Question

这是一个我想打印 TRUE 的示例，因为Drug在字符串中。如果disease在字符串中，我也希望它也打印 TRUE 。对于所有其他情况，我想打印 FALSE

mystring =  "Drug   CID006338583    AC1O3UYX    Stitch  1.515E-3    1.000E0 4.989E-2    5.235E-1    4   63  PTGES,SLC15A1,KLK8,IL7R"
check_if_in_mystring = ['Drug', 'Disease']

if check_if_in_mystring in mystring:
    print("TRUE")
else:
    print("FALSE")

Answer 1

您的示例是正确的，您可以对字符串使用“ in”运算符，只需在“ check_if_in_mystring”上添加循环即可遍历要检查的每个元素，如下面的示例。

mystring =  "Drug   CID006338583    AC1O3UYX    Stitch  1.515E-3    1.000E0 4.989E-2    5.235E-1    4   63  PTGES,SLC15A1,KLK8,IL7R"
check_if_in_mystring = ['Drug', 'Disease']

for element in check_if_in_mystring:
    if element in mystring:
        print("TRUE for {}".format(element))
    else:
        print("FALSE for {}".format(element))

输出：

TRUE for Drug
FALSE for Disease

Answer 2

此代码段将检测check_if_in_mystring中是否有mystring：

any(word in mystring for word in check_if_in_mystring)

对于问题的确切行为：

print(str(any(word in mystring for word in check_if_in_mystring)).upper())

Answer 3

尝试一下：

>>> my_string = "Drug   CID006338583    AC1O3UYX    Stitch  
1.515E-3    1.000E0 4.989E-2    5.235E-1    4   63  PTGES,SLC15A1,KLK8,IL7R"

>>> print("TRUE" if "Drug" in my_string or "Disease" in my_string else 
"FALSE")

TRUE

>>> my_string = "Some other random string"

>>> print("TRUE" if "Drug" in my_string or "Disease" in my_string else 
"FALSE")

FALSE

>>> my_string = "Has the word Disease"

>>> print("TRUE" if "Drug" in my_string or "Disease" in my_string else 
"FALSE")

TRUE

字符串支持in运算符，因此您只需检查Drug中是否包含Disease或my_string。

Answer 4

将字符串转换为列表并签入列表可以是一种解决方案，如下所述。

mystring =  "Drug   CID006338583    AC1O3UYX    Stitch  1.515E-3    1.000E0 4.989E-2    5.235E-1    4   63  PTGES,SLC15A1,KLK8,IL7R"
check_if_in_mystring = ['Drug', 'Disease']
mystring1 = mystring.split() #convertin string to list, seprated by white space
for item in check_if_in_mystring:
    if item in mystring1:
        print(item)
        print("TRUE")
    else:
        print(item)
        print("FALSE")

输出：打印工作及其在mystring中的存在

Drug
TRUE
Disease
FALSE

Answer 5

亚历克斯·泰勒（Alex Taylor）处在正确的轨道上，但是any子句可以大大简化：

>>> mystring =  "Drug   CID006338583    AC1O3UYX    Stitch  1.515E-3    1.000E0 4.989E-2    5.235E-1    4   63  PTGES,SLC15A1,KLK8,IL7R"
>>> check_if_in_mystring = ['Drug', 'Disease']
>>> any(word in mystring for word in check_if_in_mystring)
True
>>> mystring =  "Poodle   CID006338583    AC1O3UYX    Stitch  1.515E-3    1.000E0 4.989E-2    5.235E-1    4   63  PTGES,SLC15A1,KLK8,IL7R"
>>> any(word in mystring for word in check_if_in_mystring)
False

这些方法在check_if_in_mystring中的单词列表中循环，一旦第一个any(...)表达式为True，word in mystring返回True。如果for word in check_if_in_mystring循环结束而没有找到True值，则any(...)返回False。

Answer 6

您还可以使用：

mystring.contains(anotherstring)

检查列表中的任何元素是否存在于字符串中

6 个答案: