Python: Find index of fist list in list oneliner

Question

Here's the solution to find y list index in x big list

def getsubidx(x, y):
    l1, l2 = len(x), len(y)
    for i in range(l1):
        if x[i:i+l2] == y:
            return i

Is it possible to transform this to oneliner using itertools, filter or something else?

jamylak · Answer 1 · 2014-12-25T22:00:26.833

1

def getsubidx(x, y):
    return next(i for i in range(len(x)) if x[i:i+len(y)] == y)

You may want an error to be raised if no match, if you don't you can return -1

return next((i for i in range(len(x)) if x[i:i+len(y)] == y), -1)

edited Dec 25 '14 at 22:00

answered Dec 25 '14 at 21:18

jamylak

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I assumed OP would want an error, if OP doesn't -1 can be a default value – jamylak Dec 25 '14 at 22:00

score 0 · Answer 2 · answered Dec 25 '14 at 21:55

0

def getsubidx(x, y):
    l1, l2 = len(x), len(y)
    for i in range(l1-l2):   here it should be l1-l2
        if x[i:i+l2] == y:        here you may get list out of index
            return i

one liner:

def getsubidx(x, y):
    return [i for i in range(len(x)-len(y)) if x[i:i+len(y)] == y]

answered Dec 25 '14 at 21:55

Hackaholic

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This finds every index in a list , not what the question shows – jamylak Dec 25 '14 at 22:01
Also you won't **ever** get a list out of index for a slice like that `>>> [][9000:9001]` `[]` – jamylak Dec 25 '14 at 22:02

score 0 · Answer 3 · edited May 23 '17 at 12:28

0

Is it possible to transform this to oneliner using itertools, filter or something else?

Here's "somethng else", you could use index(subseq, seq) function:

index_of_y_in_x = index(y, x)

edited May 23 '17 at 12:28

Community

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answered Dec 26 '14 at 03:23

jfs

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Python: Find index of fist list in list oneliner

3 Answers3