How about
pdl> $a=pdl(1,2,3)
pdl> p pdl($a(*3),$a(:,*3))->flat
[1 1 1 2 2 2 3 3 3 1 2 3 1 2 3 1 2 3]
pdl>
though you might prefer
pdl> p append($a(*3)->flat,$a(:,*3)->flat)
[1 1 1 2 2 2 3 3 3 1 2 3 1 2 3 1 2 3]
to avoid problems when the shapes of the intermediate arrays are
incompatible. The *3 inserts a dummy dimension of size 3 in the first
place, producing three identical columns, while :,*3 produces a dummy
dimension of size 3 as the second index, so you produce tree identical
columns,
pdl> p $a(*3)
[
[1 1 1]
[2 2 2]
[3 3 3]
]
pdl> p $a(:,*3)
[
[1 2 3]
[1 2 3]
[1 2 3]
]
->flat makes them 1D and append puts one after the other.
Regards,
Luis
On Sat, Jul 13, 2019 at 02:54:34PM -0400, Robert Ryley wrote:
I've been working on translating some NumPY examples into PDL. There
has been some positive reinforcement on Perl Monks as can be seen
here:
https://www.perlmonks.org/?node_id=1233413
I am stuck on the following NumPy example:
Q. Create the following pattern without hardcoding. Use only numpy
functions and the below input array a.
Input:
a = np.array([1,2,3])`
Desired Output:
array([1, 1, 1, 2, 2, 2, 3, 3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3])
Solution
np.r_[np.repeat(a, 3), np.tile(a, 3)]
#> array([1, 1, 1, 2, 2, 2, 3, 3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3])
Does PDL have something equivalent to NumPy's "tile" function? I've
looked through the docs and do not see anything obvious to solve the
problem.
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