How to set new values to list, with same numbers occurring multiple times?-CodePudding

I'm trying to set new values based on an existing list with values rather randomly set. Some values appear multiple times and the new index value should reflect that. However, I seem unable to count a value that appears multiple times as 1, to then add to a counter (startval). What would be a good way of doing this?

startval = 901
old_index = [100, 145, 145, 740, 740, 740, 276, 277, 278]
# new_index = [901, 902, 902, 903, 903, 903, 904, 905, 906]
new_index = []

for x in old_index:
    new_index.append(startval)
    if old_index.count(x) != 1:
        for y in range(old_index.count(x)):
            startval *=1 #but add 1 once 
    else:
        startval  =1

CodePudding user response：

You can use the old defaultdict size trick to get contiguous integer keys corresponding to a set of (repeating) values:

from collections import defaultdict

old_index = [100, 145, 145, 740, 740, 740, 276, 277, 278]

offset = 901
index = defaultdict(lambda: len(index)   offset)

new_index = [index[v] for v in old_index]
# [901, 902, 902, 903, 903, 903, 904, 905, 906]

If you only want the same index for neighbouring duplicates, you can go with itertools.groupby and enumerate:

from itertools import groupby

new_index = [i for i, (_, g) in enumerate(groupby(old_index), offset) for _ in g]
# [901, 902, 902, 903, 903, 903, 904, 905, 906]

Some documentation: