Functional way in python to change this list into another

Question

I have a list [2,3,0,3] And want to generate a list

[1,0,1,2]

The reasoning is that 1 zero appears, no one appears, 1 two appears and 2 threes appear in the input list.

Is there a non for loopy procedural way to do this in python?

Use a [`Counter`](https://docs.python.org/3/library/collections.html#collections.Counter) — Patrick Haugh, Mar 03 '19 at 06:28
Thanks Nishant, the guy ubnutu shows Counter, so very close, but i updated my question to have a missing 1 — katiex7, Mar 03 '19 at 06:40
Your output list should be from `min - max` or `0-max`. Suppose the list is `[5,9,7]` what should be the output? — Sociopath, Mar 03 '19 at 06:40
That would be something like `[my_counter[i] for i in range(max(my_counter))]` — Patrick Haugh, Mar 03 '19 at 06:41

Austin · Accepted Answer · 2019-03-03T07:01:58.620

4

You can use collections.Counter which finds count of elements:

from collections import Counter

lst = [2,3,0,3]

c = Counter(lst)
print([c[x] for x in range(max(lst)+1)])
# [1, 0, 1, 2]

An alternate way avoiding loops:

from collections import Counter

lst = [2,3,0,3]

c = dict.fromkeys(range(max(lst)+1), 0)
c.update(Counter(lst))

print(c.values())
# # [1, 0, 1, 2]

edited Mar 03 '19 at 07:01

answered Mar 03 '19 at 06:35

Austin

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score 2 · Answer 2 · answered Mar 03 '19 at 06:33

2

You could probably use the following code:

lst = [2,3,0,1,3]
#Get list unique values using set
set_lst = sorted(set(lst))
#For each unique element, use the count method
counts = [lst.count(i) for i in set_lst]

At first, we find out all the unique elements of the list, by using a set object, which stores unique elements only. Then we traverse through the list and use the count method to get the counts of each element, which are sorted in order.

answered Mar 03 '19 at 06:33

Akash Purandare

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I think this is a more built-in method. Obviously, using a `Counter` or `itertools` could be faster – Akash Purandare Mar 03 '19 at 06:35
Have an upvote but more importantly look at Austins answer, more elegant – katiex7 Mar 03 '19 at 06:39

Mykola Zotko · Answer 3 · 2019-03-03T07:30:00.333

1

Another solution with count:

l = [2, 3, 0, 3]

assert len(l) == max(l) + 1

[l.count(num) for num, _ in enumerate(l)]
# [1, 0, 1, 2]

edited Mar 03 '19 at 07:30

answered Mar 03 '19 at 07:24

Mykola Zotko

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always fun to see other ideas, but because of the need for the assertion, i still favor Counter way, have an upvote – katiex7 Mar 03 '19 at 07:33

score 1 · Answer 4 · answered Mar 03 '19 at 08:25

If you want to avoid for keyword at any price, then you might use map instead, following way:

a = [2,3,0,3]
out = list(map(lambda x:a.count(x), list(range(max(a)+1))))
print(out) #[1, 0, 1, 2]

This solution assumes that a contains solely non-negative integers and is not empty list.

Functional way in python to change this list into another

4 Answers4