我一直在阅读其他一些 SO 文章，并了解我想要完成的事情的点点滴滴，但我尚未确定包含我的要求的答案。
我知道如何从 [1,2,3,4] 这样的简单列表中删除单个项目，但我有一个更复杂的列表，其中顺序/内容不可预测，我正在尝试删除一个元素来自:
list = [{u'role': u'OWNER', u'userByEmail': u'<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="097d6c7a7d38496c64686065276a6664" rel="noreferrer noopener nofollow">[email protected]</a>'}, {u'specialGroup': u'projectWriters', u'role': u'READER'}, {u'specialGroup': u'projectOwners', u'role': u'READER'}, {u'specialGroup': u'projectReaders', u'role': u'READER'}, {u'role': u'READER', u'userByEmail': u'<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="94e0f1e7e0a6d4f1f9f5fdf8baf7fbf9" rel="noreferrer noopener nofollow">[email protected]</a>'}, {u'view': {u'projectId': u'project-01', u'tableId': u'testing_tbl_2', u'datasetId': u'test2'}}, {u'view': {u'projectId': u'project-01', u'tableId': u'testing_tbl_3', u'datasetId': u'dtest2'}}, {u'view': {u'projectId': u'project-01', u'tableId': u'view1', u'datasetId': u'test2'}}]
我需要删除:
{u'role': u'READER', u'userByEmail': u'<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="a3d7c6d0d792e3c6cec2cacf8dc0ccce" rel="noreferrer noopener nofollow">[email protected]</a>'}
并返回不存在该键/值元素的新列表。我可以做这样的事情(我更喜欢这种格式)，但这仅返回过滤器正在查找的键/值元素。
>>> filter(lambda row: row.has_key('userByEmail') and row['userByEmail'] == '<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="ccb8a9bfb8fd8ca9a1ada5a0e2afa3a1" rel="noreferrer noopener nofollow">[email protected]</a>', list)
[{u'role': u'OWNER', u'userByEmail': u'<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="047061777035446169656d682a676b69" rel="noreferrer noopener nofollow">[email protected]</a>'}]
如何从列表中删除元素，然后打印出缺少过滤元素的新列表？
最佳答案
将条件反转为
not (...)
所以
filter(lambda row: row.has_key('userByEmail') and row['userByEmail'] == '<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="196d7c6a6d28597c74787075377a7674" rel="noreferrer noopener nofollow">[email protected]</a>', list)
变成了
filter(lambda row: not (row.has_key('userByEmail') and row['userByEmail'] == '<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="eb9f8e989fdaab8e868a8287c5888486" rel="noreferrer noopener nofollow">[email protected]</a>'), list)
关于Python - 从列表中删除键/值元素并返回新列表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/43526219/