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")
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')
的元素上进行迭代。
a = ['a', 'b', 'c']
str = "a123"
found_a_string = False
for item in a:
if item in str:
found_a_string = True
if found_a_string:
print "found a match"
else:
print "no match found"
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']
>>>
你可以使用任何:
a_string = "A string is more than its parts!"
matches = ["more", "wholesome", "milk"]
if any([x in a_string for x in matches]):
类似地，要检查是否找到列表中的所有字符串，请使用all而不是any。
为了提高速度，我会使用这样的函数:
def check_string(string, substring_list):
for substring in substring_list:
if substring in string:
return True
return False
