Python 从与子字符串匹配的列表中删除项目
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/12666897/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Removing an item from list matching a substring
提问by alvas
How do I remove an element from a list if it matches a substring?
如果元素与子字符串匹配，如何从列表中删除它？
I have tried removing an element from a list using the pop()and enumeratemethod but seems like I'm missing a few contiguous items that needs to be removed:
我尝试使用pop()andenumerate方法从列表中删除一个元素，但似乎我缺少一些需要删除的连续项目：
sents = ['@$\tthis sentences needs to be removed', 'this doesnt',
'@$\tthis sentences also needs to be removed',
'@$\tthis sentences must be removed', 'this shouldnt',
'# this needs to be removed', 'this isnt',
'# this must', 'this musnt']
for i, j in enumerate(sents):
if j[0:3] == "@$\t":
sents.pop(i)
continue
if j[0] == "#":
sents.pop(i)
for i in sents:
print i
Output:
输出：
this doesnt
@$ this sentences must be removed
this shouldnt
this isnt
#this should
this musnt
Desired output:
期望的输出：
this doesnt
this shouldnt
this isnt
this musnt
采纳答案by D.Shawley
How about something simple like:
简单的事情怎么样：
>>> [x for x in sents if not x.startswith('@$\t') and not x.startswith('#')]
['this doesnt', 'this shouldnt', 'this isnt', 'this musnt']
回答by mjgpy3
This should work:
这应该有效：
[i for i in sents if not ('@$\t' in i or '#' in i)]
If you want only things that begin with those specified sentential use the str.startswith(stringOfInterest)method
如果您只想要以指定句子开头的内容，请使用该str.startswith(stringOfInterest)方法
回答by cod3monk3y
Another technique using filter
另一种技术使用 filter
filter( lambda s: not (s[0:3]=="@$\t" or s[0]=="#"), sents)
The problem with your orignal approach is when you're on list item iand determine it should be deleted, you remove it from the list, which slides the i+1item into the iposition. The next iteration of the loop you're at index i+1but the item is actually i+2.
您的原始方法的问题在于，当您在列表项上i并确定应将其删除时，您将其从列表中删除，从而将i+1项目滑入该i位置。您在 index 处的循环的下一次迭代，i+1但该项目实际上是i+2.
Make sense?
有道理？
