Counter occurrences within the same item in python

Question

Sorry, I'm pretty sure that somebody else may have asked this question already but I did not find it. I'd like to keep track the number of times I've seen this specific item like

Input :

[88,88,27,0,88]

Desired Output :

[1,2,1,1,3]

I'm looking for something especially good in terms of performance. I'm ok with Numpy or Pandas solutions.

Check [this](https://stackoverflow.com/questions/2600191/how-to-count-the-occurrences-of-a-list-item). — Kyrylo, Mar 01 '18 at 17:20
The first item of the desired output means that it is the first time I've seen the "8" — hans glick, Mar 01 '18 at 17:22

score 3 · Answer 1 · answered Mar 01 '18 at 17:24

3

Here is a simple way using a list comprehension:

x = [8,1,2,3,1,3,3,1,2,99]
y = [x[:i].count(el) + 1 for i, el in enumerate(x)]
print(y)

Output:

[1, 1, 1, 1, 2, 2, 3, 3, 2, 1]

answered Mar 01 '18 at 17:24

user3483203

50,081
9
65
94

This can be solved in a single pass though, no slices, no repeated counting. – G_M Mar 01 '18 at 17:26
1

Thks for your answer but it seems very costly to do it – hans glick Mar 01 '18 at 17:31
I would agree with you, really just depends on the input size you are dealing with. – user3483203 Mar 01 '18 at 17:32

score 2 · Accepted Answer · answered Mar 01 '18 at 17:25

2

lst = [8,1,2,3,1,3,3,1,2,99]

cnt = {}
res = []

for x in lst:
    cnt[x] = cnt.get(x,0)+1
    res += [cnt[x]]

print(res)

Output

answered Mar 01 '18 at 17:25

Amaro Vita

437
3
9

Thks for your answer. Do you know any implementation with Numpy? – hans glick Mar 01 '18 at 17:33

G_M · Answer 3 · 2018-03-01T17:32:20.010

>>> from collections import defaultdict
... 
... 
... def solution(lst):
...     result = []
...     seen = defaultdict(int)
...     for num in lst:
...         seen[num] += 1
...         result.append(seen[num])
...     return result
... 
>>> solution([88, 88, 27, 0, 88])
[1, 2, 1, 1, 3]
>>> solution([8, 1, 2, 3, 1, 3, 3, 1, 2, 99])
[1, 1, 1, 1, 2, 2, 3, 3, 2, 1]

Without imports:

>>> def solution(lst):
...     result = []
...     seen = {}
...     for num in lst:
...         try:
...             seen[num] += 1
...         except KeyError:
...             seen[num] = 1
...         result.append(seen[num])
...     return result
... 
>>> solution([88, 88, 27, 0, 88])
[1, 2, 1, 1, 3]
>>> solution([8, 1, 2, 3, 1, 3, 3, 1, 2, 99])
[1, 1, 1, 1, 2, 2, 3, 3, 2, 1]

Thks for your answer But I would prefer not importing this collection library — hans glick, Mar 01 '18 at 17:31

score 1 · Answer 4 · answered Mar 01 '18 at 17:40

1

one with generators:

def return_count(l):
    cnt = {}
    for x in l:
        cnt[x] = cnt.get(x, 0) + 1
        yield cnt[x]

print(list(return_count([8, 1, 2, 3, 1, 3, 3, 1, 2, 99])))

answered Mar 01 '18 at 17:40

Totoro

867
9
10

Counter occurrences within the same item in python

4 Answers4