如何检查数组中的任何字符串是否存在于另一个字符串中?
例如:
a = ['a', 'b', 'c']
s = "a123"
if a in s:
print("some of the strings found in s")
else:
print("no strings found in s")
我如何替换如果a在s:行得到适当的结果?
如何检查数组中的任何字符串是否存在于另一个字符串中?
例如:
a = ['a', 'b', 'c']
s = "a123"
if a in s:
print("some of the strings found in s")
else:
print("no strings found in s")
我如何替换如果a在s:行得到适当的结果?
当前回答
只是关于如何在String中获得所有列表元素的更多信息
a = ['a', 'b', 'c']
str = "a123"
list(filter(lambda x: x in str, a))
其他回答
flog = open('test.txt', 'r')
flogLines = flog.readlines()
strlist = ['SUCCESS', 'Done','SUCCESSFUL']
res = False
for line in flogLines:
for fstr in strlist:
if line.find(fstr) != -1:
print('found')
res = True
if res:
print('res true')
else:
print('res false')
这是set的另一个解。使用set.intersection。对于一行代码。
subset = {"some" ,"words"}
text = "some words to be searched here"
if len(subset & set(text.split())) == len(subset):
print("All values present in text")
if subset & set(text.split()):
print("Atleast one values present in text")
如果您想要单词的精确匹配,那么可以考虑对目标字符串进行单词标记。我使用nltk推荐的word_tokenize:
from nltk.tokenize import word_tokenize
下面是接受答案的标记化字符串:
a_string = "A string is more than its parts!"
tokens = word_tokenize(a_string)
tokens
Out[46]: ['A', 'string', 'is', 'more', 'than', 'its', 'parts', '!']
接受的答案修改如下:
matches_1 = ["more", "wholesome", "milk"]
[x in tokens for x in matches_1]
Out[42]: [True, False, False]
在公认的答案中,单词“more”仍然是匹配的。但是,如果“mo”成为匹配字符串,接受的答案仍然找到匹配。这是我不希望看到的行为。
matches_2 = ["mo", "wholesome", "milk"]
[x in a_string for x in matches_1]
Out[43]: [True, False, False]
使用单词标记化,“mo”不再匹配:
[x in tokens for x in matches_2]
Out[44]: [False, False, False]
这是我想要的附加行为。这个答案也回答了这里的重复问题。
a = ['a', 'b', 'c']
str = "a123"
a_match = [True for match in a if match in str]
if True in a_match:
print "some of the strings found in str"
else:
print "no strings found in str"
只是关于如何在String中获得所有列表元素的更多信息
a = ['a', 'b', 'c']
str = "a123"
list(filter(lambda x: x in str, a))