[英]Adding values for missing data combinations in Pandas
I've got a pandas data frame containing something like the following:我有一个 pandas 数据框,其中包含如下内容:
person_id status year count
0 'pass' 1980 4
0 'fail' 1982 1
1 'pass' 1981 2
If I know that all possible values for each field are:如果我知道每个字段的所有可能值是:
all_person_ids = [0, 1, 2]
all_statuses = ['pass', 'fail']
all_years = [1980, 1981, 1982]
I'd like to populate the original data frame with count=0
for missing data combinations (of person_id, status, and year), ie I'd like the new data frame to contain:我想用count=0
填充原始数据框以用于缺少数据组合(person_id、status 和 year),即我希望新数据框包含:
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
Is there an efficient way to achieve this in pandas?在 pandas 中有没有一种有效的方法来实现这一点?
You can use itertools.product
to generate all combinations, then construct a df from this, merge
it with your original df along with fillna
to fill missing count values with 0
:您可以使用itertools.product
生成所有组合,然后从中构建一个 df, merge
其与您的原始 df 以及fillna
以使用0
填充缺失的计数值:
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
create a MultiIndex by MultiIndex.from_product() and then set_index()
, reindex()
, reset_index()
.通过 MultiIndex.from_product() 然后set_index()
, reindex()
, reset_index()
创建一个 MultiIndex 。
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()
You can use pyjanitor 's complete
method.您可以使用pyjanitor的complete
方法。
It accepts column names as input as well as {name: values} dictionaries with the exhaustive list of wanted values to complete:它接受列名作为输入以及 {name: values} 字典,其中包含要完成的详尽列表:
import janitor
df.complete({'person_id': [0,1,2]}, 'status', 'year').fillna(0, downcast='infer')
output: output:
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
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')\
.merge(pd.Series(all_years).to_frame('year'), how='cross')\
.merge(df1,on=['person_id','status','year'], how='left')\
.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
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