Using Numpy Array to Create Unique Array

Question

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

• ali_m: 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...
• code base 5000: I was asking if there was a built in method or a fast way to do this?
• ojy: Do you need random numbers, or just unique? If just unique, maybe np.random.permutation(np.arange(N))?

Answer 1

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()) == 1).all())
# 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)

• code base 5000: 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.
• ali_m: 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.

Answer 2

Just use replace = False.

import numpy as np
def generate():
 global x
 powerball = np.random.randint(1,27)
 numbers = np.random.choice(np.arange(1, 70), replace=False, size=(1, 5))
 x = numbers, powerball
 return x
generate()