Question

我的数据框中有一个列，其中包含以下格式的数据：

['com.atlassian.greenhopper.service.sprint.Sprint@339ba62[id=001,rapidViewId=24,state=CLOSED,name=proj_a,goal=,startDate=2015-01-01T04:00:26.231Z,endDate=2015-01-13T14:36:00.000Z,completeDate=2015-02-13T14:07:09.739Z,sequence=001]

我试图从上面的列中提取id的值，而我可以使用以下方法进行此操作：

df['id'] = df['value'].astype(str).str.split('id').str[1]
df['id'] = df['id'].str.split(',').str[0]
df['id'] = df['id'].str.split('=').str[1]

我现在遇到一个问题，其中同一字段具有多个如下所示的值：

['com.atlassian.greenhopper.service.sprint.Sprint@339ba62[id=001,rapidViewId=24,
state=CLOSED,name=proj_a,goal=,startDate=2015-01-01T04:00:26.231Z,endDate=2015-01-13T14:36:00.000Z,
completeDate=2015-02-13T14:07:09.739Z,sequence=001]', 
'com.atlassian.greenhopper.service.sprint.Sprint@10b316d8[id=002,rapidViewId=24,
state=CLOSED,name=proj_b,goal=,startDate=2016-01-01T04:00:26.231Z,
endDate= 2016-01-13T14:36:00.000Z,completeDate= 2016-02-13T14:07:09.739Z,sequence=002]', 
'com.atlassian.greenhopper.service.sprint.Sprint@2a13ba77[id=003,
rapidViewId=24,state=CLOSED,name=proj_c,goal=,
startDate= 2017-01-01T04:00:26.231Z,endDate= 2017-01-13T14:36:00.000Z,
completeDate= 2017-02-13T14:07:09.739Z,sequence=003]',
 'com.atlassian.greenhopper.service.sprint.Sprint@76d3dba0[id=004,rapidViewId=24,
state=CLOSED,name=proj_d,goal=,startDate=2018-01-01T04:00:26.231Z,
endDate= 2018-01-13T14:36:00.000Z,completeDate= 2018-02-13T14:07:09.739Z,sequence=004]', 'com.atlassian.greenhopper.service.sprint.Sprint@307a51a2[id=005,
rapidViewId=24,state=CLOSED,name=proj_e,goal=,startDate=2019-01-01T04:00:26.231Z,
endDate= 2019-01-13T14:36:00.000Z,completeDate= 2019-02-13T14:07:09.739Z,sequence=005]']

预期输出：

001,002,003,004,005

我正在尝试提取与id对应的值并将其存储在单个字段中

Answer 1

使用str.findall

例如：

df = pd.DataFrame({"value": ['com.atlassian.greenhopper.service.sprint.Sprint@339ba62[id=001,rapidViewId=24,state=CLOSED,name=proj_a,goal=,startDate=2015-01-01T04:00:26.231Z,endDate=2015-01-13T14:36:00.000Z,completeDate=2015-02-13T14:07:09.739Z,sequence=001]', 'com.atlassian.greenhopper.service.sprint.Sprint@10b316d8[id=002,rapidViewId=24,state=CLOSED,name=proj_b,goal=,startDate=2016-01-01T04:00:26.231Z,endDate= 2016-01-13T14:36:00.000Z,completeDate= 2016-02-13T14:07:09.739Z,sequence=002]', 'com.atlassian.greenhopper.service.sprint.Sprint@2a13ba77[id=003,rapidViewId=24,state=CLOSED,name=proj_c,goal=,startDate= 2017-01-01T04:00:26.231Z,endDate= 2017-01-13T14:36:00.000Z,completeDate= 2017-02-13T14:07:09.739Z,sequence=003]', 'com.atlassian.greenhopper.service.sprint.Sprint@76d3dba0[id=004,rapidViewId=24,state=CLOSED,name=proj_d,goal=,startDate=2018-01-01T04:00:26.231Z,endDate= 2018-01-13T14:36:00.000Z,completeDate= 2018-02-13T14:07:09.739Z,sequence=004]', 'com.atlassian.greenhopper.service.sprint.Sprint@307a51a2[id=005,rapidViewId=24,state=CLOSED,name=proj_e,goal=,startDate=2019-01-01T04:00:26.231Z,endDate= 2019-01-13T14:36:00.000Z,completeDate= 2019-02-13T14:07:09.739Z,sequence=005]']})
df["id"] = df["value"].str.findall(r"id\=(\d+),")
print(df)

输出：

                                               value   id
0  com.atlassian.greenhopper.service.sprint.Sprin...  001
1  com.atlassian.greenhopper.service.sprint.Sprin...  002
2  com.atlassian.greenhopper.service.sprint.Sprin...  003
3  com.atlassian.greenhopper.service.sprint.Sprin...  004
4  com.atlassian.greenhopper.service.sprint.Sprin...  005

如果DF在一个列表中包含所有值，则使用。

df = pd.DataFrame({"value": [['com.atlassian.greenhopper.service.sprint.Sprint@339ba62[id=001,rapidViewId=24,state=CLOSED,name=proj_a,goal=,startDate=2015-01-01T04:00:26.231Z,endDate=2015-01-13T14:36:00.000Z,completeDate=2015-02-13T14:07:09.739Z,sequence=001]', 'com.atlassian.greenhopper.service.sprint.Sprint@10b316d8[id=002,rapidViewId=24,state=CLOSED,name=proj_b,goal=,startDate=2016-01-01T04:00:26.231Z,endDate= 2016-01-13T14:36:00.000Z,completeDate= 2016-02-13T14:07:09.739Z,sequence=002]', 'com.atlassian.greenhopper.service.sprint.Sprint@2a13ba77[id=003,rapidViewId=24,state=CLOSED,name=proj_c,goal=,startDate= 2017-01-01T04:00:26.231Z,endDate= 2017-01-13T14:36:00.000Z,completeDate= 2017-02-13T14:07:09.739Z,sequence=003]', 'com.atlassian.greenhopper.service.sprint.Sprint@76d3dba0[id=004,rapidViewId=24,state=CLOSED,name=proj_d,goal=,startDate=2018-01-01T04:00:26.231Z,endDate= 2018-01-13T14:36:00.000Z,completeDate= 2018-02-13T14:07:09.739Z,sequence=004]', 'com.atlassian.greenhopper.service.sprint.Sprint@307a51a2[id=005,rapidViewId=24,state=CLOSED,name=proj_e,goal=,startDate=2019-01-01T04:00:26.231Z,endDate= 2019-01-13T14:36:00.000Z,completeDate= 2019-02-13T14:07:09.739Z,sequence=005]']]})
df["id"] = df["value"].apply(",".join).str.findall(r"id\=(\d+),").apply(",".join)
print(df)

输出：

                                               value                   id
0  [com.atlassian.greenhopper.service.sprint.Spri...  001,002,003,004,005

Answer 2

您可以仅在每个id=之后输入前三个字符。跳过第一个元素，因为这是第一个id=之前的内容。

df["id"] = [x[:3] for x in df["value"].astype(str).split("id=")[1:]]

熊猫-从一列中提取多个数据

2 个答案: