Create a list of tuples with adjacent list elements if a condition is true

Cleaner Pythonic approach:

>>> [(x,y) for x,y in zip(myList, myList[1:]) if y == 9][(8, 9), (4, 9), (7, 9)]

What is the code above doing:

• zip(some_list, some_list[1:]) would generate a list of pairs of adjacent elements.
• Now with that tuple, filter on the condition that the second element is equal to 9. You're done :)

Part of your issue is that myList[i:i] will always return an empty list. The end of a slice is exclusive, so when you do a_list[0:0] you're trying to take the elements of a_list that exist between index 0 and index 0.

You're on the right track, but you want to zip the list with itself.

[(x, y) for x, y in zip(myList, myList[1:]) if y==9]

You were pretty close, I'll show you an alternative way that might be more intuitive if you're just starting out:

sets = [(myList[i-1], myList[i]) for i in range(len(myList)) if myList[i] == 9]

Get the index in the range of the list lenght, and if the value at the position i is equal to 9, grab the adjacent elements.

The result is:

sets[(8, 9), (4, 9), (7, 9)]

This is less efficient than the other approaches but I decided to un-delete it to show you a different way of doing it. You can make it go a bit faster by using enumerate() instead:

sets = [(myList[i-1], j) for i, j in enumerate(myList) if j == 9]

Take note that in the edge case where myList[0] = 9 the behavior of the comprehension without zip and the behavior of the comprehension with zip is different.

Specifically, if myList = [9, 1, 8, 9, 2, 4, 9, 6, 7, 9, 8] then:

[(myList[i-1], myList[i]) for i in range(len(myList)) if myList[i] == 9]# results in: [(8, 9), (8, 9), (4, 9), (7, 9)]

while:

[(x, y) for x, y in zip(myList, myList[1:]) if y==9]# results in: [(8, 9), (4, 9), (7, 9)]

It is up to you to decide which of these fits your criteria, I'm just pointing out that they don't behave the same in all cases.