我是一个新的python用户，我刚刚熟悉使用zip函数重构数据，但是我现在面临着一个具有挑战性的数据集，我必须重组。我有3个json响应，我必须从我的端部合并，数据集在设计上是相同的，并且具有相同的长度，它们只是与它们运行的环境不同。
为简洁起见，让3个文件的值相同：qa.json | dev.json | prod.json
注意：这个外部对象是对象的数组/列表，我只是为了简洁而放置了一个对象
[
{
"elements": [
{
"keyword": "Scenario",
"name": "valid user can login site",
"steps": [
{
"name": "a valid user name and password",
"result": {
"status": "passed"
}
},
{
"name": "a valid user clicking on the login button after typing in user name and password",
"result": {
"status": "passed"
}
},
{
"name": "map should display",
"result": {
"status": "passed"
}
}
]
}
],
"keyword": "Feature",
"name": "login",
"status": "passed"
}
]
我想要实现的目标：
注意：我想将它们合并到一组中，反映状态的不同环境
[
{
"elements": [
{
"keyword": "Scenario",
"name": "valid user can login site",
"steps": [
{
"name": "a valid user name and password",
"result": {
"qa": "passed",
"prod": "passed",
"dev": "passed"
}
},
{
"name": "a valid user clicking on the login button after typing in user name and password",
"result": {
"qa": "passed",
"prod": "passed",
"dev": "passed"
}
},
{
"name": "map should display",
"result": {
"qa": "passed",
"prod": "passed",
"dev": "passed"
}
}
]
}
],
"keyword": "Feature",
"name": "login",
"qa": "passed",
"prod": "passed",
"dev": "passed"
}
]
到目前为止我做了什么：
我来自javascript背景，所以我还是熟悉python逻辑
import json
with open('qa.json') as data_file:
qa = json.load(data_file)
with open('dev.json') as data_file:
dev = json.load(data_file)
with open('prod.json') as data_file:
prod = json.load(data_file)
json_list = [{SOME STRUCT} for q, d, p in zip(qa, dev, prod)]
答案 0 :(得分：1)
我没有太多时间，所以我发布了我认为对你的问题有效的解决方案，即使它有点麻烦。如果我有时间，我会编辑它。
<强>代码
import json
data = """
[
{
"elements": [{
"keyword": "Scenario",
"name": "valid user can login site",
"steps": [{
"name": "a valid user name and password",
"result": {
"status": "passed"
}
}, {
"name": "a valid user clicking on the login button after typing in user name and password",
"result": {
"status": "passed"
}
}, {
"name": "map should display",
"result": {
"status": "passed"
}
}]
}],
"keyword": "Feature",
"name": "login",
"status": "passed"
},
{
"elements": [{
"keyword": "Scenario",
"name": "valid user can login site",
"steps": [{
"name": "a valid user name and password",
"result": {
"status": "passed"
}
}, {
"name": "a valid user clicking on the login button after typing in user name and password",
"result": {
"status": "failed"
}
}, {
"name": "map should display",
"result": {
"status": "passed"
}
}]
}],
"keyword": "Feature",
"name": "login",
"status": "passed"
}
]
"""
def get_result(envData, objIndex, elementIndex, stepIndex):
return envData[objIndex]['elements'][elementIndex]['steps'][stepIndex]['result']['status']
def set_combined_results(combinedData, objIndex, elementIndex, stepIndex, results):
resultNode = combinedData[objIndex]['elements'][elementIndex]['steps'][stepIndex]['result']
resultNode.update({ 'qa': results[0], 'prod': results[1], 'dev': results[2] })
if __name__ == '__main__':
qAData = json.loads(data)
prodData = json.loads(data)
devData = json.loads(data)
combinedData = json.loads(data)
for objIndex, obj in enumerate(combinedData):
for elementIndex, element in enumerate(obj['elements']):
for stepIndex, _ in enumerate(element['steps']):
qAResult = get_result(qAData, objIndex, elementIndex, stepIndex)
prodResult = get_result(prodData, objIndex, elementIndex, stepIndex)
