use*_*621 6 python group-by pandas pandas-groupby
按列分组以在另一列中查找最频繁的值。例子:
import pandas as pd
d = {'col1': ['green','green','green','blue','blue','blue'],'col2': ['gx','gx','ow','nb','nb','mj']}
df = pd.DataFrame(data=d)
df
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给出:
col1 col2
green gx
green gx
green ow
blue nb
blue nb
blue xv
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结果:
因为green
拥有gx
和blue
拥有nb
使用SeriesGroupBy.value_counts
并选择索引的第一个值:
df = df.groupby('col1')['col2'].apply(lambda x: x.value_counts().index[0]).reset_index()
print (df)
col1 col2
0 blue nb
1 green gx
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df = df.groupby('col1')['col2'].value_counts().reset_index(name='v')
df = df.drop_duplicates('col1')[['col1','col2']]
print (df)
col1 col2
0 blue nb
2 green gx
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或者使用Series.mode
并按位置选择第一个值Series.iat
:
df = df.groupby('col1')['col2'].apply(lambda x: x.mode().iat[0]).reset_index()
print (df)
col1 col2
0 blue nb
1 green gx
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编辑:
问题仅在于NaN
s 组:
d = {'col1': ['green','green','green','blue','blue','blue'],
'col2': [np.nan,np.nan,np.nan,'nb','nb','mj']}
df = pd.DataFrame(data=d)
f = lambda x: np.nan if x.isnull().all() else x.value_counts().index[0]
#or
#f = lambda x: next(iter(x.value_counts().index), np.nan)
#another solution
#f = lambda x: next(iter(x.mode()), np.nan)
df = df.groupby('col1')['col2'].apply(f).reset_index()
print (df)
col1 col2
0 blue nb
1 green NaN
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