Question

My problem is simple: I have a long list of elements that I want to iterate through and check every element against a condition. Depending on the outcome of the condition I would like to delete the current element of the list, and continue iterating over it as usual.

I have read a few other threads on this matter. Two solutions seam to be proposed. Either make a dictionary out of the list (which implies making a copy of all the data that is already filling all the RAM in my case). Either walk the list in reverse (which breaks the concept of the alogrithm I want to implement).

Is there any better or more elegant way than this to do it ?

def walk_list(list_of_g):
    g_index = 0
    while g_index < len(list_of_g):
        g_current = list_of_g[g_index]
        if subtle_condition(g_current):
            list_of_g.pop(g_index)
        else:
            g_index = g_index + 1

Answer 1

Here is an alternative answer for if you absolutely have to remove the items from the original list, and you do not have enough memory to make a copy – move the items down the list yourself:

def walk_list(list_of_g):
    to_idx = 0
    for g_current in list_of_g:
        if not subtle_condition(g_current):
            list_of_g[to_idx] = g_current
            to_idx += 1
    del list_of_g[to_idx:]

This will move each item (actually a pointer to each item) exactly once, so will be O(N). The del statement at the end of the function will remove any unwanted items at the end of the list, and I think Python is intelligent enough to resize the list without allocating memory for a new copy of the list.