如何在熊猫身上做到这一点:
我在单个文本列上有一个函数extract_text_features，返回多个输出列。具体来说，该函数返回6个值。
该函数可以工作，但是似乎没有任何合适的返回类型(pandas DataFrame/ numpy数组/ Python列表)，以便输出可以正确分配df。Ix [:，10:16] = df.textcol.map(extract_text_features)
所以我认为我需要回落到迭代与df.iterrows()，按此?
更新: 使用df.iterrows()迭代至少要慢20倍，因此我放弃并将该函数分解为6个不同的.map(lambda…)调用。
更新2:这个问题是在v0.11.0版本被问到的，在可用性df之前。在v0.16中改进了Apply或添加了df.assign()。因此，很多问题和答案都不太相关。
当前回答
基于user1827356的答案，你可以使用df.merge一次性完成赋值:
df.merge(df.textcol.apply(lambda s: pd.Series({'feature1':s+1, 'feature2':s-1})),
left_index=True, right_index=True)
textcol feature1 feature2
0 0.772692 1.772692 -0.227308
1 0.857210 1.857210 -0.142790
2 0.065639 1.065639 -0.934361
3 0.819160 1.819160 -0.180840
4 0.088212 1.088212 -0.911788
编辑: 请注意内存消耗大，速度慢:https://ys-l.github.io/posts/2015/08/28/how-not-to-use-pandas-apply/ !
其他回答
我通常使用zip:
>>> df = pd.DataFrame([[i] for i in range(10)], columns=['num'])
>>> df
num
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
>>> def powers(x):
>>> return x, x**2, x**3, x**4, x**5, x**6
>>> df['p1'], df['p2'], df['p3'], df['p4'], df['p5'], df['p6'] = \
>>> zip(*df['num'].map(powers))
>>> df
num p1 p2 p3 p4 p5 p6
0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1
2 2 2 4 8 16 32 64
3 3 3 9 27 81 243 729
4 4 4 16 64 256 1024 4096
5 5 5 25 125 625 3125 15625
6 6 6 36 216 1296 7776 46656
7 7 7 49 343 2401 16807 117649
8 8 8 64 512 4096 32768 262144
9 9 9 81 729 6561 59049 531441
我有一个更复杂的情况，数据集有一个嵌套结构:
import json
data = '{"TextID":{"0":"0038f0569e","1":"003eb6998d","2":"006da49ea0"},"Summary":{"0":{"Crisis_Level":["c"],"Type":["d"],"Special_Date":["a"]},"1":{"Crisis_Level":["d"],"Type":["a","d"],"Special_Date":["a"]},"2":{"Crisis_Level":["d"],"Type":["a"],"Special_Date":["a"]}}}'
df = pd.DataFrame.from_dict(json.loads(data))
print(df)
输出:
TextID Summary
0 0038f0569e {'Crisis_Level': ['c'], 'Type': ['d'], 'Specia...
1 003eb6998d {'Crisis_Level': ['d'], 'Type': ['a', 'd'], 'S...
2 006da49ea0 {'Crisis_Level': ['d'], 'Type': ['a'], 'Specia...
Summary列包含dict对象，所以我使用apply和from_dict和stack来提取每一行的dict:
df2 = df.apply(
lambda x: pd.DataFrame.from_dict(x[1], orient='index').stack(), axis=1)
print(df2)
输出:
Crisis_Level Special_Date Type
0 0 0 1
0 c a d NaN
1 d a a d
2 d a a NaN
看起来不错，但缺少TextID列。为了得到TextID列回来，我尝试了三种方法:
Modify apply to return multiple columns: df_tmp = df.copy() df_tmp[['TextID', 'Summary']] = df.apply( lambda x: pd.Series([x[0], pd.DataFrame.from_dict(x[1], orient='index').stack()]), axis=1) print(df_tmp) output: TextID Summary 0 0038f0569e Crisis_Level 0 c Type 0 d Spec... 1 003eb6998d Crisis_Level 0 d Type 0 a ... 2 006da49ea0 Crisis_Level 0 d Type 0 a Spec... But this is not what I want, the Summary structure are flatten. Use pd.concat: df_tmp2 = pd.concat([df['TextID'], df2], axis=1) print(df_tmp2) output: TextID (Crisis_Level, 0) (Special_Date, 0) (Type, 0) (Type, 1) 0 0038f0569e c a d NaN 1 003eb6998d d a a d 2 006da49ea0 d a a NaN Looks fine, the MultiIndex column structure are preserved as tuple. But check columns type: df_tmp2.columns output: Index(['TextID', ('Crisis_Level', 0), ('Special_Date', 0), ('Type', 0), ('Type', 1)], dtype='object') Just as a regular Index class, not MultiIndex class. use set_index: Turn all columns you want to preserve into row index, after some complicated apply function and then reset_index to get columns back: df_tmp3 = df.set_index('TextID') df_tmp3 = df_tmp3.apply( lambda x: pd.DataFrame.from_dict(x[0], orient='index').stack(), axis=1) df_tmp3 = df_tmp3.reset_index(level=0) print(df_tmp3) output: TextID Crisis_Level Special_Date Type 0 0 0 1 0 0038f0569e c a d NaN 1 003eb6998d d a a d 2 006da49ea0 d a a NaN Check the type of columns df_tmp3.columns output: MultiIndex(levels=[['Crisis_Level', 'Special_Date', 'Type', 'TextID'], [0, 1, '']], codes=[[3, 0, 1, 2, 2], [2, 0, 0, 0, 1]])
因此，如果apply函数将返回MultiIndex列，并且希望保留它，则可能需要尝试第三种方法。
在2020年，我使用apply()参数result_type='expand'
applied_df = df.apply(lambda row: fn(row.text), axis='columns', result_type='expand')
df = pd.concat([df, applied_df], axis='columns')
对于95%的用例来说，这是正确且最简单的方法:
>>> df = pd.DataFrame(zip(*[range(10)]), columns=['num'])
>>> df
num
0 0
1 1
2 2
3 3
4 4
5 5
>>> def example(x):
... x['p1'] = x['num']**2
... x['p2'] = x['num']**3
... x['p3'] = x['num']**4
... return x
>>> df = df.apply(example, axis=1)
>>> df
num p1 p2 p3
0 0 0 0 0
1 1 1 1 1
2 2 4 8 16
3 3 9 27 81
4 4 16 64 256
这是我过去所做的
df = pd.DataFrame({'textcol' : np.random.rand(5)})
df
textcol
0 0.626524
1 0.119967
2 0.803650
3 0.100880
4 0.017859
df.textcol.apply(lambda s: pd.Series({'feature1':s+1, 'feature2':s-1}))
feature1 feature2
0 1.626524 -0.373476
1 1.119967 -0.880033
2 1.803650 -0.196350
3 1.100880 -0.899120
4 1.017859 -0.982141
为完整性而编辑
pd.concat([df, df.textcol.apply(lambda s: pd.Series({'feature1':s+1, 'feature2':s-1}))], axis=1)
textcol feature1 feature2
0 0.626524 1.626524 -0.373476
1 0.119967 1.119967 -0.880033
2 0.803650 1.803650 -0.196350
3 0.100880 1.100880 -0.899120
4 0.017859 1.017859 -0.982141
