python中的多级JSON差异 [英] Multilevel JSON diff in python
问题描述
如果已回答此问题,请链接我以作答, 我的问题是我想获取无序的多层json的差异.
Please link me to answer if this has already been answered, my problem is i want to get diff of multilevel json which is unordered.
x=json.loads('''[{"y":2,"x":1},{"x":3,"y":4}]''')
y=json.loads('''[{"x":1,"y":2},{"x":3,"y":4}]''')
z=json.loads('''[{"x":3,"y":4},{"x":1,"y":2}]''')
import json_tools as jt
import json_delta as jd
print jt.diff(y,z)
print jd.diff(y,z)
print y==z
print x==y
输出为
[{'prev': 2, 'value': 4, 'replace': u'/0/y'}, {'prev': 1, 'value': 3, 'replace': u'/0/x'}, {'prev': 4, 'value': 2, 'replace': u'/1/y'}, {'prev': 3, 'value': 1, 'replace': u'/1/x'}]
[[[2], {u'y': 2, u'x': 1}], [[0]]]
False
True
我的问题是如何使y和z相等,或者是否存在实际差异,具体取决于JSON的非顺序.
my question is how can i get y and z to be equal or if there are actual differences depending on non-order of the JSON.
种类繁多的词典列表,但我正在寻找一种能证明层次的东西,即列表/词典的列表/词典...
kind of unordered List of dictionaries but i am looking for something which is level-proof that is list/dict of dictionaries of list/dictionaries ...
推荐答案
使用以下函数部分解决了该问题
solved it partially with following function
def diff(prev,lat):
p=prev
l=lat
prevDiff = []
latDiff = []
for d1 in p[:]:
flag = False
for d2 in l:
if len(set(d1.items()) ^ set(d2.items())) == 0:
p.remove(d1)
l.remove(d2)
flag = True
break
if not flag:
prevDiff.append(d1)
p.remove(d1)
prevDiff = prevDiff + p
latDiff = latDiff + l
resJSONdata=[]
if len(prevDiff) != 0:
resJSONdata.append({'prevCount':len(prevDiff)})
resJSONdata.append({'prev':prevDiff})
if len(latDiff) != 0:
resJSONdata.append({'latestCount':len(latDiff)})
resJSONdata.append({'latest':latDiff})
# return json.dumps(resJSONdata,indent = 4,sort_keys=True)
return resJSONdata
它不是递归地逐级执行,但出于我的目的,这解决了问题
it's not doing it recursively into level into levels but for my purpose this solved the issue
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