Python - Syntax for Boolean Expression with Type and Or

Question

varA = 1

varB = 2

Code w/ Correct Result:

if type(varA) == type('a') or type(varB) == type('a'):
    print "string involved (either varA or varB is a string)"
else:
    print "varA and varB are not strings"

Code w/ Incorrect Result:

if type(varA) or type(varB) == type('a'):
    print "string involved (either varA or varB is a string)"
else:
    print "varA and varB are not strings"

Why exactly does the 2nd set of code not return the expected result (i.e. "varA and varB are not strings")? What is the step-by-step breakdown of what Python is doing with the 2nd set of code? I found a similar question had already been answered but did not entirely understand the explanation. Python: If-else statements.

@volcano: the better form is isinstance(varA, str), to allow for subclasses. — Martijn Pieters
– Martijn Pieters, Commented Jan 18, 2014 at 23:45

score 3 · Accepted Answer · 2014-01-18 23:00:27Z

3

In the second code snippet, the condition of the if-statement is being interpreted by Python like this:

if (type(varA)) or (type(varB) == type('a')):

Moreover, it will always evaluate to True.

This is because, no matter what the value of varA is, type(varA) evaluates to True:

>>> varA = 'a'
>>> bool(type(varA))
True
>>> varA = False
>>> bool(type(varA))
True
>>>

In fact, since Python's logical operators short-circuit (stop evaluating as soon as possible), the type(varB) == type('a') part of the condition will never even be evaluated.

On a separate note, you should be using is to compare types:

if type(varA) is str or type(varB) is str:

or, you can use isinstance:

if isinstance(varA, str) or isinstance(varB, str):

edited Jan 18, 2014 at 23:00

answered Jan 18, 2014 at 22:40

user2555451

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1 Comment

AyeSea Over a year ago

Thanks for the quick answer! So since {type(varA)} is always {True}, Python will always run the code corresponding to True regardless of the actual type of {varA}.

Ryan Haining · Accepted Answer · 2014-01-18 22:55:49Z

1

Your second example does not work because it parses as

if (type(varA)) or (type(varB) == type('a')):

and type(varA) will always be a class type which is considered True, so the whole expression will be True

There are better ways to do this

if any(isinstance(v, str) for v in (varA, varB)):

any takes an iterable and evaluates to True if anything in the iterable is true. isinstance checks to see if the first argument "is a" second argument. Placing the generator expression inside of any reads as "if any v in (varA, varB) is a string): ... "

>>> var = 1
>>> isinstance(var, str) # var is an int, not a str
False
>>> isinstance(var, int)
True
>>> isinstance('a', int)
False
>>> isinstance('a', str) # 'a' is a str
True

edited Jan 18, 2014 at 22:55

answered Jan 18, 2014 at 22:49

Ryan Haining

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Comments

Cu3PO42 · Accepted Answer · 2014-01-18 22:47:42Z

0

iCodez is absolutely correct, but if you really want to do something along the lines of "list all elements and check if one of them is a string":

if str in map(type, [varA, varB]):
    print "string involved"

answered Jan 18, 2014 at 22:47

Cu3PO42

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Comments

Alexander Ejbekov · Accepted Answer · 2014-01-18 22:44:27Z

-1

Because in the second case you are not comparing both variables. Any integer above 0 would return True so you are not comparing types here.

if type(varA):

will always be True, because varA is equal to 1. You never even get to the second part of the condition.

answered Jan 18, 2014 at 22:44

Alexander Ejbekov

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4 Comments

volcano Over a year ago

Wrong, type(0) is int.

Alexander Ejbekov Over a year ago

Not what I said. type(0) is int, but it's also False according to python. Check: if 0 == False: print 123

volcano Over a year ago

type(varA) is not an integer. And any non-0 value - integer or float - evaluates to True in boolean expression. Here is my piece of code for you if -1: print "Alex Hristov is wrong" ;-)

Alexander Ejbekov Over a year ago

Look at the original post. varA = 1 . varA is clearly an integer.

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