一个如何将消除可以通过相似性和长度包含从Python列表串的元素(如果字符串X在另一个发现,较长的字符串ÿ,X必须被移除)?剪枝相似字符串由长度
IN: [('this is string that stays', 0), ('this is string', 1), ('string that stays', 2), ('i am safe', 3)]
OUT: [('this is string that stays', 0), ('i am safe', 3)]
一个如何将消除可以通过相似性和长度包含从Python列表串的元素(如果字符串X在另一个发现,较长的字符串ÿ,X必须被移除)?剪枝相似字符串由长度
IN: [('this is string that stays', 0), ('this is string', 1), ('string that stays', 2), ('i am safe', 3)]
OUT: [('this is string that stays', 0), ('i am safe', 3)]
在这里你去:
l = [('this is string that stays', 0), ('this is string', 1), ('string that stays', 2), ('i am safe', 3)]
survivors = set(s for s, _ in l)
for s1, _ in l:
if any(s1 != s2 and s1 in s2 for s2 in survivors):
survivors.discard(s1)
survivors
是你想要的,但它不包含输入的元组号码 - 改变这应该是为读者:-P练习。
谢谢。我想我自己解决它:https://pzt.me/13gn – SomeOne
试试这个:
IN = [('this is string that stays', 0), ('this is string', 1), ('string that stays', 2), ('i am safe', 3)]
OUT=[]
def check_item(liste, item2check):
for item, _ in liste:
if item2check in item and len(item2check) < len(item):
return True
return False
for item, rank in IN:
if not check_item(IN, item):
OUT.append((item, rank))
# or in a list-comprehension :
OUT = [(item, rank) for item, rank in IN if not check_item(IN, item)]
print OUT
>>> [('this is string that stays', 0), ('i am safe', 3)]
,如果你不介意的顺序(N * N)
>>> s=[('this is string that stays', 0), ('this is string', 1), ('string that stays', 2), ('i am safe', 3)]
>>> s=[i[0] for i in s]
>>> result=[s[i] for i in range(len(s)) if not any(s[i] in s[j] for j in range(i)+range(i+1,len(s)-i))]
>>> result
['this is string that stays', 'i am safe']
如果你关心效率,我建议你每个字符串分割成的序列字(或甚至字符)并且制作树数据结构,例如trie(http://community.topcoder.com/tc?module=Static &d1 = tutorials &d2 = usingTries),其允许在每个子序列上快速查找
所有其他答案都提供了很好的解决方案。我只想补充您尝试记:
for i in range(0, len(d)):
for j in range(1, len(d)):
if d[j][0] in d[i][0] and len(d[i][0]) > len(d[j][0]):
del d[j]
失败与列表索引超出范围,因为遍历目录,而你删除。这里有一种方法可以防止这个问题:
d = [('this is string that stays', 0), ('this is string', 1), ('string that stays', 2), ('i am safe', 3)]
to_be_removed = list()
for i in range(0, len(d)):
for j in range(0, len(d)):
if i != j and d[j][0] in d[i][0] and len(d[i][0]) > len(d[j][0]):
to_be_removed.append(j)
for m, n in enumerate(to_be_removed):
del d[n - m]
print d
给出你的例子,''这是字符串''在** **中是**,这是一个字符串''。是因为第一个字符串中的每个单词出现在第二个字符串中?如果这不是拼写错误,请在**另一个字符串中指定一个字符串的含义是**。 –
更正了示例。 – SomeOne
好。你有什么想法吗?给我们一些讨论。 –