两个 list 怎么比较后求出多出来的那值？

[css3]

前提是，里边的元素可能是乱序的

a = [{'key': 1, 'value': 2}, {'key': 3, 'value': 4}]

b = [{'key': 5, 'value': 6}, {'key': 1, 'value': 2}, {'key': 3, 'value': 41} ] # value 不一样的忽略掉

# 想要的结果：
c = [{'key': 5, 'value': 6}]

[hehheh]

把 key,val 转成 tuple，然后整个 list 打包成 set 然后 intersection

[Cooky]

两边各自合成一个字典做比较

[fdppzrl]

c.addAll(a.removeall(b))
c.addAll(b.removeall(a))
Java 大概的写法就酱

[ranlan]

b = [{'key': 5, 'value': 6}, {'key': 1, 'value': 2}, {'key': 3, 'value': 41} ]中应该是 {'key': 3, 'value': 41} 这个元素应该是 {'key': 3, 'value': 4}吧？
新手的解法
c = [i for i in b if i not in a]

[css3]

@ranlan 不是

[yesterdaysun]

ak = set(map(lambda x: x['key'], a))
bk = set(map(lambda x: x['key'], a))
c = list(filter(lambda x: x['key'] not in bk, a)) + list(filter(lambda x: x['key'] not in ak, b))
print(c)

[ranlan]

不还意思我理解错了
应该是这样
a1 = [x['key'] for x in a]

c = [i for i in b if i['key'] not in a1]

[css3]

@yesterdaysun
@ranlan
已采纳多谢🤪

[johnnyluck]

d = set([x['key'] for x in a]) ^ set([y['key'] for y in b])
c = [x for x in (a+b) if x['key'] in d]

[20015jjw]

^ 双关了

[xhxhx]

array_diff