在 Pandas 中为缺失数据组合添加值

问题描述：

我有一个包含以下内容的熊猫数据框：

person_id   status    year    count
0           'pass'    1980    4
0           'fail'    1982    1
1           'pass'    1981    2

如果我知道每个字段的所有可能值都是：

all_person_ids = [0, 1, 2]
all_statuses = ['pass', 'fail']
all_years = [1980, 1981, 1982]

我想用count=0缺失的数据组合（person_id、status 和 year）填充原始数据框，即我希望新的数据框包含：

person_id   status    year    count
0           'pass'    1980    4
0           'pass'    1981    0
0           'pass'    1982    0
0           'fail'    1980    0
0           'fail'    1981    0
0           'fail'    1982    2
1           'pass'    1980    0
1           'pass'    1981    2
1           'pass'    1982    0
1           'fail'    1980    0
1           'fail'    1981    0
1           'fail'    1982    0
2           'pass'    1980    0
2           'pass'    1981    0
2           'pass'    1982    0
2           'fail'    1980    0
2           'fail'    1981    0
2           'fail'    1982    0

有没有一种有效的方法可以在熊猫中实现这一点？

解决方案 1：

您可以使用itertools.product来生成所有组合，然后从中构建一个 df，merge并用您的原始 dffillna填充缺失的计数值0：

In [77]:
import itertools
all_person_ids = [0, 1, 2]
all_statuses = ['pass', 'fail']
all_years = [1980, 1981, 1982]
combined = [all_person_ids, all_statuses, all_years]
df1 = pd.DataFrame(columns = ['person_id', 'status', 'year'], data=list(itertools.product(*combined)))
df1

Out[77]:
    person_id status  year
0           0   pass  1980
1           0   pass  1981
2           0   pass  1982
3           0   fail  1980
4           0   fail  1981
5           0   fail  1982
6           1   pass  1980
7           1   pass  1981
8           1   pass  1982
9           1   fail  1980
10          1   fail  1981
11          1   fail  1982
12          2   pass  1980
13          2   pass  1981
14          2   pass  1982
15          2   fail  1980
16          2   fail  1981
17          2   fail  1982

In [82]:    
df1 = df1.merge(df, how='left').fillna(0)
df1

Out[82]:
    person_id status  year  count
0           0   pass  1980      4
1           0   pass  1981      0
2           0   pass  1982      0
3           0   fail  1980      0
4           0   fail  1981      0
5           0   fail  1982      1
6           1   pass  1980      0
7           1   pass  1981      2
8           1   pass  1982      0
9           1   fail  1980      0
10          1   fail  1981      0
11          1   fail  1982      0
12          2   pass  1980      0
13          2   pass  1981      0
14          2   pass  1982      0
15          2   fail  1980      0
16          2   fail  1981      0
17          2   fail  1982      0

解决方案 2：

通过 MultiIndex.from_product() 创建一个 MultiIndex ，然后set_index()，，reindex()。reset_index()

import pandas as pd
import io

all_person_ids = [0, 1, 2]
all_statuses = ['pass', 'fail']
all_years = [1980, 1981, 1982]
df = pd.read_csv(io.BytesIO("""person_id   status    year    count
0           pass    1980    4
0           fail    1982    1
1           pass    1981    2"""), delim_whitespace=True)
names = ["person_id", "status", "year"]

mind = pd.MultiIndex.from_product(
    [all_person_ids, all_statuses, all_years], names=names)
df.set_index(names).reindex(mind, fill_value=0).reset_index()

解决方案 3：

您可以使用pyjanitor的complete方法。

它接受列名作为输入以及 {name: values} 字典，其中包含要完成的所需值的详尽列表：

import janitor
df.complete({'person_id': [0,1,2]}, 'status', 'year').fillna(0, downcast='infer')

输出：

    person_id  status  year  count
0           0  'fail'  1980      0
1           0  'fail'  1981      0
2           0  'fail'  1982      1
3           0  'pass'  1980      4
4           0  'pass'  1981      0
5           0  'pass'  1982      0
6           1  'fail'  1980      0
7           1  'fail'  1981      0
8           1  'fail'  1982      0
9           1  'pass'  1980      0
10          1  'pass'  1981      2
11          1  'pass'  1982      0
12          2  'fail'  1980      0
13          2  'fail'  1981      0
14          2  'fail'  1982      0
15          2  'pass'  1980      0
16          2  'pass'  1981      0
17          2  'pass'  1982      0

解决方案 4：

all_person_ids = [0, 1, 2]
all_statuses = ['pass', 'fail']
all_years = [1980, 1981, 1982]


pd.Series(all_person_ids).to_frame('person_id').merge(pd.Series(all_statuses).to_frame('status'), how='cross')\n    .merge(pd.Series(all_years).to_frame('year'), how='cross')\n    .merge(df1,on=['person_id','status','year'], how='left')\n    .fillna(0)

    person_id status  year  count
0           0   pass  1980    4.0
1           0   pass  1981    0.0
2           0   pass  1982    0.0
3           0   fail  1980    0.0
4           0   fail  1981    0.0
5           0   fail  1982    1.0
6           1   pass  1980    0.0
7           1   pass  1981    2.0
8           1   pass  1982    0.0
9           1   fail  1980    0.0
10          1   fail  1981    0.0
11          1   fail  1982    0.0
12          2   pass  1980    0.0
13          2   pass  1981    0.0
14          2   pass  1982    0.0
15          2   fail  1980    0.0
16          2   fail  1981    0.0
17          2   fail  1982    0.0