Python, comparison sublists and making a list

Question

I have a list that contains a lot of sublists. i.e.

mylst = [[1, 343, 407, 433, 27], 
         [1, 344, 413, 744, 302], 
         [1, 344, 500, 600, 100], 
              


        

Answer 1

This seems to have worked for me. I'm not sure if its more Pythonic in any way though and you'll be looping through the list multiple times so there's some things you can definitely do to optimize it more.

def seg(mylist):
    # converted list to set in case there are any duplicates
    segments = set()

    for entry_index in range(len(mylist)):
        for c in range(len(mylist)):
            first = mylist[entry_index]
            comparison = mylist[c]

            # ignore comparing the same items
            if entry_index == c:
               continue

            # ignore cases where the first item does not match
            if first[0] != comparison[0]:
                continue

            # ignore cases where the second item differs by more than 2
            if abs(first[1] - comparison[1]) > 2:
                continue

            # add cases where the third and fourth items differ by less than 10
            if abs(first[2] - comparison[2]) < 10 or abs(first[3] - comparison[3]) < 10:
                segments.add(entry_index)

            elif abs(first[2] - comparison[3]) < 10 or abs(first[3] - comparison[2]) < 10:
                segments.add(entry_index)

    return segments

Answer 2

You will have to catch the duplicate indexes but this should be a lot more efficient:

gr = []
it = iter(mylst)
prev = next(it)

for ind, ele in enumerate(it):
    if ele[0] == prev[0] and abs(ele[1] - prev[1]) <= 2:
        if any(abs(ele[i] - prev[i]) < 10 for i in (2, 3)):
            gr.extend((ind, ind+1))
    prev = ele

Based on your logic 6 and 7 should not appear as they don't meet the criteria:

     [2, 346, 953, 995, 43], 
     [3, 346, 967, 1084, 118], 

Also for 10 to appear it should be <= 2 not < 2 as per your description.

You could use an OrderedDict to remove the dupes and keep the order:

from collections import OrderedDict

print(OrderedDict.fromkeys(gr).keys())
[0, 1, 3, 4, 10, 11, 12]