Repeat NumPy array without replicating data?

You can't do this; a NumPy array must have a consistent stride along each dimension, while your strides would need to go one way most of the time but sometimes jump backwards.

The closest you can get is either a 1000-row 2D array where every row is a view of your first array, or a flatiter object, which behaves kind of like a 1D array. (flatiters support iteration and indexing, but you can't take views of them; all indexing makes a copy.)

Setup:

import numpy as npa = np.arange(10)

2D view:

b = np.lib.stride_tricks.as_strided(a, (1000, a.size), (0, a.itemsize))

flatiter object:

c = b.flat

broadcast_to was added in numpy 1.10, which allows you effectively repeat an array with a little less effort.

Copying the style of the accepted answer:

import numpy as nparr = np.arange(10)repeated = np.broadcast_to(arr, (1000, arr.size))

I'm not 100% sure what you mean by 'not replicating the data 1000 times'. If you are looking for a numpy method to build b from a in one fell swoop (rather than looping), you can use:

a = np.arange(1000)b = np.tile(a,1000)

Otherwise, I would do something like:

a = np.arange(1000)ii = [700,2000,10000] # The indices you want of the tiled arrayb = a[np.mod(ii,a.size)]

b is not a view of a in this case because of the fancy indexing (it makes a copy), but at least it returns a numpy array and doesn't create the 1000*1000x1 array in memory and just contains the elements you want.

As far as them being immutable (see Immutable numpy array?), you would need to switch the flag for each separately since copies don't retain the flag setting.