CODEWITHSUNDEEP
pythonnumpy
Rikudo Pain
Rikudo Pain

Reputation: 461

Add Condition on generated numpy

I have piece of code :

V = numpy.floor(3*np.random.rand(5,5))
print V

It create random result of array in 5x5 table, how to add condition "1" only generate x times, "2" only generate y times, else are "0". Thanks

Upvotes: 0

Views: 60

Answers (2)

Lunaweaver
Lunaweaver

Reputation: 206

How about the following?

import numpy as np
shape = (5, 5)
area = shape[0] * shape[1]
np.random.permutation([1]*x + [2]*y + [0]*(area-x-y)).reshape(shape)

Seems pretty simple. You take a random permutation of [1, ... 1, 2, ... 2, 0, ... 0] and then you just turn it into a square. I'm not too sure but it also seems less computationally expensive and is also extensible to n numbers or dimensions quite easily.

Upvotes: 1

Jaroslaw Matlak
Jaroslaw Matlak

Reputation: 574

Try this:

import numpy as np

def newArray( x, y, n):
    if x + y > n ** 2:
        print "Values error!"
        return
    res = [[0 for col in range(n)] for row in range(n)]
    # Create roulette
    roulette = range(0, n ** 2)

    printkxtimes(res, roulette, 1, x, n)
    printkxtimes(res, roulette, 2, y, n)
    print res

# This function draws random element from roulette, 
# gets the position in array and sets value of this position to k. 
# Then removes this element from roulette to prevent drawing it again
def printkxtimes(array, roulette, k, x, n):
    for i in xrange(0, x):
        r = int(np.floor(roulette.__len__()*np.random.rand(1))[0])
        array[roulette[r] / n][roulette[r] % n] = k
        roulette.pop(r)

newArray(10,2,5)

A little explanation of roulette :

Every element of table res can be represented equivalently with a number from range(0, n^2) :

z = row*n + column <=> row = int(z/n) , column= z%n

We can then represent the list of positions in table res as the table [0,1,...,n^2-1]

Upvotes: 1

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