Python嵌套dict比较算法

我有一个比较两个python dict的程序，对于深度为1的非嵌套字典，它工作得非常好。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，它适用于深度为1且不嵌套的dict。我应该如何改变算法来处理嵌套和更深入的dict，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

---

下面采用了一种输出差异的较短形式（您可以根据自己的情况进行更改）

层次结构由输出中的开始/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
>>> 

最好记住以下几点：

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

---

使用递归比较两个字典：

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

---

你试过什么代码？dict1={'a'：3，'b'：{'b'：4，'a'：5}}和dict2={'a'：{'a'：3，'b'：4}，'a'：5}的预期输出是什么？@Totem将我的代码添加到主位置，而不是删除你的问题，你可以考虑自己发布一个答案，然后把它标记为可以接受的。- 1请显示一个示例函数的用法。如果出现错误，请发布完整的回溯。这些是写一个好问题的基本信息。在没有进一步解释的情况下，你不应该使用“doeswork”或“notwork”。如果没有发布完整的回溯，你永远不应该说“它引发了一个错误”。不给学分的复制粘贴被否决