[英]How to compare two dictionaries with lists inside?
我有两本字典,例如:
first = { 'test' : [1, 2, 3] }
second = { 'test' : [3, 2, 1] }
first == second # False
有没有办法比较这两个字典并忽略里面列表中值的顺序?
我尝试使用OrderedDict
但它没有用:
from collections import OrderedDict
first = OrderedDict({ 'test' : [1, 2, 3] })
second = OrderedDict({ 'test' : [3, 2, 1] })
first == second # False
对包含的列表使用列表的Counter
s(如果有[1,2] == [2,1,2]
则使用set
s):
from collections import Counter
first = { 'test' : [1, 2, 3] }
second = { 'test' : [3, 2, 1] }
third = { 'test' : [3, 2, 1], "meh":"" }
def comp_dict(d1,d2):
return d1.keys() == d2.keys() and all(Counter(d1[k]) == Counter(d2[k]) for k in d1)
# or
# return d1.keys() == d2.keys() and all(set(d1[k]) == set(d2[k]) for k in d1)
print(comp_dict(first, second))
print(comp_dict(first, third))
Output:
True
False
我找到了这个解决方案:
import unittest
class TestClass(unittest.TestCase):
def test_count_two_dicts(self, d1, d2):
self.assertCountEqual(d1, d2)
if __name__ == '__main__':
first = { 'test' : [1, 2, 3] }
second = { 'test' : [3, 2, 1] }
t = TestClass()
t.test_count_two_dicts(d1=first, d2=second)
如果你想比较两个字典(真或假)
first = { 'test' : [1, 2, 3] }
second = { 'test' : [3, 2, 1] }
is_equal = first == second
is_equal
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