Question

我有一个pandas数据帧：

apple   banana  carrot  diet coke
1         1       1         0
0         1       0         0
1         0       0         0
1         0       1         1
0         1       1         0
0         1       1         0

我想将此转换为以下内容：

[['apple', 'banana', 'carrot'],
 ['banana'],
 ['apple'],
 ['apple', 'carrot', 'diet coke'],
 ['banana', 'carrot'],
 ['banana', 'carrot']]

我该怎么办？非常感谢。

Answer 1

因为生命很短暂，我可能会像

那样直截了当

>>> fruit = [df.columns[row.astype(bool)].tolist() for row in df.values]
>>> pprint.pprint(fruit)
[['apple', 'banana', 'carrot'],
 ['banana'],
 ['apple'],
 ['apple', 'carrot', 'diet coke'],
 ['banana', 'carrot'],
 ['banana', 'carrot']]

这是有效的，因为我们可以使用布尔数组（row.astype(bool)）来仅选择行为True的df.columns元素。

Answer 2

@DSM解决方案非常棒，但只有在您的值为1或0时才有效。如果您需要将其与其他值进行比较，您可以尝试：

[df.columns[df.ix[i,:]==1].tolist() for i in range(len(df.index))]

In [156]: [df.columns[df.ix[i,:]==1].tolist() for i in range(len(df.index))]
Out[156]:
[['apple', 'banana', 'carrot'],
 ['banana'],
 ['apple'],
 ['apple', 'carrot', 'dietcoke'],
 ['banana', 'carrot'],
 ['banana', 'carrot']]

修改

虽然你可以修改一下@DSM解决方案：

In [177]: [df.columns[row == 1].tolist() for row in df.values] Out[177]: [['apple', 'banana', 'carrot'], ['banana'], ['apple'], ['apple', 'carrot', 'dietcoke'], ['banana', 'carrot'], ['banana', 'carrot']]

一些性能测试：

In [179]: %timeit [df.columns[row == 1].tolist() for row in df.values] The slowest run took 4.03 times longer than the fastest. This could mean that an intermediate result is being cached 1000 loops, best of 3: 212 us per loop In [180]: %timeit [df.columns[row.astype(bool)].tolist() for row in df.values] 10000 loops, best of 3: 186 us per loop In [181]: %timeit [df.columns[df.ix[i,:]==1].tolist() for i in range(len(df.index))] 100 loops, best of 3: 2.4 ms per loop

Answer 3

In [24]: import pandas as pd

In [25]: import io

In [26]: data = """                                          
apple   banana  carrot  dietcoke
1         1       1         0
0         1       0         0
1         0       0         0
1         0       1         1
0         1       1         0
0         1       1         0
"""

In [27]: df = pd.read_csv(io.StringIO(data), delimiter='\s+')

In [28]: df
Out[28]: 
   apple  banana  carrot  dietcoke
0      1       1       1         0
1      0       1       0         0
2      1       0       0         0
3      1       0       1         1
4      0       1       1         0
5      0       1       1         0

In [29]: [[df.columns[i] for i,field in enumerate(record) if field == 1] for j,*record in df.itertuples()]
Out[29]: 
[['apple', 'banana', 'carrot'],
 ['banana'],
 ['apple'],
 ['apple', 'carrot', 'dietcoke'],
 ['banana', 'carrot'],
 ['banana', 'carrot']]

不使用列表推导和扩展元组解包的解决方案如下所示：

In [32]: result = []

In [33]: for record in df.itertuples():
   ....:     row = []
   ....:     for i,field in enumerate(record[1:]):
   ....:         if field == 1:
   ....:             row.append(df.columns[i])
   ....:     result.append(row)
   ....: 

In [34]: result
Out[34]: 
[['apple', 'banana', 'carrot'],
 ['banana'],
 ['apple'],
 ['apple', 'carrot', 'dietcoke'],
 ['banana', 'carrot'],
 ['banana', 'carrot']]

Answer 4

您可以在Pedro提及的情况下进行编辑和创建，或者只使用stack()和groupby()列出，

df
Out[14]: 
   apple  banana  carrot  diet_coke
0      1       1       1          0
1      0       1       0          0
2      1       0       0          0
3      1       0       1          1
4      0       1       1          0
5      0       1       1          0

df.stack()
Out[15]: 
0  apple        1
   banana       1
   carrot       1
   diet_coke    0
1  apple        0
   banana       1
   carrot       0
   diet_coke    0
2  apple        1
   banana       0
   carrot       0
   diet_coke    0
3  apple        1
   banana       0
   carrot       1
   diet_coke    1
4  apple        0
   banana       1
   carrot       1
   diet_coke    0
5  apple        0
   banana       1
   carrot       1
   diet_coke    0
dtype: int64



df.stack()[df.stack().values ==1].reset_index()
Out[20]: 
    level_0    level_1  0
0         0      apple  1
1         0     banana  1
2         0     carrot  1
3         1     banana  1
4         2      apple  1
5         3      apple  1
6         3     carrot  1
7         3  diet_coke  1
8         4     banana  1
9         4     carrot  1
10        5     banana  1
11        5     carrot  1


newdf.groupby(['level_0'])['level_1'].apply(list)
Out[27]: 
level_0
0       [apple, banana, carrot]
1                      [banana]
2                       [apple]
3    [apple, carrot, diet_coke]
4              [banana, carrot]
5              [banana, carrot]
Name: level_1, dtype: object

将pandas数据帧转换为列表

4 个答案: