Extend and forward fill numpy array

Question

I'd like to duplicate each line of an array N times. Is there a quick way to do so?

Example (N=3):

# INPUT
a=np.arange(9).reshape(3,3)
array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])
# OUTPUT 
array([[0, 1, 2],
       [0, 1, 2],
       [0, 1, 2],
       [3, 4, 5],
       [3, 4, 5],
       [3, 4, 5],
       [6, 7, 8],
       [6, 7, 8],
       [6, 7, 8]])

"duplicate", not "multiply", perhaps?

Andrew Jaffe
– Andrew Jaffe

2013-09-02 11:40:21 +00:00
Commented Sep 2, 2013 at 11:40 — Andrew Jaffe
– Andrew Jaffe, Commented Sep 2, 2013 at 11:40
Indeed. I edited my phrasing.

bioslime
– bioslime

2013-09-02 11:41:27 +00:00
Commented Sep 2, 2013 at 11:41 — bioslime
– bioslime, Commented Sep 2, 2013 at 11:41

Daniel · Accepted Answer · 2013-09-02 13:33:08Z

3

This is a job for np.repeat:

np.repeat(a,3,axis=0)
array([[0, 1, 2],
       [0, 1, 2],
       [0, 1, 2],
       [3, 4, 5],
       [3, 4, 5],
       [3, 4, 5],
       [6, 7, 8],
       [6, 7, 8],
       [6, 7, 8]])

Note that this is much faster then the other method.

N=100
%timeit np.repeat(a,N,axis=0)
100000 loops, best of 3: 4.6 us per loop

%timeit rows, cols = a.shape;b=np.hstack(N*(a,));b.reshape(N*rows, cols)
1000 loops, best of 3: 257 us per loop

N=100000
%timeit np.repeat(a,N,axis=0)
100 loops, best of 3: 3.93 ms per loop

%timeit rows, cols = a.shape;b=np.hstack(N*(a,));b.reshape(N*rows, cols)
1 loops, best of 3: 245 ms per loop

Also np.tile is useful in similar situations.

answered Sep 2, 2013 at 13:33

Daniel

19.6k7 gold badges64 silver badges74 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Janne Karila · Accepted Answer · 2013-09-02 11:42:39Z

1

>>> N = 3
>>> rows, cols = a.shape
>>> b=np.hstack(N*(a,))
>>> b
array([[0, 1, 2, 0, 1, 2, 0, 1, 2],
       [3, 4, 5, 3, 4, 5, 3, 4, 5],
       [6, 7, 8, 6, 7, 8, 6, 7, 8]])
>>> b.reshape(N*rows, cols)
array([[0, 1, 2],
       [0, 1, 2],
       [0, 1, 2],
       [3, 4, 5],
       [3, 4, 5],
       [3, 4, 5],
       [6, 7, 8],
       [6, 7, 8],
       [6, 7, 8]])

edited Sep 2, 2013 at 11:42

answered Sep 2, 2013 at 11:33

Janne Karila

25.3k6 gold badges59 silver badges97 bronze badges

1 Comment

bioslime Over a year ago

Awesome, but what about big N. If i have N=1000 is it possible to generate such a tuple (a,a,a,...) automatically?

Collectives™ on Stack Overflow

Extend and forward fill numpy array

2 Answers 2

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related