How to repeat list's elements with each different n times and in Python?

Question

The idea is to repeat list's elements with each different n times as below.

ls = [7, 3, 11, 5, 2, 3, 4, 4, 2, 3]
id_list_fname = ['S11', 'S15', 'S16', 'S17', 'S19', 'S3', 'S4', 'S5', 'S6', 'S9']
all_ls = []
for id, repeat in zip(id_list_fname, ls):
    res = [ele for ele in[id] for i in range(repeat)]
    all_ls.append(res)

Consequently, I would like the result to be a single flat list, which I achieved as below.

def flatten(lst):
    for item in lst:
        if isinstance(item, list):
            yield from flatten(item)
        else:
            yield item

final_output = list(flatten(all_ls))

Output of the final_output:

['S11', 'S11', 'S11', 'S11', 'S11', 'S11', 'S11', 'S15', 'S15', 'S15',
 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16',
 'S16', 'S17', 'S17', 'S17', 'S17', 'S17', 'S19', 'S19', 'S3', 'S3',
 'S3', 'S4', 'S4', 'S4', 'S4', 'S5', 'S5', 'S5', 'S5', 'S6', 'S6',
 'S9', 'S9', 'S9']

But I wonder there exist much more compact approach or technique such as itertools to have the repeat elements as what I achieved using the code snippet above.

Hi @lain, thanks for your interest in this OP. The example input and desired output already provided above. — mpx, Jan 31 '21 at 06:25
The problem is that `all_ls.append(res)` appends the sub-list of `res` values as a single element, rather than appending all those values separately. For that, `.extend` works, as explained in the first linked duplicate. Otherwise, the trick is to use a nested comprehension to encapsulate the work of the current list comprehension plus the `for id, repeat in zip(id_list_fname, ls):` loop, as in the second linked duplicate. Incidentally, the `ele for ele in[id]` part is useless; the shown list comprehension is equivalent to `[id for i in range(repeat)]`. — Karl Knechtel, Oct 04 '22 at 02:07

score 6 · Answer 1 · answered Jan 31 '21 at 06:29

You can use itertools.chain and itertools.repeat

from itertools import chain, repeat

ls = [7, 3, 11, 5, 2, 3, 4, 4, 2, 3]
id_list_fname = ['S11', 'S15', 'S16', 'S17', 'S19', 'S3', 'S4', 'S5', 'S6', 'S9']
res = list(chain.from_iterable(repeat(j, times = i) for i, j in zip(ls, id_list_fname)))

print(res)

Output

['S11', 'S11', 'S11', 'S11', 'S11', 'S11', 'S11', 'S15', 'S15', 'S15', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S17', 'S17', 'S17', 'S17', 'S17', 'S19', 'S19', 'S3', 'S3', 'S3', 'S4', 'S4', 'S4', 'S4', 'S5', 'S5', 'S5', 'S5', 'S6', 'S6', 'S9', 'S9', 'S9']

mhawke · Answer 2 · 2021-01-31T06:34:33.213

3

With respect to your code, use list.extend() rather than list.append() when adding items into all_ls. This adds the items in the list to the existing list instead of adding the list to the existing list.

ls = [7, 3, 11, 5, 2, 3, 4, 4, 2, 3]
id_list_fname = ['S11', 'S15', 'S16', 'S17', 'S19', 'S3', 'S4', 'S5', 'S6', 'S9']
all_ls=[]
for s,repeat in zip(id_list_fname ,ls):
    res =  [ele for ele in [s] for i in range(repeat)]
    all_ls.extend(res)

Output

>>> all_ls
['S11', 'S11', 'S11', 'S11', 'S11', 'S11', 'S11', 'S15', 'S15', 'S15', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S17', 'S17', 'S17', 'S17', 'S17', 'S19', 'S19', 'S3', 'S3', 'S3', 'S4', 'S4', 'S4', 'S4', 'S5', 'S5', 'S5', 'S5', 'S6', 'S6', 'S9', 'S9', 'S9']

A better way is with a list comprehension - it can be done in one line:

>>> [item for n,s in zip(ls, id_list_fname) for item in [s]*n]
['S11', 'S11', 'S11', 'S11', 'S11', 'S11', 'S11', 'S15', 'S15', 'S15', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S16', 'S17', 'S17', 'S17', 'S17', 'S17', 'S19', 'S19', 'S3', 'S3', 'S3', 'S4', 'S4', 'S4', 'S4', 'S5', 'S5', 'S5', 'S5', 'S6', 'S6', 'S9', 'S9', 'S9']

edited Jan 31 '21 at 06:34

answered Jan 31 '21 at 06:30

mhawke

84,695
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The dict comprehension can be replaced by `res = [id] * repeat`. `id` should not be used as a variable name in Python though, you are overwriting the built-in function `id` – Iain Shelvington Jan 31 '21 at 06:32
@IainShelvington: what dict comprehension? Fair enough with `id`, just copied from OP. Updated. – mhawke Jan 31 '21 at 06:37
1

@IainShelvington: also, minimal changes to OP's code helps OP to understand the difference. I have added a better list comprehension to perform same task in one line. – mhawke Jan 31 '21 at 06:38
Thanks @mhawke, extra knowledge obained from your post. – mpx Jan 31 '21 at 06:39
@mhawke I meant list comprehension, apologies – Iain Shelvington Jan 31 '21 at 06:40
@IainShelvington: oh, reasonable mistake. Thanks for your feedback. – mhawke Jan 31 '21 at 06:42

Mad Physicist · Accepted Answer · 2021-02-04T13:58:24.910

1

Numpy has the repeat function:

np.repeat(id_list_fname, ls)

If you absolutely need a list:

np.repeat(id_list_fname, ls).tolist()

If you don't want to use numpy, keep in mind that python allows nested comprehensions, which are essentially multiple for loops creating a single flat list. Your original loop can be written as a flat one-liner:

[id for id, repeat in zip(id_list_fname, ls) for _ in range(repeat)]

edited Feb 04 '21 at 13:58

answered Feb 04 '21 at 13:49

Mad Physicist

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How to repeat list's elements with each different n times and in Python?

3 Answers3