python嵌套的dict比较算法

Question

我有一个比较两个python dicts的程序，并且对具有1的深度的非嵌套词典的工作正常。该算法非常简单：

for key in dict1, if not in dict2
    print "-" key
for key in dict2, if not in dict1
    print "+" key
for dict2Value, if not equal to dict1Value
    print "+" dict2value
    print "-" dict1value

正如我所说的那样，它可以很好地效果，其深度为1，并且不嵌套。我应该如何将算法更改为与嵌套并更深入的命令一起工作，我已经被困了一段时间了。

我的代码：

def print_diff(dict1, dict2):
    for n in dict1:
        if n not in dict2:
            print('-   "' + str(n) + '":')
    for n in dict2:
        if n not in dict1:
            print('+   "' + str(n) + '":')
            continue
        if dict2[n] != dict1[n]:
            if type(dict2[n]) not in (dict, list):
                print('-   "' + str(n) + '" : "' + str(dict1[n]))
                print('+   "' + str(n) + '" : "' + str(dict2[n]))
            else:
                if type(dict2[n]) == dict:
                    print_diff(dict1[n], dict2[n])
                    continue
    return

Answer 1

以下采用了输出差异的较短形式（您可以更改以遵循自己的差异）。

层次结构用输出中的开始/停止表示。

>>> def dictdiff(d1, d2):
    s1, s2 = set(d1), set(d2)
    in1, in12, in2 = s1 - s2, s1 & s2, s2 - s1
    if in1:  print('Keys only in 1:', sorted(in1))
    if in2:  print('Keys only in 2:', sorted(in2))
    sameval = {key for key in in12 if d1[key] == d2[key]}
    if sameval: print('Keys with equal values:', sorted(sameval))
    diffval = in12 - sameval
    diffdict = {key for key in diffval if type(d1[key]) == type(d2[key]) == dict}
    diffother = diffval - diffdict
    if diffother: print('Keys with different values (that are not both dicts):', sorted(diffother))
    for key in sorted(diffdict):
        print('## START dictdiff for common key', key)
        dictdiff(d1[key], d2[key])
        print('## STOP  dictdiff for common key', key)

>>> d1 = dict(a=1, b=2, c=dict(x=1, y=2), d=1, f=dict(s=1, t=3))
>>> d2 = dict(a=1, b=3, c=dict(x=1, y=2), e=1, f=dict(s=1, t=2))
>>> dictdiff(d1, d2)
Keys only in 1: ['d']
Keys only in 2: ['e']
Keys with equal values: ['a', 'c']
Keys with different values (that are not both dicts): ['b']
## START dictdiff for common key f
Keys with equal values: ['s']
Keys with different values (that are not both dicts): ['t']
## STOP  dictdiff for common key f
>>> 

也可能会记住以下内容：

>>> d1['c'] is d2['c']
False
>>> d1['c'] == d2['c']
True
>>> 

Answer 2

使用递归比较2个字典：

d1= {'a':{'b':{'cs':10},'d':{'cs':20}}}
d2= {'a':{'b':{'cs':30} ,'d':{'cs':20}},'newa':{'q':{'cs':50}}}
def findDiff(d1, d2, path=""):
    for k in d1.keys():
        if not d2.has_key(k):
            print path, ":"
            print "keys not in d2: " + k, "n"
        else:
            if type(d1[k]) is dict:
                if path == "":
                    path = k
                else:
                    path = path + "->" + k
                findDiff(d1[k],d2[k], path)
            else:
                if d1[k] != d2[k]:
                    print path, ":"
                    print " - ", k," : ", d1[k]
                    print " + ", k," : ", d2[k] 
print "comparing d1 to d2:"
print findDiff(d1,d2)
print "comparing d2 to d1:"
print findDiff(d2,d1)

输出：

comparing d1 to d2:
a->b :
 -  cs  :  10
 +  cs  :  30
None
comparing d2 to d1:
a->b :
 -  cs  :  30
 +  cs  :  10
a :
keys not in d2: newa
None