devResult = get_result(devData, objIndex, elementIndex, stepIndex)
combinedResults = (qAResult, prodResult, devResult)
set_combined_results(combinedData, objIndex, elementIndex, stepIndex, combinedResults)
qAAggregateResult = qAData[objIndex]['status']
prodAggregateResult = prodData[objIndex]['status']
devAggregateResult = devData[objIndex]['status']
del combinedData[objIndex]['status']
combinedData[objIndex]['qa'] = qAAggregateResult
combinedData[objIndex]['prod'] = prodAggregateResult
combinedData[objIndex]['dev'] = devAggregateResult
print(json.dumps(combinedData, indent=True))
<强>输出
[
{
"keyword": "Feature",
"name": "login",
"elements": [
{
"keyword": "Scenario",
"name": "valid user can login site",
"steps": [
{
"result": {
"qa": "passed",
"status": "passed",
"dev": "passed",
"prod": "passed"
},
"name": "a valid user name and password"
},
{
"result": {
"qa": "passed",
"status": "passed",
"dev": "passed",
"prod": "passed"
},
"name": "a valid user clicking on the login button after typing in user name and password"
},
{
"result": {
"qa": "passed",
"status": "passed",
"dev": "passed",
"prod": "passed"
},
"name": "map should display"
}
]
}
],
"dev": "passed",
"prod": "passed",
"qa": "passed"
},
{
"keyword": "Feature",
"name": "login",
"elements": [
{
"keyword": "Scenario",
"name": "valid user can login site",
"steps": [
{
"result": {
"qa": "passed",
"status": "passed",
"dev": "passed",
"prod": "passed"
},
"name": "a valid user name and password"
},
{
"result": {
"qa": "failed",
"status": "failed",
"dev": "failed",
"prod": "failed"
},
"name": "a valid user clicking on the login button after typing in user name and password"
},
{
"result": {
"qa": "passed",
"status": "passed",
"dev": "passed",
"prod": "passed"
},
"name": "map should display"
}
]
}
],
"dev": "failed",
"prod": "failed",
"qa": "failed"
}
]
答案 1 :(得分：0)
由于list中的每个对象都是dict，您可以使用dict.update方法更新dict。
f.e。
a = [{'one': 1}, {'three': 3}]
b = [{'one': 1}, {'two': 2}] # {'one': 1} is duplicate with same value
c = [{'a': 'aaa'}, {'two': 22}] # {'two': 22} is duplicate with different value
for x, y, z in zip(a, b, c):
x.update(y)
x.update(z)
现在x将为[{'a': 'aaa', 'one': 1}, {'three': 3, 'two': 22}]
对于你提到的那些json文件，它将是;
import json
from pprint import pprint
qa = '''
[
{
"elements": [
{
"keyword": "Scenario",
"name": "valid user can login site",
"steps": [
{
"name": "a valid user name and password",
"result": {
"qa": "passed"
}
},
{
"name": "a valid user clicking on the login button after typing in user name and password",
"result": {
"qa": "passed"
}
},
{
"name": "map should display",
"result": {
"qa": "passed"
}
}
]
}
],
"keyword": "Feature",
"name": "login",
"qa": "passed"
}
]
'''
dev = '''
[
{
"elements": [
{
"keyword": "Scenario",
"name": "valid user can login site",
"steps": [
{
"name": "a valid user name and password",
"result": {
"dev": "passed"
}
},
{
"name": "a valid user clicking on the login button after typing in user name and password",
"result": {
"dev": "passed"
}
},
{
"name": "map should display",
"result": {
"dev": "passed"
}
}
]
}
],
"keyword": "Feature",
"name": "login",
"dev": "passed"
}
]
'''
prod = '''
[
{
"elements": [
{
"keyword": "Scenario",
"name": "valid user can login site",
"steps": [
{
"name": "a valid user name and password",
"result": {
"prod": "passed"
}
},
{
