[英]Convert multiple string stored in a variable into a single list in python
我希望每个人都做得很好。 我需要一点帮助,我需要从变量中获取所有字符串并需要存储到 python 中的单个列表中。 例如 - 我有 json 文件,我从中获取 id,并且当我运行 print(id) 时,所有 id 都存储到一个名为 id 的变量中,如下所示
17298626-991c-e490-bae6-47079c6e2202
17298496-19bd-2f89-7b5f-881921abc632
17298698-3e17-7a9b-b337-aacfd9483b1b
172986ac-d91d-c4ea-2e50-d53700480dd0
172986d0-18aa-6f51-9c62-6cb087ad31e5
172986f4-80f0-5c21-3aee-12f22a5f4322
17298712-a4ac-7b36-08e9-8512fa8322dd
17298747-8cc6-d9d0-8d05-50adf228c029
1729875c-050f-9a99-4850-bb0e6ad35fb0
1729875f-0d50-dc94-5515-b4891c40d81c
17298761-c26b-3ce5-e77e-db412c38a5b4
172987c8-2b5d-0d94-c365-e8407b0a8860
1729881a-e583-2b54-3a52-d092020d9c1d
1729881c-64a2-67cf-d561-6e5e38ed14cb
172987ec-7a20-7eb6-3ebe-a9fb621bb566
17298813-7ac4-258b-d6f9-aaf43f9147b1
17298813-f1ef-d28a-0817-5f3b86c3cf23
17298828-b62b-9ee6-248b-521b0663226e
17298825-7449-2fcb-378e-13671cb4688a
我希望将所有这些值存储到一个列表中。 有人可以帮我解决这个问题。
下面是我正在使用的代码:
import json
with open('requests.json') as f:
data = json.load(f)
print(type(data))
for i in data:
if 'traceId' in i:
id = i['traceId']
newid = id.split()
#print(type(newid))
print(newid)
下面是我的 json 文件看起来像:
[
{
"id": "376287298-hjd8-jfjb-khkf-6479280283e9",
"submittedTime": 1591692502558,
"traceId": "17298626-991c-e490-bae6-47079c6e2202",
"userName": "ABC",
"onlyChanged": true,
"description": "Not Required",
"startTime": 1591694487929,
"result": "NONE",
"state": "EXECUTING",
"paused": false,
"application": {
"id": "16b22a09-a840-f4d9-f42a-64fd73fece57",
"name": "XYZ"
},
"applicationProcess": {
"id": "dihihdosfj9279278yrie8ue",
"name": "Deploy",
"version": 12
},
"environment": {
"id": "fkjdshkjdshglkjdshgldshldsh03r937837",
"name": "DEV"
},
"snapshot": {
"id": "djnglkfdglki98478yhgjh48yr844h",
"name": "DEV_snapshot"
},
},
{
"id": "17298495-f060-3e9d-7097-1f86d5160789",
"submittedTime": 1591692844597,
"traceId": "17298496-19bd-2f89-7b5f-881921abc632",
"userName": "UYT,
"onlyChanged": true,
"startTime": 1591692845543,
"result": "NONE",
"state": "EXECUTING",
"paused": false,
"application": {
"id": "osfodsho883793hgjbv98r3098w",
"name": "QA"
},
"applicationProcess": {
"id": "owjfoew028r2uoieroiehojehfoef",
"name": "EDC",
"version": 5
},
"environment": {
"id": "16cf69c5-4194-e557-707d-0663afdbceba",
"name": "DTESTU"
},
}
]
从我试图获取traceId的地方。
您可以使用如下简单的split
方法:
ids = '''17298626-991c-e490-bae6-47079c6e2202 17298496-19bd-2f89-7b5f-881921abc632 17298698-3e17-7a9b-b337-aacfd9483b1b 172986ac-d91d-c4ea-2e50-d53700480dd0 172986d0-18aa-6f51-9c62-6cb087ad31e5 172986f4-80f0-5c21-3aee-12f22a5f4322 17298712-a4ac-7b36-08e9-8512fa8322dd 17298747-8cc6-d9d0-8d05-50adf228c029 1729875c-050f-9a99-4850-bb0e6ad35fb0 1729875f-0d50-dc94-5515-b4891c40d81c 17298761-c26b-3ce5-e77e-db412c38a5b4 172987c8-2b5d-0d94-c365-e8407b0a8860 1729881a-e583-2b54-3a52-d092020d9c1d 1729881c-64a2-67cf-d561-6e5e38ed14cb 172987ec-7a20-7eb6-3ebe-a9fb621bb566 17298813-7ac4-258b-d6f9-aaf43f9147b1 17298813-f1ef-d28a-0817-5f3b86c3cf23 17298828-b62b-9ee6-248b-521b0663226e 17298825-7449-2fcb-378e-13671cb4688a'''
l = ids.split(" ")
print(l)
这将给出以下结果,我假设所需的分隔符是简单的空间你可以适当调整:
['17298626-991c-e490-bae6-47079c6e2202', '17298496-19bd-2f89-7b5f-881921abc632', '17298698-3e17-7a9b-b337-aacfd9483b1b', '172986ac-d91d-c4ea-2e50-d53700480dd0', '172986d0-18aa-6f51-9c62-6cb087ad31e5', '172986f4-80f0-5c21-3aee-12f22a5f4322', '17298712-a4ac-7b36-08e9-8512fa8322dd', '17298747-8cc6-d9d0-8d05-50adf228c029', '1729875c-050f-9a99-4850-bb0e6ad35fb0', '1729875f-0d50-dc94-5515-b4891c40d81c', '17298761-c26b-3ce5-e77e-db412c38a5b4', '172987c8-2b5d-0d94-c365-e8407b0a8860', '1729881a-e583-2b54-3a52-d092020d9c1d', '1729881c-64a2-67cf-d561-6e5e38ed14cb', '172987ec-7a20-7eb6-3ebe-a9fb621bb566', '17298813-7ac4-258b-d6f9-aaf43f9147b1', '17298813-f1ef-d28a-0817-5f3b86c3cf23', '17298828-b62b-9ee6-248b-521b0663226e', '17298825-7449-2fcb-378e-13671cb4688a']
您会得到列表列表,因为每次迭代您只读取 1 个id
,所以您需要做的是启动一个空列表和 append 每个 id 以下列方式:
l = []
for i in data
if 'traceId' in i:
id = i['traceId']
l.append(id)
您可以将 append 的ids
变量添加到列表中,例如,
#list declaration
l1=[]
#this must be in your loop
l1.append(ids)
我假设您将id
作为str
类型值。 使用id.split()
将在一个 Python 列表中返回所有 id 的列表,因为在您的示例中每个 id 都用空格分隔。
id = """17298626-991c-e490-bae6-47079c6e2202 17298496-19bd-2f89-7b5f-881921abc632
17298698-3e17-7a9b-b337-aacfd9483b1b 172986ac-d91d-c4ea-2e50-d53700480dd0
172986d0-18aa-6f51-9c62-6cb087ad31e5 172986f4-80f0-5c21-3aee-12f22a5f4322
17298712-a4ac-7b36-08e9-8512fa8322dd 17298747-8cc6-d9d0-8d05-50adf228c029
1729875c-050f-9a99-4850-bb0e6ad35fb0 1729875f-0d50-dc94-5515-b4891c40d81c
17298761-c26b-3ce5-e77e-db412c38a5b4 172987c8-2b5d-0d94-c365-e8407b0a8860
1729881a-e583-2b54-3a52-d092020d9c1d 1729881c-64a2-67cf-d561-6e5e38ed14cb
172987ec-7a20-7eb6-3ebe-a9fb621bb566 17298813-7ac4-258b-d6f9-aaf43f9147b1
17298813-f1ef-d28a-0817-5f3b86c3cf23 17298828-b62b-9ee6-248b-521b0663226e
17298825-7449-2fcb-378e-13671cb4688a"""
id_list = id.split()
print(id_list)
Output:
['17298626-991c-e490-bae6-47079c6e2202', '17298496-19bd-2f89-7b5f-881921abc632',
'17298698-3e17-7a9b-b337-aacfd9483b1b', '172986ac-d91d-c4ea-2e50-d53700480dd0',
'172986d0-18aa-6f51-9c62-6cb087ad31e5', '172986f4-80f0-5c21-3aee-12f22a5f4322',
'17298712-a4ac-7b36-08e9-8512fa8322dd', '17298747-8cc6-d9d0-8d05-50adf228c029',
'1729875c-050f-9a99-4850-bb0e6ad35fb0', '1729875f-0d50-dc94-5515-b4891c40d81c',
'17298761-c26b-3ce5-e77e-db412c38a5b4', '172987c8-2b5d-0d94-c365-e8407b0a8860',
'1729881a-e583-2b54-3a52-d092020d9c1d', '1729881c-64a2-67cf-d561-6e5e38ed14cb',
'172987ec-7a20-7eb6-3ebe-a9fb621bb566', '17298813-7ac4-258b-d6f9-aaf43f9147b1',
'17298813-f1ef-d28a-0817-5f3b86c3cf23', '17298828-b62b-9ee6-248b-521b0663226e',
'17298825-7449-2fcb-378e-13671cb4688a']
split()
默认使用空格作为分隔符。 如果需要,您可以使用sep
参数来使用任何其他分隔符。
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