List of increasing lists

Question

Given integers n >= m, I want to build the list of all increasing lists of length m with elements the set {1,...,n}.

For instance, if n= 4 and m=2, I want [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]].

Is there a nice way to do that with very few lines of Python code?

I would like to build an efficient function

def listOfLists(n, m):
    ????
    return ? # the list of all increasing lists of m elements in {1, 2, ..., n}

Thanks!

Remark: I have basically done this, but my code is ridiculously long https://sagecell.sagemath.org/?z=eJxVkcFuwyAQRO98xR5Bdqo66ikq-YJIvfRm5eDIuF2ZbBB21H5-FpYWh4PFmHmzHjy6Ccj9rp_34J2eWyBzUMArvQQLc384Zx1YeEd6NlkiywA76LL6-Ubv2ItnsJbPGiA-C5Ik9p0t3hScjI19ghHeJbBC4mw6Dq1UgST0PyO69R4pu5QauZPHZf2YTvxcNLUQSiuceMgRqA4pZC8tZx7Vv8p-ukUegQRxoC-nu5qSnS9DCI5GjXIhl2HBJdEVjk_wBel2xcHL56Qim7RM_yWmOzd1UGm_-UPbyplUKkSkVW9bv7WwN-YBIpR8dQ==&lang=python&interacts=eJyLjgUAARUAuQ==

"Is there a nice way to do that with very few lines of Python code?" Have you tried anything at all? — juanpa.arrivillaga, May 30 '20 at 19:57
@juanpa.arrivillaga Yes I have... It works but it is painfully long, which is why I am asking. — Julien, May 30 '20 at 19:58
@juanpa.arrivillaga Here is my code https://sagecell.sagemath.org/?z=eJxVkcFuhCAQhu88xSS9QNSmNj2Z2ido0svezB4wYDtBR4Juto_fAeyqnBj4v59_BmMHIPu7Xm5-tNKVQKoRwCseQguua66p9lyMlqRTqUQuPVRQp-r-g6NlLV6hbfmuAOI7n52yvGo3bTSOwqI9wQjv2XCHsrKo2XSnNiibPjyCXW-BkkoIwz31esHla_jUU2-0pBK8agCeYJgD-GqN3S6gyYDBydKCM8O5sYFjfADtMTbvLs_BcZjuJe-jGQISBE3fVtbplQfnnrX3lozEPLIUKdI7HE5wjzRPqMccN7Z6cEv0v2P8FbU_tM3n8IfHoSRSCB-QVnmey1sJr0r9Ab9liHA=&lang=python&interacts=eJyLjgUAARUAuQ== — Julien, May 30 '20 at 20:02

score 2 · Accepted Answer · answered May 30 '20 at 20:09

2

Use combinations from itertools

In [29]: from itertools import combinations

In [30]: def get_list(n, m):
    ...:     return [i for i in combinations(range(1, n+1), m)]

In [31]: get_list(4, 2)
Out[31]: [(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]

answered May 30 '20 at 20:09

Osman Mamun

2,864
1
16
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Why to use List Comprehension? Why not just return the itertools.combinations() function? – João Victor May 30 '20 at 20:11
2

that will return a generator, either you can use `list()` like you did or list comp like I did – Osman Mamun May 30 '20 at 20:12

João Victor · Answer 2 · 2020-05-30T20:13:33.110

1

You can use itertools:

from itertools import combinations

m = 2
n = 4

list(combinations(list(range(1, n+1)), m))

EDITED: If you want a function, just return it:

from itertools import combinations

m = 2
n = 4

def listOfLists(n, m):
    return list(combinations(list(range(1, n+1)), m))

listOfLists(n, m)

edited May 30 '20 at 20:13

answered May 30 '20 at 20:07

João Victor

407
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score 0 · Answer 3 · answered May 30 '20 at 20:04

0

I think this might work :

n=4
m=2
result=[]
for i in range(1,n):
    for j in range(i+1,n+1):
        result.append([i,j])

answered May 30 '20 at 20:04

fireball.1

1,413
2
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This I can do, but thanks. In the end, I want a function that returns the list of lists for any given (n, m). – Julien May 30 '20 at 20:06

List of increasing lists

3 Answers3