Finding consecutive values within a list

By : Baz
Source: Stackoverflow.com
Question!

I have a list of values:

a = [1,3,4,5,2]

I now want the following function:

does_segment_exist(a, [1,3,4]) #True
does_segment_exist(a, [3,4,5]) #True
does_segment_exist(a, [4,5,2]) #True
does_segment_exist(a, [1,4,5]) #False
does_segment_exist(a, [1,3]) #True
does_segment_exist(a, [1,4]) #False

So the values must be found in consecutive order.

I there a clever way of doing this in Python 3?

By : Baz


Answers
You can use a rolling window iterator, in this case one from an old version of the itertools docs:

from itertools import islice

def window(seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result
    for elem in it:
        result = result[1:] + (elem,)
        yield result

def does_segment_exist(iterable, sublist):
    return tuple(sublist) in window(iterable, len(sublist))

print(does_segment_exist([1,3,4,5,2], [3,4,5]))

If you only need it to work on lists, not any iterable, you can use:

def does_segment_exist(seq, sublist):
    # seq and sublist must both be lists
    n = len(sublist)
    return sublist in (seq[i:i+n] for i in range(len(seq) + 1 - n))

A basic implementation of the method mentioned by Raymond:

def does_segment_exist(seq, sublist):
    first = sublist[0]
    i = 0
    n = len(sublist)
    while True:
        try:
            i = seq.index(first, i)
        except ValueError:
            return False
        if sublist == seq[i:i+n]:
            return True
        i += 1

print(does_segment_exist([1,3,4,5,2], [3,4,5]))

The advantage of this method is that it doesn't have to slice for every index up to the first match, just for indexes corresponding to matches for the first value in the segment.

By : agf


There are many ways to do this and they are all isomorphic to substring search algorithms.

The easiest way is the naïve search using list.index() to find a common start point and then using a slice to check for a full match. If there is not a match, repeat the search until you hit the end of the list.



This should work with Python 2.5 and newer:

def does_segment_exist(sequence, segment):
    n, m = len(sequence), len(segment)
    return any(segment == sequence[i:i+m] for i in range(n+1-m))
By : Cito


This video can help you solving your question :)
By: admin