Python: How to test multiple values inside an if [multiple values] in list:

Question

I have returned a list of href values from a HTML document. I want to go though every link within this list and test to see if they contain any of the values within my IMAGE_FORMAT tuple.

IMAGE_FORMAT = (
    '.png',
    '.jpg',
    '.jpeg',
    '.gif',
)

At present I am simply testing for '.jpg' e.g if '.jpg' in link.get('href'):

I'd like to extend this code to something along the lines of if [any value inside IMAGEFORMAT] in link.get('href'):

What would be the most efficient or cleanest way or doing so?

DSM · Accepted Answer · 2012-05-30 23:19:57Z

6

If you really want in, then maybe

href = link.get('href')
if any(end in href for end in IMAGE_FORMAT):
    # do something
    pass

but if you actually want ends with, then use .endswith:

>>> IMAGE_FORMAT = ('.png','.gif','.jpg','.jpeg')
>>> 'fred.gif'.endswith(IMAGE_FORMAT)
True

Depends on how you want to treat 'fred.gif.gz', etc. Also note you might want to work with href.lower() if you don't care about case.

answered May 30, 2012 at 23:19

DSM

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2 Comments

Karl Knechtel Over a year ago

I always find myself tempted to write any('fred.gif'.endswith(ending) for ending in IMAGE_FORMAT), neglecting that .endswith already accepts *args to do what I want. x.x

Kristian Roebuck Over a year ago

You mention If you really want in, then maybe... would there be a nicer way of writing the same code?

cheeken · Accepted Answer · 2012-05-31 02:01:20Z

1

Try any against list comprehension.

any(e in href for e in IMAGE_FORMAT)

Or, in English, "are any of the items in my image formats in my URI?" Bare in mind how in functions with strings, though.

edited May 31, 2012 at 2:01

answered May 30, 2012 at 23:19

cheeken

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1 Comment

kindall Over a year ago

No need for the list. any(e in href for e in IMAGE_FORMAT) will do fine.

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