How to check if a variable is equal to one string or another string?

Python

Python Problem Overview

if var is 'stringone' or 'stringtwo':
    dosomething()

This does not work! I have a variable and I need it to do something when it is either of the values, but it will not enter the if statement. In Java if (var == "stringone" || "stringtwo") works. How do I write this in Python?

Python Solutions

Solution 1 - Python

This does not do what you expect:

if var is 'stringone' or 'stringtwo':
    dosomething()

It is the same as:

if (var is 'stringone') or 'stringtwo':
    dosomething()

Which is always true, since 'stringtwo' is considered a "true" value.

There are two alternatives:

if var in ('stringone', 'stringtwo'):
    dosomething()

Or you can write separate equality tests,

if var == 'stringone' or var == 'stringtwo':
    dosomething()

Don't use is, because is compares object identity. You might get away with it sometimes because Python interns a lot of strings, just like you might get away with it in Java because Java interns a lot of strings. But don't use is unless you really want object identity.

>>> 'a' + 'b' == 'ab'
True
>>> 'a' + 'b' is 'abc'[:2]
False # but could be True
>>> 'a' + 'b' is 'ab'
True  # but could be False

Solution 2 - Python

if var == 'stringone' or var == 'stringtwo':
    do_something()

or more pythonic,

if var in ['string one', 'string two']:
    do_something()

Solution 3 - Python

if var == 'stringone' or var == 'stringtwo':
    dosomething()

'is' is used to check if the two references are referred to a same object. It compare the memory address. Apparently, 'stringone' and 'var' are different objects, they just contains the same string, but they are two different instances of the class 'str'. So they of course has two different memory addresses, and the 'is' will return False.

Solution 4 - Python

Two separate checks. Also, use == rather than is to check for equality rather than identity.

 if var=='stringone' or var=='stringtwo':
     dosomething()

Solution 5 - Python

for a in soup("p",{'id':'pagination'})[0]("a",{'href': True}):
        if createunicode(a.text) in ['<','<']:
            links.append(a.attrMap['href'])
        else:
            continue

It works for me.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionRahul SharmaView Question on Stackoverflow
Solution 1 - PythonDietrich EppView Answer on Stackoverflow
Solution 2 - PythoninspectorG4dgetView Answer on Stackoverflow
Solution 3 - PythonLingfeng XiongView Answer on Stackoverflow
Solution 4 - PythonAndrew JaffeView Answer on Stackoverflow
Solution 5 - PythonYaseer ArafatView Answer on Stackoverflow