Generating a list of random lists

Question

I'm new to Python, so I might be doing basic errors, so apologies first.

Here is the kind of result I'm trying to obtain :

foo = [
    ["B","C","E","A","D"],
    ["E","B","A","C","D"],
    ["D","B","A","E","C"],
    ["C","D","E","B","A"]
]

So basically, a list of lists of randomly permutated letters without repeat.

Here is the look of what I can get so far :

foo = ['BDCEA', 'BDCEA', 'BDCEA', 'BDCEA']

The main problem being that everytime is the same permutation. This is my code so far :

import random
import numpy as np
letters = ["A", "B", "C", "D", "E"]
nblines = 4
foo = np.repeat(''.join(random.sample(letters, len(letters))), nblines)

Help appreciated. Thanks

I state the result I'm looking for, give the state of my research and code to reproduce... Why vote for closing (at least without explaining) ? -_- — François M.
– François M., Commented Sep 15, 2016 at 13:11
Is there any particular reason that you want to do this with Numpy? BTW, the sample foo at the top of your question is a normal Python list of lists, not a Numpy array. — PM 2Ring
– PM 2Ring, Commented Sep 15, 2016 at 13:14
@fmalaussena Are repetitions allowed? Could the same string present in the final result more than once? — gowrath
– gowrath, Commented Sep 15, 2016 at 13:23
No particular reason, it just so happen that I found this while searching for answers, and thought it could solve my problem. I thought I could transform the Numpy array in a list later. — François M.
– François M., Commented Sep 15, 2016 at 13:24
It's probably a Good Idea to get familiar with core Python (and the major standard libraries that are supplied with it) before attempting to learn Numpy. — PM 2Ring
– PM 2Ring, Commented Sep 15, 2016 at 13:42

Albert P · Accepted Answer · 2016-09-15 13:32:28Z

1

The problem with your code is that the line

foo = np.repeat(''.join(random.sample(letters, len(letters))), nblines)

will first create a random permutation, and then repeat that same permutation nblines times. Numpy.repeat does not repeatedly invoke a function, it repeats elements of an already existing array, which you created with random.sample.

Another thing is that numpy is designed to work with numbers, not strings. Here is a short code snippet (without using numpy) to obtain your desired result:

[random.sample(letters,len(letters)) for i in range(nblines)]

Result: similar to this:

foo = [
    ["B","C","E","A","D"],
    ["E","B","A","C","D"],
    ["D","B","A","E","C"],
    ["C","D","E","B","A"]
]

I hope this helped ;)

PS: I see that others gave similar answers to this while I was writing it.

answered Sep 15, 2016 at 13:32

Albert P

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score 1 · Accepted Answer · 2016-09-15 13:25:37Z

1

np.repeat repeats the same array. Your approach would work if you changed it to:

[''.join(random.sample(letters, len(letters))) for _ in range(nblines)]
Out: ['EBCAD', 'BCEAD', 'EBDCA', 'DBACE']

This is a short way of writing this:

foo = []
for _ in range(nblines):
    foo.append(''.join(random.sample(letters, len(letters))))

foo
Out: ['DBACE', 'CBAED', 'ACDEB', 'ADBCE']

edited Sep 15, 2016 at 13:25

answered Sep 15, 2016 at 13:12

user2285236

3 Comments

François M. Over a year ago

Thanks. "this can generate the same sequence" > if you mean that "BCDAE" could appear twice, this is a behaviour I want to be possible. Moreover, do you have a link where I could understand the syntax of this for loop with _, please ? I'm coming from R and I've never seen that.

user2285236 Over a year ago

Yes I meant BCDAE could appear twice. The one with _ is a list comprehension. It is a compact way of writing loops. I used _ as a variable because even though we are iterating over range(nblines) (with 0, 1, 2, 3) we don't need those numbers. We just want to do the same thing 4 times. _ is generally used to indicate that that enumerator is redundant.

François M. Over a year ago

Alright, thank you very much. I actually tried the for loop inside the list, but by putting it before the expression and not after, so I wasn't that far.

PM 2Ring · Accepted Answer · 2016-09-15 13:39:03Z

Here's a plain Python solution using a "traditional" style for loop.

from random import shuffle

nblines = 4
letters = list("ABCDE")
foo = []
for _ in range(nblines):
    shuffle(letters)
    foo.append(letters[:])

print(foo)

typical output

[['E', 'C', 'D', 'A', 'B'], ['A', 'B', 'D', 'C', 'E'], ['A', 'C', 'B', 'E', 'D'], ['C', 'A', 'E', 'B', 'D']]

The random.shuffle function shuffles the list in-place. We append a copy of the list to foo using letters[:], otherwise foo would just end up containing 4 references to the one list object.

Here's a slightly more advanced version, using a generator function to handle the shuffling. Each time we call next(sh) it shuffles the lst list stored in the generator and returns a copy of it. So we can call next(sh) in a list comprehension to build the list, which is a little neater than using a traditional for loop. Also, list comprehesions can be slightly faster than using .append in a traditional for loop.

from random import shuffle

def shuffler(seq):
    lst = list(seq)
    while True:
        shuffle(lst)
        yield lst[:]

sh = shuffler('ABCDE')
foo = [next(sh) for _ in range(10)]

for row in foo:
    print(row)

typical output

['C', 'B', 'A', 'E', 'D']
['C', 'A', 'E', 'B', 'D']
['D', 'B', 'C', 'A', 'E']
['E', 'D', 'A', 'B', 'C']
['B', 'A', 'E', 'C', 'D']
['B', 'D', 'C', 'E', 'A']
['A', 'B', 'C', 'E', 'D']
['D', 'C', 'A', 'B', 'E']
['D', 'C', 'B', 'E', 'A']
['E', 'D', 'A', 'C', 'B']

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