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}] 推荐学习书目 Learn Python the Hard Way Python Sites PyPI - Python Package Index http://diveintopython.org/toc/index.html Pocoo 值得关注的项目 PyPy Celery Jinja2 Read the Docs gevent pyenv virtualenv Stackless Python Beautiful Soup 结巴中文分词 Green Unicorn Sentry Shovel Pyflakes pytest Python 编程 pep8 Checker Styles PEP 8 Google Python Style Guide Code Style from The Hitchhiker's Guide
This topic created in 2381 days ago, the information mentioned may be changed or developed.
 | 1 hehheh Nov 6, 2019 把 key,val 转成 tuple，然后整个 list 打包成 set 然后 intersection |
 | 2 Cooky Nov 6, 2019 via Android 两边各自合成一个字典做比较 |
 | 3 fdppzrl Nov 6, 2019 via Android c.addAll(a.removeall(b)) c.addAll(b.removeall(a)) Java 大概的写法就酱 |
 | 4 ranlan Nov 6, 2019 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] |
 | 6 yesterdaysun Nov 6, 2019 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) |
| 7 ranlan Nov 6, 2019 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] |
 | 8 css3 OP |
 | 9 johnnyluck Nov 6, 2019 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] |
 | 10 20015jjw Nov 7, 2019 via Android ^ 双关了 |
 | 11 xhxhx Nov 7, 2019 array_diff |
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