我如何在 Python 中检查某个字母的字符串？
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How would I check a string for a certain letter in Python?
提问by Noah R
How would I tell Python to check the below for the letter x and then print "Yes"? The below is what I have so far...
我如何告诉 Python 检查下面的字母 x 然后打印“是”？以下是我到目前为止...
dog = "xdasds"
if "x" is in dog:
print "Yes!"
采纳答案by ide
Use the inkeyword without is.
使用in不带的关键字is。
if "x" in dog:
print "Yes!"
If you'd like to check for the non-existence of a character, use not in:
如果您想检查字符是否不存在，请使用not in：
if "x" not in dog:
print "No!"
回答by pyfunc
inkeyword allows you to loop over a collection and check if there is a member in the collection that is equal to the element.
in关键字允许您遍历集合并检查集合中是否有与元素相等的成员。
In this case string is nothing but a list of characters:
在这种情况下，字符串只不过是一个字符列表：
dog = "xdasds"
if "x" in dog:
print "Yes!"
You can check a substring too:
您也可以检查子字符串：
>>> 'x' in "xdasds"
True
>>> 'xd' in "xdasds"
True
>>>
>>>
>>> 'xa' in "xdasds"
False
Think collection:
认为收藏：
>>> 'x' in ['x', 'd', 'a', 's', 'd', 's']
True
>>>
You can also test the set membership over user defined classes.
您还可以在用户定义的类上测试集合成员资格。
For user-defined classes which define the __contains__ method, x in y is true if and only if y.__contains__(x) is true.
对于定义 __contains__ 方法的用户定义类，当且仅当 y.__contains__(x) 为真时，y 中的 x 为真。
回答by Foo Bah
If you want a version that raises an error:
如果您想要一个引发错误的版本：
"string to search".index("needle")
If you want a version that returns -1:
如果你想要一个返回 -1 的版本：
"string to search".find("needle")
This is more efficient than the 'in' syntax
这比“in”语法更有效
