I am trying to make a numpy array that looks like this:
[a b c ]
[ a b c ]
[ a b c ]
[ a b c ]
So this involves updat
import numpy as np
def using_tile_and_stride():
arr = np.tile(np.array([10,20,30,0,0,0], dtype='float'), (4,1))
row_stride, col_stride = arr.strides
arr.strides = row_stride-col_stride, col_stride
return arr
In [108]: using_tile_and_stride()
Out[108]:
array([[ 10., 20., 30., 0., 0., 0.],
[ 0., 10., 20., 30., 0., 0.],
[ 0., 0., 10., 20., 30., 0.],
[ 0., 0., 0., 10., 20., 30.]])
Other, slower alternatives include:
import numpy as np
import numpy.lib.stride_tricks as stride
def using_put():
arr = np.zeros((4,6), dtype='float')
a, b, c = 10, 20, 30
nrows, ncols = arr.shape
ind = (np.arange(3) + np.arange(0,(ncols+1)*nrows,ncols+1)[:,np.newaxis]).ravel()
arr.put(ind, [a, b, c])
return arr
def using_strides():
return np.flipud(stride.as_strided(
np.array([0, 0, 0, 10, 20, 30, 0, 0, 0], dtype='float'),
shape=(4, 6), strides = (8, 8)))
If you use using_tile_and_stride, note that the array is only appropriate for read-only purposes. Otherwise, if you were to try to modify the array, you might be surprised when multiple array locations change simultaneously:
In [32]: arr = using_tile_and_stride()
In [33]: arr[0, -1] = 100
In [34]: arr
Out[34]:
array([[ 10., 20., 30., 0., 100.],
[ 100., 10., 20., 30., 0.],
[ 0., 0., 10., 20., 30.],
[ 30., 0., 0., 10., 20.]])
You could work around this by returning np.ascontiguousarray(arr) instead of just arr, but then using_tile_and_stride would be slower than using_put. So if you intend to modify the array, using_put would be a better choice.