Change Data Structure of an array (new ID and replace Duplicates in loop)

November 15, 2022

I’m new in the "python game", so maybe the following question is for some of you very easy to answer:

I have an array like this:

store=([9,4,5],[9,4,1],[1,2,3],[9,4,1],[3,7,5],[2,4,1])

I want to "loop" the array and creat a new structure:

store_new=([0,1,2],[0,1,3],[3,4,5],[0,1,3],[5,6,2],[4,1,3])

The first value starts with 0, the second is 1 and so on. If there is a duplicate it should take the same new value as before. For example for "9"(old) to "0"(new)….and for the next "9"(old) again to "0" (new). Is there a nice way to handle this problem?

I tried something with “enumerate” in a “for loop”…but this don’t really work.

>Solution :

You can use itertools.count and dict.setdefault in a list comprehension:

store=([9,4,5],[9,4,1],[1,2,3],[9,4,1],[3,7,5],[2,4,1])

from itertools import count

d = {}
c = count()
out = tuple([d[x] if x in d else d.setdefault(x, next(c)) for x in l]
            for l in store)

Output:

([0, 1, 2], [0, 1, 3], [3, 4, 5], [0, 1, 3], [5, 6, 2], [4, 1, 3])

With a classical python loop:

out = []
d = {}
max_val = 0
for l in store:
    out.append([])
    for x in l:
        if x in d:
            out[-1].append(d[x])
        else:
            d[x] = max_val
            out[-1].append(max_val)
            max_val += 1

out = tuple(out)