如何在 Python 3.8 中的公共 id 中创建包含另一行年龄的变量？

Question

df = pd.DataFrame({'id': [1,1,2,2,2,3,3,4,4,4,5,5,6,6],
                   'rank' : [1,2,1,2,3,1,2,1,2,3,1,2,1,2],
                   'age' : [20,2,30,28,4,26,3,22,23,1,34,29,56,25],
                   'sex' : ['female','female','male','female','male','female','male','female','male','male','male','female','female','female'],
                   'care' : [np.nan,np.nan,3,np.nan,np.nan,2,np.nan,np.nan,3,np.nan,np.nan,np.nan,2,np.nan]})

id	rank	age	sex	care	new
1	1	20	female	NA	NA
1	2	2	female	NA	NA
2	1	30	male	3	4
2	2	28	female	NA	NA
2	3	4	male	NA	NA
3	1	26	female	2	3
3	2	3	male	NA	NA
4	1	22	female	NA	NA
4	2	23	male	3	1
4	3	1	male	NA	NA
5	1	34	male	NA	NA
5	2	29	female	NA	NA
6	1	56	female	2	25
6	2	25	female	NA	NA

大家好，上表包含以下变量：

• id : 家庭识别号码
• rank : 户内人的排名，例如1表示户主，2配偶3子女
• age : 人的年龄
• sex : 人的性别
• care ：包含家庭中接受护理服务的人的排名

所以我想创建一个名为 new 的变量，其中包含在家中接受护理的人的年龄。我正在尝试在 Python 3.8 中执行此操作

我用这个命令在 Stata 中做了这个练习：

rangestat new = age, by(id) int(rank care care)

并且在 rtudio 中也这样做：

df <- df %>%                             
  group_by(id) %>%
  mutate(new = age[match(care, rank)]) %>%
  ungroup

我希望我已经理解了自己，也没有造成不便，我提前谢谢你

Answer 1

试试这个班轮：

result = pd.merge(
    df,
    df[['id', 'rank', 'age']].rename(
        columns={'rank': 'care', 'age': 'new'},
    ),
    how='left',
)

结果正是你想要的：

>>> result
   id  rank  age     sex  care   new
0   1     1   20  female     2     2
1   1     2    2  female  <NA>  <NA>
2   2     1   30    male     3     4
3   2     2   28  female  <NA>  <NA>
4   2     3    4    male  <NA>  <NA>
5   3     1   26  female     2     3
6   3     2    3    male  <NA>  <NA>
7   4     1   22  female  <NA>  <NA>
8   4     2   23    male     3     1
9   4     3    1    male  <NA>  <NA>

Answer 2

你想要这样的东西吗？

new = pd.Series([2,np.nan,4,np.nan,np.nan,3,np.nan,np.nan,1,np.nan], name="new")
new_df = pd.concat([df, new], axis=1)
print(new_df)

"""
   id  rank  age     sex  care  new
0   1     1   20  female   2.0  2.0
1   1     2    2  female   NaN  NaN
2   2     1   30    male   3.0  4.0
3   2     2   28  female   NaN  NaN
4   2     3    4    male   NaN  NaN
5   3     1   26  female   2.0  3.0
6   3     2    3    male   NaN  NaN
7   4     1   22  female   NaN  NaN
8   4     2   23    male   3.0  1.0
9   4     3    1    male   NaN  NaN
"""