Creating a numpy array with repetitive pattern

Question

I am trying to create an array of size 6*n, such that for every batch of 6 cells in the array I will have the following integer values:

a = [n-2, n-1,n,n,n+1,n+1,n+2,n+3]

The banal way that I can think of is using this routine:

a = []
for i in xrange(n):
    np.append(a,[n-2, n-1,n,n,n+1,n+1,n+2,n+3])

But is there a smarter-faster way of doing it?

Your example is still broken, which makes it harder to understand exactly what you want. If you want a single list you could use a.extend([n-2, n-1,n,n,n+1,n+1,n+2,n+3]), or if you want nested lists you could use a.append([n-2, n-1,n,n,n+1,n+1,n+2,n+3]) — ali_m
– ali_m, Commented Aug 31, 2015 at 17:34
Are you sure about 6? The length of your pattern is 8. Your code gives an array with length 8*n. And do you mean to have i in your loop or not? — askewchan
– askewchan, Commented Aug 31, 2015 at 18:04

Ashwini Chaudhary · Accepted Answer · 2015-08-31 17:23:07Z

5

You can use numpy.tile:

>>> n = 6
>>> arr = np.array([n-2, n-1, n, n, n+1, n+1, n+2, n+3])
>>> np.tile(arr, n)
array([4, 5, 6, 6, 7, 7, 8, 9, 4, 5, 6, 6, 7, 7, 8, 9, 4, 5, 6, 6, 7, 7, 8,
       9, 4, 5, 6, 6, 7, 7, 8, 9, 4, 5, 6, 6, 7, 7, 8, 9, 4, 5, 6, 6, 7, 7,
       8, 9])
# Reshape to get the desired output
>>> np.tile(arr, n).reshape(n, arr.size)
array([[4, 5, 6, 6, 7, 7, 8, 9],
       [4, 5, 6, 6, 7, 7, 8, 9],
       [4, 5, 6, 6, 7, 7, 8, 9],
       [4, 5, 6, 6, 7, 7, 8, 9],
       [4, 5, 6, 6, 7, 7, 8, 9],
       [4, 5, 6, 6, 7, 7, 8, 9]])

answered Aug 31, 2015 at 17:23

Ashwini Chaudhary

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1 Comment

askewchan Over a year ago

I don't think the reshape is needed. OP's example gives a 1d (8*n,) shaped array.

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