Pandas -如何在列中扁平化分层索引
Pandas -如何在列中扁平化分层索引
提问于 2013-01-25 02:03:11
我在轴1(列)中有一个具有分层索引的数据帧(来自groupby.agg操作)：
USAF WBAN year month day s_PC s_CL s_CD s_CNT tempf
sum sum sum sum amax amin
0 702730 26451 1993 1 1 1 0 12 13 30.92 24.98
1 702730 26451 1993 1 2 0 0 13 13 32.00 24.98
2 702730 26451 1993 1 3 1 10 2 13 23.00 6.98
3 702730 26451 1993 1 4 1 0 12 13 10.04 3.92
4 702730 26451 1993 1 5 3 0 10 13 19.94 10.94我想将其展平，使其看起来像这样(名称并不重要-我可以重命名)：
USAF WBAN year month day s_PC s_CL s_CD s_CNT tempf_amax tmpf_amin
0 702730 26451 1993 1 1 1 0 12 13 30.92 24.98
1 702730 26451 1993 1 2 0 0 13 13 32.00 24.98
2 702730 26451 1993 1 3 1 10 2 13 23.00 6.98
3 702730 26451 1993 1 4 1 0 12 13 10.04 3.92
4 702730 26451 1993 1 5 3 0 10 13 19.94 10.94我该怎么做呢？(我尝试了很多，但都无济于事。)
根据建议，这里是字典形式的头部
{('USAF', ''): {0: '702730',
1: '702730',
2: '702730',
3: '702730',
4: '702730'},
('WBAN', ''): {0: '26451', 1: '26451', 2: '26451', 3: '26451', 4: '26451'},
('day', ''): {0: 1, 1: 2, 2: 3, 3: 4, 4: 5},
('month', ''): {0: 1, 1: 1, 2: 1, 3: 1, 4: 1},
('s_CD', 'sum'): {0: 12.0, 1: 13.0, 2: 2.0, 3: 12.0, 4: 10.0},
('s_CL', 'sum'): {0: 0.0, 1: 0.0, 2: 10.0, 3: 0.0, 4: 0.0},
('s_CNT', 'sum'): {0: 13.0, 1: 13.0, 2: 13.0, 3: 13.0, 4: 13.0},
('s_PC', 'sum'): {0: 1.0, 1: 0.0, 2: 1.0, 3: 1.0, 4: 3.0},
('tempf', 'amax'): {0: 30.920000000000002,
1: 32.0,
2: 23.0,
3: 10.039999999999999,
4: 19.939999999999998},
('tempf', 'amin'): {0: 24.98,
1: 24.98,
2: 6.9799999999999969,
3: 3.9199999999999982,
4: 10.940000000000001},
('year', ''): {0: 1993, 1: 1993, 2: 1993, 3: 1993, 4: 1993}}回答 15
Stack Overflow用户
发布于 2015-12-14 16:00:21
pd.DataFrame(df.to_records()) # multiindex become columns and new index is integers onlyStack Overflow用户
发布于 2013-01-25 02:54:14
Andy Hayden的答案无疑是最简单的方法--如果您想避免重复的列标签，则需要稍微调整一下
In [34]: df
Out[34]:
USAF WBAN day month s_CD s_CL s_CNT s_PC tempf year
sum sum sum sum amax amin
0 702730 26451 1 1 12 0 13 1 30.92 24.98 1993
1 702730 26451 2 1 13 0 13 0 32.00 24.98 1993
2 702730 26451 3 1 2 10 13 1 23.00 6.98 1993
3 702730 26451 4 1 12 0 13 1 10.04 3.92 1993
4 702730 26451 5 1 10 0 13 3 19.94 10.94 1993
In [35]: mi = df.columns
In [36]: mi
Out[36]:
MultiIndex
[(USAF, ), (WBAN, ), (day, ), (month, ), (s_CD, sum), (s_CL, sum), (s_CNT, sum), (s_PC, sum), (tempf, amax), (tempf, amin), (year, )]
In [37]: mi.tolist()
Out[37]:
[('USAF', ''),
('WBAN', ''),
('day', ''),
('month', ''),
('s_CD', 'sum'),
('s_CL', 'sum'),
('s_CNT', 'sum'),
('s_PC', 'sum'),
('tempf', 'amax'),
('tempf', 'amin'),
('year', '')]
In [38]: ind = pd.Index([e[0] + e[1] for e in mi.tolist()])
In [39]: ind
Out[39]: Index([USAF, WBAN, day, month, s_CDsum, s_CLsum, s_CNTsum, s_PCsum, tempfamax, tempfamin, year], dtype=object)
In [40]: df.columns = ind
In [46]: df
Out[46]:
USAF WBAN day month s_CDsum s_CLsum s_CNTsum s_PCsum tempfamax tempfamin \
0 702730 26451 1 1 12 0 13 1 30.92 24.98
1 702730 26451 2 1 13 0 13 0 32.00 24.98
2 702730 26451 3 1 2 10 13 1 23.00 6.98
3 702730 26451 4 1 12 0 13 1 10.04 3.92
4 702730 26451 5 1 10 0 13 3 19.94 10.94
year
0 1993
1 1993
2 1993
3 1993
4 1993Stack Overflow用户
发布于 2017-05-31 08:25:38
df.columns = ['_'.join(tup).rstrip('_') for tup in df.columns.values]页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/14507794
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