如何将标题映射到熊猫中的列

Question

我有一个数据框架，例如：

A    B    C 
1    0    0
1    1    0
0    1    0
0    0    1

我想拥有：

 A    B    C  label
 1    0    0   A
 1    1    0   AB
 0    1    0   B
 0    0    1   C

我试图通过地图或申请进行操作，但我无法弄清楚。

Answer 1

这是一个惯用和性能解决方案

df['label'] = np.where(df, df.columns, '').sum(axis=1)
   A  B  C label
0  1  0  0     A
1  1  1  0    AB
2  0  1  0     B
3  0  0  1     C

Answer 2

使用 dot

df.assign(label=df.dot(df.columns))
   A  B  C label
0  1  0  0     A
1  1  1  0    AB
2  0  1  0     B
3  0  0  1     C

---

使用基础numpy阵列

相同的东西

df.assign(label=df.values.dot(df.columns.values))
   A  B  C label
0  1  0  0     A
1  1  1  0    AB
2  0  1  0     B
3  0  0  1     C

Answer 3

或使用melt和groupby

df1 = df.reset_index().melt('index')
df1 = df1[df1.value==1]
df['label'] = df1.groupby('index').variable.sum()
df
Out[976]: 
   A  B  C label
0  1  0  0     A
1  1  1  0    AB
2  0  1  0     B
3  0  0  1     C

或

df['label'] = df.T.apply(lambda x: ''.join(x.index[x==1]),axis=0)
df
Out[984]: 
   A  B  C label
0  1  0  0     A
1  1  1  0    AB
2  0  1  0     B
3  0  0  1     C

Answer 4

In [101]: df['label'] = df.apply(lambda x: ''.join(df.columns[x.astype(bool)].tolist()), axis=1)
In [102]: df
Out[102]:
   A  B  C label
0  1  0  0     A
1  1  1  0    AB
2  0  1  0     B
3  0  0  1     C

ps我肯定会选择 @ted的解决方案，因为它更好，而且要好得多...更快

Answer 5

df = df.assign(label=[''.join([df.columns[n] for n, bool in enumerate(row) if bool]) 
                      for _, row in df.iterrows()])
>>> df
   A  B  C label
0  1  0  0     A
1  1  1  0    AB
2  0  1  0     B
3  0  0  1     C

时间

# Set-up:
df_ = pd.concat([df] * 10000)
%%timeit
# Solution by @Wen 
df1 = df_.reset_index().melt('index')
df1 = df1[df1.value==1]
df['label'] = df1.groupby('index').variable.sum()
# 10 loops, best of 3: 47.6 ms per loop
%%timeit
# Solution by @MaxU
df_['label'] = df_.apply(lambda x: ''.join(df_.columns[x.astype(bool)].tolist()), axis=1)
# 1 loop, best of 3: 4.99 s per loop
%%timeit
# Solution by @TedPetrou
df_['label'] = np.where(df_, df_.columns, '').sum(axis=1)
# 100 loops, best of 3: 12.5 ms per loop
%%timeit
# Solution by @Alexander
df_['label'] = [''.join([df_.columns[n] for n, bool in enumerate(row) if bool]) for _, row in df_.iterrows()]
# 1 loop, best of 3: 3.75 s per loop
%%time
# Solution by @PiRSquared
df_['label'] = df_.dot(df_.columns)
# CPU times: user 18.1 ms, sys: 706 µs, total: 18.8 ms
# Wall time: 18.9 ms