Compare each list's element with others in the list

Question

So, there is a list like

list=["one","two","ne","three"....]

I wonder, how do I compare each element of the list with others by using endswith() method? in this list, for example, list[0] endwith list[2].

I couldn't get how to make a comparison itself. I'm trying something like:

aa=list
flg=False
for i in range(len(ll)-1):
    aa.append(ll[i+1])
    if ll[i].endswith(aa[i]):
        flg=True

however, it's good only for the first element, not with each one.

Answer 1

Using sets:

words = {"one","two","ne","three"}

[x for x in words if any(word.endswith(x) for word in words - {x})]
Out[69]: ['ne']

Basically, for each element, remove it from the words set, and then test if any of the other words in the truncated set end with that word.

Answer 2

You are only doing a single pass through the list, even though you have stated your goal as " Compare each element of the list with the others ". This necessitates making one traversal of the list per element in the list . If you have n items, you will traverse the list n times for every item, for a total of n^2 traversals.

Hence, you need two for loops in your solution: one to traverse the list once and select the element that will be compared, and inside that loop, another that will check that element against the others.

for n in ll:
    for m in ll:
        if m.endswith(n) and m != n:
            print(m, "ends with", n)