[英]Matching a substring from a substring list in a list of strings
[英]Removing an item from list matching a substring
如果它與子字符串匹配,如何從列表中刪除元素?
我嘗試使用pop()
和enumerate
方法從列表中刪除一個元素,但似乎我缺少一些需要刪除的連續項目:
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
輸出:
this doesnt
@$ this sentences must be removed
this shouldnt
this isnt
#this should
this musnt
期望的輸出:
this doesnt
this shouldnt
this isnt
this musnt
像這樣簡單的東西怎么樣:
>>> [x for x in sents if not x.startswith('@$\t') and not x.startswith('#')]
['this doesnt', 'this shouldnt', 'this isnt', 'this musnt']
這應該有效:
[i for i in sents if not ('@$\t' in i or '#' in i)]
如果您只想要以指定句子開頭的內容,請使用str.startswith(stringOfInterest)
方法
[i for i in sents if i.startswith('#')]
另一種使用filter
的技術
filter( lambda s: not (s[0:3]=="@$\t" or s[0]=="#"), sents)
您的原始方法的問題是,當您在列表項i
上並確定它應該被刪除時,您將其從列表中刪除,這會將i+1
項滑入i
位置。 您在索引i+1
處的循環的下一次迭代,但該項目實際上是i+2
。
說得通?
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