Numpy: Array of `arange`s

Question

Is there a way to take...

>>> x = np.array([0, 8, 10, 15, 50]).reshape((-1, 1)); ncols = 5

...and turn it into...

array([[ 0,  1,  2,  3,  4],
       [ 8,  9, 10, 11, 12],
       [10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19],
       [50, 51, 52, 53, 54]])

I was able to do it with np.apply_along_axis...

>>> def myFunc(a, ncols):
        return np.arange(a, (a+ncols))

>>> np.apply_along_axis(myFunc, axis=1, arr=x)

and with for loops...

>>> X = np.zeros((x.size,ncols))
>>> for a,b in izip(xrange(x.size),x):
        X[a] = myFunc(b, ncols)

but they are too slow. Is there a faster way?

Thanks in advance.

To be able to answer, we really need a more detailed explanation of what exactly you're trying to do. Or perhaps a complete code example that works but is too slow. — NPE
– NPE, Commented Jan 25, 2013 at 7:38
@JanneKarila: Yes, generic is what i'm going for. This is just a simple example. In real life, the values in x and the number of columns could be anything. But given values in x and ncol, return array of aranges of shape (x.size,ncol). — Noob Saibot
– Noob Saibot, Commented Jan 25, 2013 at 7:38

NPE · Accepted Answer · 2013-01-25 09:21:44Z

6

The following will do it:

In [9]: x = np.array([0, 8, 10, 15, 50]).reshape((-1, 1))

In [10]: ncols = 5

In [11]: x + np.arange(ncols)
Out[11]: 
array([[ 0,  1,  2,  3,  4],
       [ 8,  9, 10, 11, 12],
       [10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19],
       [50, 51, 52, 53, 54]])

It adds a row vector to a column vector and relies on broadcasting to do the rest.

This should be as fast as anything: producing a 1000x1000 matrix takes ~1.6ms:

In [17]: %timeit np.arange(1000).reshape((-1, 1)) + np.arange(1000)
1000 loops, best of 3: 1.61 ms per loop

edited Jan 25, 2013 at 9:21

answered Jan 25, 2013 at 7:44

NPE

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1 Comment

Noob Saibot Over a year ago

Geez...Thanks, @NPE. One of these days, i'll get the hang of this.

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