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
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")
其他回答
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"
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。
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))
在另一个字符串列表中查找多个字符串的一种紧凑方法是使用set.intersection。这比大型集或列表中的列表理解执行得快得多。
>>> astring = ['abc','def','ghi','jkl','mno']
>>> bstring = ['def', 'jkl']
>>> a_set = set(astring) # convert list to set
>>> b_set = set(bstring)
>>> matches = a_set.intersection(b_set)
>>> matches
{'def', 'jkl'}
>>> list(matches) # if you want a list instead of a set
['def', 'jkl']
>>>
这取决于上下文 假设你想检查单个文字(任何单个单词a,e,w,..等)就足够了
original_word ="hackerearcth"
for 'h' in original_word:
print("YES")
如果你想检查original_word中的任何一个字符: 利用
if any(your_required in yourinput for your_required in original_word ):
如果您想要original_word中的所有输入,请使用所有输入 简单的
original_word = ['h', 'a', 'c', 'k', 'e', 'r', 'e', 'a', 'r', 't', 'h']
yourinput = str(input()).lower()
if all(requested_word in yourinput for requested_word in original_word):
print("yes")