Creating a numpy matrix without replacement in rows

Question

I can easily use np.random.choice with replace=False to generate a 1D array without replacement. However, I need to generate a matrix such that each row is without replacement, i.e.[[1, 2, 3], [2, 3, 1], [3, 1, 2]]. np.random.randint is the obvious choice, but there's no provision for non-repetition in this function. I'd also rather not iteratively append new rows to a matrix, as I imagine this is rather inefficient. So, what is the most efficient way to generate my matrix?

Answer 1

You can first create your whole array and then use np.random.shuffle for shuffling your rows.

n = 3
data = np.arange(n*n).reshape(n, n) % n + 1
for row in data:
    np.random.shuffle(row)

Here, the modulo operator is used for producing numbers from 0 to (n-1) in each row.

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As an alternative,

data = np.empty(shape=(3, 3))

reserves the memory for an empty array that can afterwards be filled more efficiently:

for i in range(len(data)):
    data[i,:] = np.random.choice(3, 3, replace=False) + 1

Answer 2

You could use list comprehension:

data = np.array([np.random.choice([1,2,3], 3, replace=False) for _ in range(3)])