Array of index values for unique elements in list

Question

I have the following list:

x = np.array([1, 1, 2, 2, 2])

with np.unique values of [1, 2]

How do I generate the following list:

[1, 2, 1, 2, 3]

i.e. a running index from 1 for each of the unique elements in the list x.

Answer 1

you can use pandas.cumcount() after grouping by the value itself, it does exactly that:

Number each item in each group from 0 to the length of that group - 1.

try this:

import numpy as np
import pandas as pd
x = np.array([1, 1, 2, 2, 2])
places = list(pd.Series(x).groupby(by=x).cumcount().values + 1)
print(places)

Output:

[1, 2, 1, 2, 3]

• ajrlewis: Thanks. Is there a way to do this with out pandas?
• Adam.Er8: @alex_lewis I tried with native numpy function but couldn't get any nice solution. the other way I think of is using list.index in a loop, which will be 100x times slower :( maybe someon else will come up with something numpy only.

Answer 2

Just use return_counts=True of np.unique with listcomp and np.hstack. It is still faster pandas solution

c = np.unique(x, return_counts=True)[1]
np.hstack([np.arange(item)+1 for item in c])
Out[869]: array([1, 2, 1, 2, 3], dtype=int64)

• Adam.Er8: this will only work if the values are consecutive. for x = np.array([1, 1, 2, 2, 2, 1]) you'll get [1 2 3 1 2 3] while you should be getting [1 2 1 2 3 3]

Answer 3

I'm not sure, if this is any faster or slower solution, but if you need just a list result with no pandas, you could try this

arr = np.array([1, 1, 2, 2, 2])
from collections import Counter
ranges = [range(1,v+1) for k,v in Counter(arr).items()]
result = []
for l in ranges:
 result.extend(list(l))
print(result)

[1, 2, 1, 2, 3]

(or make your own counter with dict instead of Counter())