"name": "a valid user clicking on the login button after typing in user name and password",
"result": {
"prod": "passed"
}
},
{
"name": "map should display",
"result": {
"prod": "passed"
}
}
]
}
],
"keyword": "Feature",
"name": "login",
"prod": "passed"
}
]
'''
qa = json.loads(qa)
dev = json.loads(dev)
prod = json.loads(prod)
for q, p, d in zip(qa, dev, prod):
# update all keys but 'elements'
q.update({k: v for k, v in p.items() if k != 'elements'})
q.update({k: v for k, v in d.items() if k != 'elements'})
# update the three 'result' dict
for i in range(3):
q['elements'][0]['steps'][i]['result'].update(p['elements'][0]['steps'][i]['result'])
q['elements'][0]['steps'][i]['result'].update(d['elements'][0]['steps'][i]['result'])
pprint(qa)
输出;
[{'dev': 'passed',
'elements': [{'keyword': 'Scenario',
'name': 'valid user can login site',
'steps': [{'name': 'a valid user name and password',
'result': {'dev': 'passed',
'prod': 'passed',
'qa': 'passed'}},
{'name': 'a valid user clicking on the login button '
'after typing in user name and password',
'result': {'dev': 'passed',
'prod': 'passed',
'qa': 'passed'}},
{'name': 'map should display',
'result': {'dev': 'passed',
'prod': 'passed',
'qa': 'passed'}}]}],
'keyword': 'Feature',
'name': 'login',
'prod': 'passed',
'qa': 'passed'}]
答案 2 :(得分：0)
这是一个递归合并实现。它将合并任意类型的Python对象，这样除了dicts之外的任何东西在合并时必须相等。
对于dicts，如果相同的键与存在键的那些键中的可合并值相关联，则允许合并。在合并之前，任何等于参数dict的{{1}}个键都会以索引为后缀：
fieldname现在可以在from copy import deepcopy
from itertools import chain
def merge(field_name, *objs):
# Make sure all objs are of same type
types = list(set(map(type, objs)))
if not len(set(map(type, objs))) == 1:
raise Exception('Cannot merge objects of different types: {types}'.format(types=types))
first = objs[0]
# for any random objects, make sure they are equal!
if not isinstance(first, (list, tuple, dict)):
if not len(set(objs)) == 1:
raise Exception("Cannot merge non-equal objects that aren't dicts: {objs}".format(objs=objs))
return deepcopy(first)
# for lists, tuples: zip 'em and merge the zipped elements
if isinstance(first, (list, tuple)):
return [merge(field_name, *zipped) for zipped in zip(*objs)]
# dicts
result_dict = {}
keys = list(set(chain.from_iterable(d.keys() for d in objs)))
try:
keys.remove(field_name)
except ValueError:
pass
for k in keys:
# merge values from all dicts where key is present
result_dict[k] = merge(field_name, *(d[k] for d in objs if k in d))
for i, d in enumerate(objs):
if field_name in d:
result_dict['{f}_{i}'.format(f=field_name, i=i)] = d[field_name]
return result_dict
上使用dev, qa, prod，这是OP提供的结构的dicts：
>>> from pprint import pprint
>>> pprint(merge('status', qa, dev, prod))
[{'elements': [{'keyword': 'Scenario',
'name': 'valid user can login site',
'steps': [{'name': 'a valid user name and password',
'result': {'status_0': 'passed',
'status_1': 'passed',
'status_2': 'passed'}},
{'name': 'a valid user clicking on the login button after typing in user name and password',
'result': {'status_0': 'passed',
'status_1': 'passed',
'status_2': 'passed'}},
{'name': 'map should display',
'result': {'status_0': 'passed',
'status_1': 'passed',
'status_2': 'passed'}}]}],
'keyword': 'Feature',
'name': 'login',
'status_0': 'passed',
'status_1': 'passed',
'status_2': 'passed'}]
这当然不是完全通用的，例如它只是“深度”合并序列lists或tuples，但它应该足够适合从json加载的数据结构。我希望它可以帮助您找到解决方案。