "Fantasy":[0,0,1],
                   "Vestern":[1,1,0]})
#print(df)

#30k rows
df = pd.concat([df] * 10000, ignore_index=True)

def csm(df):
    cols = df.columns.tolist()[1:]
    df['genres'] = df.apply(lambda x: "|".join(str(z) for z in [i for i in cols if x[i] !=0]) ,axis=1)
    return df

def jez1(df):
    df['new'] = df.iloc[:, 1:].dot(df.columns[1:] + '|').str.rstrip('|')
    return df

def jez2(df):
    arr = df.iloc[:, 1:].values * df.columns[1:].values
    df['new'] = ['|'.join(y for y in x if y) for x in arr]
    return df

根据另一个具有非空值的列，使用lambda在数据框中创建一列

2 个答案:

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