我有两本字典，但为了简化起见，我就选这两本:
>>> x = dict(a=1, b=2)
>>> y = dict(a=2, b=2)
现在，我想比较x中的每个键值对在y中是否有相同的对应值，所以我这样写:
>>> for x_values, y_values in zip(x.iteritems(), y.iteritems()):
if x_values == y_values:
print 'Ok', x_values, y_values
else:
print 'Not', x_values, y_values
它的工作原理是返回一个元组，然后比较是否相等。
我的问题:
这对吗?还有更好的办法吗?最好不是在速度上，我说的是代码优雅。
更新:我忘了提到，我必须检查有多少键，值对是相等的。
当前回答
迟回复总比不回复好!
比较Not_Equal比比较Equal更有效。因此，如果一个字典中的任何键值在另一个字典中找不到，那么两个字典就不相等。下面的代码考虑到您可能会比较默认的dict，因此使用get而不是getitem[]。
在get调用中使用一种随机值作为默认值，等于要检索的键-以防dicts在一个dict中有None作为值，而该键在另一个dict中不存在。此外，为了提高效率，get !=条件是在not in条件之前检查的，因为您同时对两边的键和值进行检查。
def Dicts_Not_Equal(first,second):
""" return True if both do not have same length or if any keys and values are not the same """
if len(first) == len(second):
for k in first:
if first.get(k) != second.get(k,k) or k not in second: return (True)
for k in second:
if first.get(k,k) != second.get(k) or k not in first: return (True)
return (False)
return (True)
其他回答
Code
def equal(a, b):
type_a = type(a)
type_b = type(b)
if type_a != type_b:
return False
if isinstance(a, dict):
if len(a) != len(b):
return False
for key in a:
if key not in b:
return False
if not equal(a[key], b[key]):
return False
return True
elif isinstance(a, list):
if len(a) != len(b):
return False
while len(a):
x = a.pop()
index = indexof(x, b)
if index == -1:
return False
del b[index]
return True
else:
return a == b
def indexof(x, a):
for i in range(len(a)):
if equal(x, a[i]):
return i
return -1
Test
>>> a = {
'number': 1,
'list': ['one', 'two']
}
>>> b = {
'list': ['two', 'one'],
'number': 1
}
>>> equal(a, b)
True
我正在使用这个解决方案，在Python 3中完美地为我工作
import logging
log = logging.getLogger(__name__)
...
def deep_compare(self,left, right, level=0):
if type(left) != type(right):
log.info("Exit 1 - Different types")
return False
elif type(left) is dict:
# Dict comparison
for key in left:
if key not in right:
log.info("Exit 2 - missing {} in right".format(key))
return False
else:
if not deep_compare(left[str(key)], right[str(key)], level +1 ):
log.info("Exit 3 - different children")
return False
return True
elif type(left) is list:
# List comparison
for key in left:
if key not in right:
log.info("Exit 4 - missing {} in right".format(key))
return False
else:
if not deep_compare(left[left.index(key)], right[right.index(key)], level +1 ):
log.info("Exit 5 - different children")
return False
return True
else:
# Other comparison
return left == right
return False
它比较dict、list和其他单独实现“==”操作符的类型。 如果你需要比较其他不同的东西，你需要在“If树”中添加一个新的分支。
希望这能有所帮助。
下面的代码将帮助您比较python中的dict列表
def compate_generic_types(object1, object2):
if isinstance(object1, str) and isinstance(object2, str):
return object1 == object2
elif isinstance(object1, unicode) and isinstance(object2, unicode):
return object1 == object2
elif isinstance(object1, bool) and isinstance(object2, bool):
return object1 == object2
elif isinstance(object1, int) and isinstance(object2, int):
return object1 == object2
elif isinstance(object1, float) and isinstance(object2, float):
return object1 == object2
elif isinstance(object1, float) and isinstance(object2, int):
return object1 == float(object2)
elif isinstance(object1, int) and isinstance(object2, float):
return float(object1) == object2
return True
def deep_list_compare(object1, object2):
retval = True
count = len(object1)
object1 = sorted(object1)
object2 = sorted(object2)
for x in range(count):
if isinstance(object1[x], dict) and isinstance(object2[x], dict):
retval = deep_dict_compare(object1[x], object2[x])
if retval is False:
print "Unable to match [{0}] element in list".format(x)
return False
elif isinstance(object1[x], list) and isinstance(object2[x], list):
retval = deep_list_compare(object1[x], object2[x])
if retval is False:
print "Unable to match [{0}] element in list".format(x)
return False
else:
retval = compate_generic_types(object1[x], object2[x])
if retval is False:
print "Unable to match [{0}] element in list".format(x)
return False
return retval
def deep_dict_compare(object1, object2):
retval = True
if len(object1) != len(object2):
return False
for k in object1.iterkeys():
obj1 = object1[k]
obj2 = object2[k]
if isinstance(obj1, list) and isinstance(obj2, list):
retval = deep_list_compare(obj1, obj2)
if retval is False:
print "Unable to match [{0}]".format(k)
return False
elif isinstance(obj1, dict) and isinstance(obj2, dict):
retval = deep_dict_compare(obj1, obj2)
if retval is False:
print "Unable to match [{0}]".format(k)
return False
else:
retval = compate_generic_types(obj1, obj2)
if retval is False:
print "Unable to match [{0}]".format(k)
return False
return retval
如果你想知道在两个字典中有多少值是匹配的，你应该这样说:)
也许是这样的:
shared_items = {k: x[k] for k in x if k in y and x[k] == y[k]}
print(len(shared_items))
这是我的答案，使用递归的方式:
def dict_equals(da, db):
if not isinstance(da, dict) or not isinstance(db, dict):
return False
if len(da) != len(db):
return False
for da_key in da:
if da_key not in db:
return False
if not isinstance(db[da_key], type(da[da_key])):
return False
if isinstance(da[da_key], dict):
res = dict_equals(da[da_key], db[da_key])
if res is False:
return False
elif da[da_key] != db[da_key]:
return False
return True
a = {1:{2:3, 'name': 'cc', "dd": {3:4, 21:"nm"}}}
b = {1:{2:3, 'name': 'cc', "dd": {3:4, 21:"nm"}}}
print dict_equals(a, b)
希望有帮助!
