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.
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3 Answers
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]
2 Comments
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.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)
1 Comment
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]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())
