so say I do this
x = np.arange(0, 3)
which gives
array([0, 1, 2])
but what can I do like
x = np.arange(0, 3)*repeat(N=3)times
to get
array([0, 1, 2, 0, 1, 2, 0, 1, 2])
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3 Answers
I've seen several recent questions about resize. It isn't used often, but here's one case where it does just what you want:
In [66]: np.resize(np.arange(3),3*3)
Out[66]: array([0, 1, 2, 0, 1, 2, 0, 1, 2])
There are many other ways of doing this.
In [67]: np.tile(np.arange(3),3)
Out[67]: array([0, 1, 2, 0, 1, 2, 0, 1, 2])
In [68]: (np.arange(3)+np.zeros((3,1),int)).ravel()
Out[68]: array([0, 1, 2, 0, 1, 2, 0, 1, 2])
np.repeat doesn't repeat in the way we want
In [70]: np.repeat(np.arange(3),3)
Out[70]: array([0, 0, 0, 1, 1, 1, 2, 2, 2])
but even that can be reworked (this is a bit advanced):
In [73]: np.repeat(np.arange(3),3).reshape(3,3,order='F').ravel()
Out[73]: array([0, 1, 2, 0, 1, 2, 0, 1, 2])
3 Comments
Runner Bean
great answer but I accepted the other answer as you are way to good at programming in python and the other guy needs some answers accepted to catch up on your exceptional stats! But thanks for the helpful answer :-)
Alex-droid AD
I think that
resize is the exact answer to the original question
AER
@RunnerBean thanks for the score, but probs should've given it to hpaulj. That way other people get to know the best answer too!
EDIT: Refer to hpaulj's answer. It is frankly better.
The simplest way is to convert back into a list and use:
list(np.arange(0,3))*3
Which gives:
>> [0, 1, 2, 0, 1, 2, 0, 1, 2]
Or if you want it as a numpy array:
np.array(list(np.arange(0,3))*3)
Which gives:
>> array([0, 1, 2, 0, 1, 2, 0, 1, 2])
how about this one?
arr = np.arange(3)
res = np.hstack((arr, ) * 3)
Output
array([0, 1, 2, 0, 1, 2, 0, 1, 2])
Not much overhead I would say.