我创建了一个程序来比较两个python词典并输出两者的差异。它适用于深度为2或更小的字典。我该怎么做才能更深入地处理dicts,还有嵌套的dicts?
我遇到的另一个问题是,当我通过我的get_json()函数传递一个json数组时,它将作为列表返回。而且程序正在用列表而不是字典来查找。我该如何解决这个问题呢?
我的节目:
#!/usr/bin/env python2
import json
def get_json():
file_name = raw_input("Enter name of JSON File: ")
with open(file_name) as json_file:
json_data = json.load(json_file)
return json_data
def print_diff(json1, json2):
for n in json1:
if n not in json2:
print('- "' + str(n) + '":')
for n in json2:
if n not in json1:
print('+ "' + str(n) + '":')
continue
if json2[n] != json1[n]:
if type(json2[n]) not in (dict, list):
print('- "' + str(n) + '" : "' + str(json1[n]))
print('+ "' + str(n) + '" : "' + str(json2[n]))
else:
if type(json2[n]) == dict:
print_diff(json1[n], json2[n])
continue
return
def main():
file1 = get_json()
print(type(file1))
file2 = get_json()
print(type(file2))
print_diff(file1, file2)
if __name__ == "__main__":
main()
dict 1的例子:
{
"widget": {
"debug": "on",
"window": {
"title": "Sample Konfabulator Widget",
"name": "main_window",
"width": 500,
"height": 500
},
"image": {
"src": "Images/Sun.png",
"name": "sun1",
"hOffset": 250,
"vOffset": 250,
"alignment": "center"
},
"text": {
"data": "Click Here",
"size": 36,
"style": "bold",
"name": "text1",
"hOffset": 250,
"vOffset": 100,
"alignment": "center",
"onMouseUp": "sun1.opacity = (sun1.opacity / 100) * 90;"
}
}
}
dict 2的例子:
{
"widget": {
"debug": "on",
"window": {
"title": "Sample Konfabulator Widget",
"name": "main_window",
"width": 500,
"height": 500
},
"image": {
"src": "Images/Sun.png",
"name": "sun2",
"hOffset": 100,
"vOffset": 100,
"alignment": "center"
},
"text": {
"data": "Click Here",
"size": 36,
"style": "bold",
"name": "text1",
"hOffset": 250,
"vOffset": 100,
"alignment": "center",
"onMouseUp": "sun1.opacity = (sun1.opacity / 100) * 90;"
}
}
}
示例输出:
Enter name of JSON File: JSON1.json
<type 'dict'>
Enter name of JSON File: JSON2.json
<type 'dict'>
- "vOffset" : "250
+ "vOffset" : "100
- "name" : "sun1
+ "name" : "sun2
- "hOffset" : "250
+ "hOffset" : "100
答案 0 :(得分:2)
我编写了以下代码,可以比较两个不同深度的词典,并将差异打印到控制台输出。请注意,如果密钥在第一个字典中找到而在第二个字典中找不到,则只打印未找到的密钥(它不会在躺着的树下打印)。我希望这是预期的。此代码适用于单向比较,因此您需要进行两次调用以便以其他方式比较字典。
def findDiff(d1, d2, path=""):
for k in d1.keys():
if not d2.has_key(k):
print path, ":"
print k + " as key not in d2", "\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 s1 to s2:"
print findDiff(s1,s2)
print "comparing s2 to s1:"
print findDiff(s2,s1)
输出:
comparing s1 to s2:
widget->text :
data as key not in d2
widget->text->window->image :
- vOffset : 250
+ vOffset : 100
widget->text->window->image :
- name : sun1
+ name : sun2
widget->text->window->image :
- hOffset : 250
+ hOffset : 100
None
comparing s2 to s1:
widget->text->window->image :
- vOffset : 100
+ vOffset : 250
widget->text->window->image :
- name : sun2
+ name : sun1
widget->text->window->image :
- hOffset : 100
+ hOffset : 250
None
请注意,在您的问题中,您没有考虑密钥是否有差异,密钥路径未打印。我在我的代码中打印它。我已从widget-&gt;文本中删除了数据元素以进行测试。所以,请忽略此控制台输出
答案 1 :(得分:0)
没有阅读你所有的东西,
def check_dict(first, second):
for key in first:
for keyTwo in second:
if type(first[key]) == dict and type(second[keyTwo]) == dict:
return check_dict(first[key], second[keyTwo])
if not first[key] == second[keyTwo]:
return false
目前没有python解释器的访问权限,但是这样做的事情应该有效。这只是一个想法的开始,但希望这是足以引发某些事情的信息。