How to check if only the specified elements remain in a list?

Question

So lets say I have a list like this

color = ['blue', 'green', 'red', 'white', 'hurley', 'maladroit']

What I want to do here is detect when the list 'color' only consists of elements that aren't colors which are 'hurley' and 'maladroit'.

So something like:

#If the list color has at least one color in it (E.g)color = ['blue','white','maladroit']
    print('Nothing to see here')

#If the list only consists of elements that aren't colors 
#(E.g)color = ['hurley','maladroit']
    print("I don't see colors.... I actually legit don't...")

Oh no, the colors where just examples I used for this code, sorry if that was misleading. I'll look into filtering methods though, thank you. — Joshua Kim, Jan 06 '17 at 12:47
Possible duplicate of [Python - Intersection of two lists](http://stackoverflow.com/questions/642763/python-intersection-of-two-lists) — Peter Wood, Jan 06 '17 at 13:08
@PeterWood sort of, although that question doesn't solve the test for True/False that the OP asked about. It's a relatively small next step but these answers should solve the problem of how to test the response as well . — Phil Sheard, Jan 06 '17 at 13:13

Jean-François Fabre · Accepted Answer · 2017-01-06T12:56:39.917

2

Let's create a set of available colors

allowed = {'blue', 'green', 'red', 'white'}

then check the negation of "at least one element of the list is a real color"

print(not any(c in allowed for c in color))

yields True for ['hurley', 'maladroit'], False if there's at least a color in the list

should be very fast performance-wise because:

it's using a set for testing
it's not creating other temporary list
it's using built-in any function

EDIT: even simpler and faster using isdisjoint method while we're using a set (thx PM 2Ring for letting me find out):

print(allowed.isdisjoint(color))

edited Jan 06 '17 at 12:56

answered Jan 06 '17 at 12:46

Jean-François Fabre

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at least the `set` existed :) thanks. Didn't take time to dig this out. – Jean-François Fabre Jan 06 '17 at 12:58
This looks very interesting, will look into this and play around with it, thank you. – Joshua Kim Jan 06 '17 at 13:03

UpSampler · Answer 2 · 2017-01-09T06:16:14.663

2

Inspired by the answers before: Create a set of allowed colors and check wether there are any allowed colors in the difference.

allowed_colors = {'red', 'green', 'blue', 'white'}

color = ['blue', 'green', 'red', 'white', 'hurley', 'maladroit']

color_set = set(color)
if len(color_set - allowed_colors) < len(color_set):
    print('Nothing to see here')
else:
    print("I don't see colors.... I actually legit don't...")

Edit: Solution was not correct. Works as expected now. Though isdisjoint is the most elegant solution, if you know set theory, as pointed out by Jean-François.

edited Jan 09 '17 at 06:16

answered Jan 06 '17 at 13:09

UpSampler

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You can write code blocks by skipping a line and putting 4 spaces in front of it the line you want to start writing code in. Way better formatting that way, just letting you know :) – Joshua Kim Jan 06 '17 at 13:13
`if set(color) - allowed_colors:` is enough, no need to call `len` – Jérôme Jan 06 '17 at 13:18

score 1 · Answer 3 · edited May 23 '17 at 11:53

Inspired by a related answer (https://stackoverflow.com/a/642919/2034487), you can use a set()'s intersection() method:

List containing colours:

$ approved_colours = set(['blue','green'])
$ list_with_colours = ['blue', 'green', 'red', 'white', 'hurley', 'maladroit']
$ if approved_colours.intersection(list_with_colours):
$     print("There are colours present")
$ else:
$     print("There are no colours present")
> "There are colours present"

Do the same without any any colours:

$ list_without_colours = ['hurley', 'maladroit']
$ if approved_colours.intersection(list_without_colours):
$     print("There are colours present")
$ else:
$     print("There are no colours present")
> "There are no colours present"

Obviously you would put this method into a wrapper to test a variable in real life. I'm writing long-form to demonstrate both results.

Learning a lot about the intersection function from this answer. Didn't know something like this existed, will test around with this as well. Thank you! — Joshua Kim, Jan 06 '17 at 13:15

score 0 · Answer 4 · answered Jan 06 '17 at 12:59

0

color = ['blue', 'green', 'red', 'white', 'hurley', 'maladroit']

you can combine both these

color.count('hurley') 
color.count('maladroit')

or these

'hurley' in color
'maladroit' in color

answered Jan 06 '17 at 12:59

Chandan Rai

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What if the list has a million entries? – Peter Wood Jan 06 '17 at 13:06
for large number of items `code` def intersect(a, b): return list(set(a) & set(b)) print intersect(color, allowedcolor) `code` – Chandan Rai Jan 06 '17 at 13:21

How to check if only the specified elements remain in a list?

4 Answers4