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up vote 1 down vote favorite

Can you create a numpy array with all unique values in it?

myArray = numpy.random.random_integers(0,100,2500)
myArray.shape = (50,50)

So here I have a given random 50x50 numpy array, but I could have non-unique values. Is there a way to ensure every value is unique?

Thank you

Update:

I have created a basic function to generate a list and populate a unique integer. 

        dist_x = math.sqrt(math.pow((extent.XMax - extent.XMin), 2))
        dist_y = math.sqrt(math.pow((extent.YMax - extent.YMin),2))
        col_x = int(dist_x / 100)
        col_y = int(dist_y / 100)
        if col_x % 100 > 0:
            col_x += 1
        if col_y % 100 > 0:
            col_y += 1
        print col_x, col_y, 249*169
        count = 1
        a = []

        for y in xrange(1, col_y + 1):
            row = []
            for x in xrange(1, col_x + 1):
                row.append(count)
                count += 1
            a.append(row)
            del row

        numpyArray = numpy.array(a)

Is there a better way to do this?

Thanks

share|improve this question
2  
You seem to be asking for 2500 unique random integers between 0 and 100. It should be pretty obvious why that's not going to happen... – ali_m Aug 18 '14 at 17:44
    
I was asking if there was a built in method or a fast way to do this? – josh1234 Aug 18 '14 at 17:59
1  
Do you need random numbers, or just unique? If just unique, maybe np.random.permutation(np.arange(N))? – ojy Aug 18 '14 at 18:04
up vote 4 down vote accepted

The most convenient way to get a unique random sample from a set is probably np.random.choice with replace=False.

For example:

import numpy as np

# create a (5, 5) array containing unique integers drawn from [0, 100]
uarray = np.random.choice(np.arange(0, 101), replace=False, size=(5, 5))

# check that each item occurs only once
print(np.bincount(uarray.ravel()).sum() == uarray.size)
# True

If replace=False the set you're sampling from must, of course, be at least as big as the number of samples you're trying to draw:

np.random.choice(np.arange(0, 101), replace=False, size=(50, 50))
# ValueError: Cannot take a larger sample than population when 'replace=False'

If all you're looking for is a random permutation of the integers between 1 and the number of elements in your array, you could also use np.random.permutation like this:

nrow, ncol = 5, 5
uarray = (np.random.permutation(nrow * ncol) + 1).reshape(nrow, ncol)
share|improve this answer
    
So if I know the number of columns/rows I should be able to do this? numpyArray = numpy.random.choice(numpy.arange(1, (col_x*col_y)+1), replace=0, size=(col_x, col_y)). This should lead to me getting the correct size with allowed unique values in each column/row location in the matrix. – josh1234 Aug 18 '14 at 18:18
    
Well, if you just want a random permutation of the elements in the (inclusive) range [1, col_x * col_y], you might as well just use (np.random.permutation(col_x * col_y) + 1).reshape(col_x, col_y), as I think @oja was hinting. – ali_m Aug 18 '14 at 18:22
    
Thank you for your help! Both answers work well. – josh1234 Aug 19 '14 at 10:07

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