生成随机列表的列表

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

Answer 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.

Answer 2

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']    

Answer 3

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']