Bry*_*Fok 13 python group-by dataframe pandas cumsum
我在DataFrame中有一个包含值的列:
[1, 1, -1, 1, -1, -1]
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我该如何将它们分组呢?
[1,1] [-1] [1] [-1, -1]
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jez*_*ael 28
您可以groupby按自定义使用Series:
df = pd.DataFrame({'a': [1, 1, -1, 1, -1, -1]})
print (df)
a
0 1
1 1
2 -1
3 1
4 -1
5 -1
print ((df.a != df.a.shift()).cumsum())
0 1
1 1
2 2
3 3
4 4
5 4
Name: a, dtype: int32
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for i, g in df.groupby([(df.a != df.a.shift()).cumsum()]):
print (i)
print (g)
print (g.a.tolist())
a
0 1
1 1
[1, 1]
2
a
2 -1
[-1]
3
a
3 1
[1]
4
a
4 -1
5 -1
[-1, -1]
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WeN*_*Ben 10
使用groupby从itertools来自杰斯数据
from itertools import groupby
[ list(group) for key, group in groupby(df.a.values.tolist())]
Out[361]: [[1, 1], [-1], [1], [-1, -1]]
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Series.diff是标记组边界的另一种方法(a!=a.shiftmeans a.diff!=0):
consecutives = df['a'].diff().ne(0).cumsum()
# 0 1
# 1 1
# 2 2
# 3 3
# 4 4
# 5 4
# Name: a, dtype: int64
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要将这些组转换为一系列列表(请参阅列表列表的其他答案),请使用groupby.agg或进行聚合groupby.apply:
df['a'].groupby(consecutives).agg(list)
# a
# 1 [1, 1]
# 2 [-1]
# 3 [1]
# 4 [-1, -1]
# Name: a, dtype: object
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