CODEWITHSUNDEEP
python
Adam Michael Wood
Adam Michael Wood

Reputation: 1788

Double list comprehension in Python - finding borders coordinates

Given (x,y), I want the collection:

[(0,0),(0,y),(1,0),(1,y),(2,0),(2,y)...(x-1,0),(x-1,y),(x,0),(x,y),
             (0,1),(x,1),(0,2),(x,2)...(0,y-1),(x,y-1)]

(I don't really care if it is a list, set, or any other type of collection.)

I've experimented with several permutations of list comps, nothing has really worked.

I found a BAD solution::

all_points = list(itertools.product([x for x in range(x+1)], [y for y in range(y+1)]))
border = [xy for xy in all_points if xy[0]==0 or xy[0]==x or xy[1]==0 or xy[1]==y]

But I really hate this solution and am wondering if there is a more direct approach.

EDIT

The BAD solution can be made better, as mentioned below in comments:

all_points = list(itertools.product(range(x+1), range(y+1))
border = [xy for xy in all_points if xy[0]==0 or xy[0]==x or xy[1]==0 or xy[1]==y]

But the problem remains --- i'm just getting all the coords and then dropping the ones that aren't in the comp...

EDIT

The BAD solution can be made better still...

border = [xy for xy in itertools.product(range(x+1), range(y+1)) if xy[0]==0 or xy[0]==x or xy[1]==0 or xy[1]==y]

But I don't know how I feel about this...

EDIT -- what I really want to know is...

Is there a way to do some kind of (I dunno) recursive or loopy list comprehension that returns the desired results by directly building the list?

I can solve the practical problem of finding the coords with the bad solution. But I want to grok list comps better.

Upvotes: 2

Views: 925

Answers (2)

Copperfield
Copperfield

Reputation: 8510

as list comprehension perhaps like this

def border(x,y):
    return [ (a,b) for a in range(x+1) for b in range(y+1) if 0 in (a,b) or x==a or y==b ]

But I rather produce directly what I need instead of searching for some esoteric and/or potentially inefficient way of doing it, is better to be clear than clever.

Like this

def border(x,y):
    for p in range(x+1):
        yield (p,0)
        yield (p,y)
    for p in range(1,y):
        yield (0,p)
        yield (x,p)

and this one is way more efficient as it don't waste time producing unnecessary stuff just to be discarded

Upvotes: 1

rassar
rassar

Reputation: 5660

If you really want a list comprehension here's one.

l = sorted({j for i in [[[(i, y), (i, 0)] for i in range(x+1)] + [[(x, i), (0, i)] for i in range(y+1)]][0] for j in i})

This will return a sorted set of tuples.

Upvotes: 3

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