Python: If-else statements

问题

The following piece of code is showing an error:

if ((type(varA) or type(varB) ) == type('t')):
    print "string involved"
elif varA<varB:
    print "RANDOM"

the error is for this case:

Test Values: varA = 0, varB = adios

output:

RANDOM

while this other piece of code

if ((type(varA) == type('t')) or (type(varB)== type('t'))):
    print "string involved"
elif varA<varB:
    print "RANDOM"`

For the following test values:

Test Values: varA = 6, varB = adios

ouput is as follows:

string involved

What is the difference between these two "if" conditions? I am finding them to be of the same logic!

回答1:

The problem is this expression:

if ((type(varA) or type(varB) ) == type('t')):

Programming languages don't work like English. The above first evaluates type(varA) or type(varB), which will yield the type of varA - because or returns the first truthy value and any type is truthy.

Then it will check to see if that is the same as type('t') - that is, str. Which means that it will only be true when varA is a string, and the type of varB is completely ignored.

What you want is probably this:

if type(varA) == type('t') or type(varB) == type('t'):

But there are more idiomatic/Pythonic ways of doing that; see Óscar López's answer for some examples.

回答2:

This is wrong:

if ((type(varA) or type(varB) ) == type('t')):

It should be:

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

Equivalently:

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

Or a bit shorter:

if str in ((type(varA), type(varB)):

回答3:

If you want the test to work well for subclasses of str as well it's a good idea to use isintance instead of type

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

来源：https://stackoverflow.com/questions/19485061/python-if-else-statements

标签

python

if-statement