Check if a string doesn't contain any substring from list of substrings in Python

Question

I'm trying to do the following logic:

bitmap_id = 'HelloWorld'
if 'SLIDE' not in bitmap_id and 'ICON' not in bitmap_id  and 'BOT' not in bitmap_id:
    print bitmap_id

So if bitmap_id is 'ICON_helloworld', then nothing should be printed.

I'm pretty sure you all agree that it's too long and looks ugly, so I tried to do it as shown below, but it's not working.

if any(s not in bitmap_id for s in ['SLIDE', 'ICON', 'BOT']):
    print bitmap_id

What am I doing wrong?

Have you tried declaring the table out of the command line, and just make a for loop for each string in the table ? — Peut22
– Peut22, Commented Aug 5, 2015 at 8:29
It looks to me like in your first statement you are using 'and' but in your second statement you have an implicit 'or'. — ypnos
– ypnos, Commented Aug 5, 2015 at 8:30
@ypnos thanks heaps for hint, I'm new to Python and didn't know of the all function :) your comment was sufficient for me to find the answer quickly — sayo9394
– sayo9394, Commented Aug 5, 2015 at 8:35

DeepSpace · Accepted Answer · 2015-08-05 08:35:21Z

15

Use either

if all(s not in bitmap_id for s in ['SLIDE', 'ICON', 'BOT']):
     print bitmap_id

or

if not any(s in bitmap_id for s in ['SLIDE', 'ICON', 'BOT']):
    print bitmap_id

edited Aug 5, 2015 at 8:35

answered Aug 5, 2015 at 8:31

DeepSpace

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Andrey · Accepted Answer · 2015-08-05 08:30:15Z

3

You need actually:

if all(s not in bitmap_id for s in ['SLIDE', 'ICON', 'BOT']):
    print bitmap_id

if any will be false only if all of the of the strings are in bitmap_id, for example 'bitmap_id = 'ICON_SLIDE_BOT'. You want opposite, that none of the strings are there or all strings are not there.

answered Aug 5, 2015 at 8:30

Andrey

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2 Comments

sayo9394 Over a year ago

you got the answer over DeepSpace by one minute! Sorry for the late acceptance! actually forgot bout this question as you guys answered it so fast I had to wait to accept! cheers,

Andrey Over a year ago

@sayo9394 I actually like his answer more, he has 2 options.

Muhammad Tahir · Accepted Answer · 2015-08-05 08:34:06Z

2

You can do something like this:

bitmap_id = 'HelloWorld'
blacklist = ['SLIDE', 'ICON', 'BOT']
if not filter(lambda x: x in bitmap_id, blacklist):
    print bitmap_id

answered Aug 5, 2015 at 8:34

Muhammad Tahir

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