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

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]