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按列分组,然后根据条件筛选

pandas-groupby group-by dataframe pandas python

Zmnako Awrahman beloncfy · 技术社区 · 8 年前

我有一个df,我想根据分组筛选出一个列。我想按组合分组(( cc , odd , tree1 和 tree2 )如果是第4天,则保留,否则放弃

df = pd.DataFrame()
df['cc'] = ['BB', 'BB', 'BB', 'BB','BB', 'BB','BB', 'BB', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ']
df['odd'] = [3434, 3434, 3434, 3434, 3435, 3435, 3435, 3435, 3434, 3434, 3434, 3434, 3435, 3435, 3435, 3435, 3434, 3434, 3434, 3434, 3435, 3435, 3435, 3435]
df['tree1'] = ['ASP', 'ASP', 'ASP', 'ASP', 'SAP', 'SAP', 'SAP', 'SAP', 'ASP', 'ASP', 'ASP', 'ASP', 'SAP', 'SAP', 'SAP', 'SAP', 'ASP', 'ASP', 'ASP', 'ASP', 'SAP', 'SAP', 'SAP', 'SAP']
df['tree2'] = ['ATK', 'ATK','ATK','ATK','ATK','ATK','ATK','ATK', 'ATK', 'ATK','ATK','ATK','ATK','ATK','ATK','ATK', 'ATK', 'ATK','ATK','ATK','ATK','ATK','ATK','ATK']
df['day'] = [1, 2, 3, 4, 3, 4, 5, 6, 2, 3, 4, 5, 1, 3, 5, 7, 1, 2, 6, 8, 2, 4, 6, 8]
df

我试过了,但这会删除日值小于4的任何行

df_grouped = df.groupby(['cc', 'odd', 'tree1', 'tree2']).filter(df['day'] > 4)

我知道这个错误 TypeError: 'Series' object is not callable

试过这个

df_grouped = df.groupby(['cc', 'odd', 'tree1', 'tree2']).filter(lambda x: x['day'] > 4)

我知道这个错误 TypeError: filter function returned a Series, but expected a scalar bool .

我寻找并试图解决这些错误,但提出的解决方案对我不起作用。我想得到如下df:

df1 = pd.DataFrame()
df1['cc'] = ['BB', 'BB','BB', 'BB', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ']
df1['odd'] = [3435, 3435, 3435, 3435, 3434, 3434, 3434, 3434, 3435, 3435, 3435, 3435, 3434, 3434, 3434, 3434, 3435, 3435, 3435, 3435]
df1['tree1'] = ['SAP', 'SAP', 'SAP', 'SAP', 'ASP', 'ASP', 'ASP', 'ASP', 'SAP', 'SAP', 'SAP', 'SAP', 'ASP', 'ASP', 'ASP', 'ASP', 'SAP', 'SAP', 'SAP', 'SAP']
df1['tree2'] = ['ATK','ATK','ATK','ATK', 'ATK', 'ATK','ATK','ATK','ATK','ATK','ATK','ATK', 'ATK', 'ATK','ATK','ATK','ATK','ATK','ATK','ATK']
df1['day'] = [3, 4, 5, 6, 2, 3, 4, 5, 1, 3, 5, 7, 1, 2, 6, 8, 2, 4, 6, 8]
df1

我试着用 any 但我做不到,它只会回来 True 或 False 而不是筛选数据帧。

2 回复 | 直到 8 年前

cs95 abhishek58g 8 年前

既然我明白了你的意思,我们试试 transform + any :

df[df.assign(key=df.day > 4)
     .groupby(['cc', 'odd', 'tree1', 'tree2']).key.transform('any')
]

或者,

df[df.day.gt(4).groupby([df.cc, df.odd, df.tree1, df.tree2]).transform('any')]

    cc   odd tree1 tree2  day
4   BB  3435   SAP   ATK    3
5   BB  3435   SAP   ATK    4
6   BB  3435   SAP   ATK    5
7   BB  3435   SAP   ATK    6
8   DD  3434   ASP   ATK    2
9   DD  3434   ASP   ATK    3
10  DD  3434   ASP   ATK    4
11  DD  3434   ASP   ATK    5
12  DD  3435   SAP   ATK    1
13  DD  3435   SAP   ATK    3
14  DD  3435   SAP   ATK    5
15  DD  3435   SAP   ATK    7
16  ZZ  3434   ASP   ATK    1
17  ZZ  3434   ASP   ATK    2
18  ZZ  3434   ASP   ATK    6
19  ZZ  3434   ASP   ATK    8
20  ZZ  3435   SAP   ATK    2
21  ZZ  3435   SAP   ATK    4
22  ZZ  3435   SAP   ATK    6
23  ZZ  3435   SAP   ATK    8

EdChum Arthur G 8 年前

你想要的是:

In[116]:
df_grouped = df.groupby(['cc', 'odd', 'tree1', 'tree2']).filter(lambda x: (x['day'] > 4).any())
df_grouped

Out[116]: 
    cc   odd tree1 tree2  day
4   BB  3435   SAP   ATK    3
5   BB  3435   SAP   ATK    4
6   BB  3435   SAP   ATK    5
7   BB  3435   SAP   ATK    6
8   DD  3434   ASP   ATK    2
9   DD  3434   ASP   ATK    3
10  DD  3434   ASP   ATK    4
11  DD  3434   ASP   ATK    5
12  DD  3435   SAP   ATK    1
13  DD  3435   SAP   ATK    3
14  DD  3435   SAP   ATK    5
15  DD  3435   SAP   ATK    7
16  ZZ  3434   ASP   ATK    1
17  ZZ  3434   ASP   ATK    2
18  ZZ  3434   ASP   ATK    6
19  ZZ  3434   ASP   ATK    8
20  ZZ  3435   SAP   ATK    2
21  ZZ  3435   SAP   ATK    4
22  ZZ  3435   SAP   ATK    6
23  ZZ  3435   SAP   ATK    8

所以这将过滤掉组中没有 'day' 值大于4

计时 :

%timeit df[df.day.gt(4).groupby([df.cc, df.odd, df.tree1, df.tree2]).transform('any')]
%timeit df.groupby(['cc', 'odd', 'tree1', 'tree2']).filter(lambda x: (x['day'] > 4).any())
%timeit df[df.assign(key=df.day > 4).groupby(['cc', 'odd', 'tree1', 'tree2']).key.transform('any')]
100 loops, best of 3: 5.9 ms per loop
100 loops, best of 3: 5.42 ms per loop
100 loops, best of 3: 3.62 ms per loop

所以@coldspeed的第一个方法是这里最快的