How do I initialize a NumPy array with a different default for each column?
I am trying to initialize a NumPy matrix of size (x, y) where y is very large.
The first column of the matrix is the identifier (integer) and the rest of the triplets (int8), where each member of the triplet should have a different default value.
i.e. if default values [2,5,9]
I would like to initialize the following matrix:
0 2 5 9 2 5 9 2 5 9 ...
0 2 5 9 2 5 9 2 5 9 ...
0 2 5 9 2 5 9 2 5 9 ...
0 2 5 9 2 5 9 2 5 9 ...
...
The fastest way to initialize a matrix is
defaults = [2, 5, 9]
mat = numpy.zeros(shape=(x,y),
dtype=['i'] + ['int8'] * (y - 1))
# fill the triplets with default values
for i in range(1, y/3):
j = i * 3
mat[:, j] = defaults[0]
mat[:, j+1] = defaults[1]
mat[:, j+2] = defaults[2]
What is the fastest way to initialize such a matrix?
Thank!
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You can use np.tile
with changing the array of values, for example:
>>> b=np.array([2,5,9])
>>> b=b.reshape(3,1)
>>> np.tile(b,3)
array([[2, 2, 2],
[5, 5, 5],
[9, 9, 9]])
Then you can use np.dstack
to rotate the array and then use np.hstack
to add columns of zeros:
>>> np.hstack((np.zeros((3,1)),np.dstack(new)[0]))
array([[ 0., 2., 5., 9.],
[ 0., 2., 5., 9.],
[ 0., 2., 5., 9.]])
Or you can repeat the non-zero part again with tile
:
>>> np.hstack((np.zeros((3,1)),np.tile(np.dstack(new)[0],4)))
array([[ 0., 2., 5., 9., 2., 5., 9., 2., 5., 9., 2., 5., 9.],
[ 0., 2., 5., 9., 2., 5., 9., 2., 5., 9., 2., 5., 9.],
[ 0., 2., 5., 9., 2., 5., 9., 2., 5., 9., 2., 5., 9.]])
EDIT:
For clarification only, a simple gasket is the following:
defaults = [2, 5, 9] np.hstack((np.zeros((x,1)), np.tile(defaults, (x,y))))
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I would do it like this:
np.tile([0] + [2, 5, 9] * 4, (3, 1))
Here I used list addition and list multiplication to create the first row and then used np.tile
more than three rows to replicate. np.tile
automatically converts the list to an array before iterating vertically three times. You can wrap this in a function that looks something like this if you want,
def make_array(triple, n_triple, n_row):
return np.tile([0] + list(triple) * n_triple, (n_row, 1))
I forced it to triple
be a list here, but if you just pass the list to a variable triple
when calling this function, you won't need it.
Good luck.
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