从列表列表中删除第一个元素，然后是第一个和第二个元素[保留]

Question

My aim is to remove the 1st element of a list and then the first and second element of a list etc. So a list entered like this: [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]] will return a list like this: [[1,2],[1,3],[2,3] I have tried many things but none work well

Answer 1

How about something like this:

list1 = [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]]
list2 = []

keep_index = 1
increment = 0
for i in range(len(list1)):
    if i == keep_index:
        list2.append(list1[i])
        increment += 2 if keep_index == 2 else 1
        keep_index += increment
print(list2)

And as @schwobaseggl commented, this code could be what you are looking for:

import itertools
itertools.combinations([1, 2, 3], 2) # combinations of 2 elements (without repeats)

Answer 2

import itertools

def f( l ): 
    def elements_to_remove():
        for i in itertools.count( 1 ): 
            yield from itertools.repeat( i, i )

    def _g():
        es = elements_to_remove()

        skip = None
        e = next( es )

        while True:
            skip = (yield not skip)[0] == e
            e    = next( es ) if skip else e

    g = _g()
    next( g )
    return list( filter( g.send, l ) )

inputs = [] 
inputs.append( [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]] )
inputs.append( [[0,0],[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]] )
inputs.append( [[2,0],[1,0]] )

for input in inputs:
    print( f"{input} -> {f(input)}" )

I think f is a function you are looking for based on your description. Here is the output of the above script:

[[1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3], [3, 1], [3, 2], [3, 3]] -> [[1, 2], [1, 3], [2, 3]]
[[0, 0], [1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3], [3, 1], [3, 2], [3, 3]] -> [[0, 0], [1, 2], [1, 3], [2, 3]]
[[2, 0], [1, 0]] -> [[2, 0]]

. Note that the input is NOT modified in place; you'd need to adapt the code a little if you want that.