How to compare items in list?

Question

This is my list

my_list = [18, 19, 20, 27, 28, 29, 38, 39, 40]

My goal is to sort or create new lists like these

list1 = [18, 19, 20]
list2 = [27, 28, 29]
list3 = [38, 39, 40]

And then compare each of them and only leave items which are equal like that ,

list1 x == list2 x - 10 or list1 x == list3 x - 10
list1 x == list2 x + 10 or list1 x == list3 x + 10
list2 x == list1 x - 10 or list2 x == list3 x - 10
list2 x == list1 x + 10 or list2 x == list2 x + 10
list3 x == list1 x - 10 or list3 x == list2 x - 10
list3 x == list1 x + 10 or list3 x == list2 x + 10

So the final list will be like

final = [18,19,28,29,38,39]

I have wrote those codes but can’t achieve the goal. Those were just unsuccessful attempts.

N1

for lst in my_list:
    ind = my_list.index(lst)
    if int(my_list[ind]) == any(int(x)-10 for x in my_list) or int(my_list[ind]) == any(int(x)+10 for x in my_list):
        print(ind)

N2

for id in my_list:
    ind = my_list.index(id)
    try:
        if id == (my_list[ind+1])-1 and id :
            print(id)
    except IndexError:
            print("out of", id)

Thank you in advance.

Asked By: Bary Simmons

||

Source

Answer 1

It sounds like you want to take a list as input and generate a new list that only contains elements of the original that are 10 away from another element in the original.

my_list = [18, 19, 20, 27, 28, 29, 38, 39, 40]
new_list = list(filter(lambda x: x + 10 in my_list or x - 10 in my_list, my_list))

Here is a simple way to do just that using python’s filter method.

Results:

[18, 19, 28, 29, 38, 39]

Answered By: h0r53

How to compare items in list?

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