read about basic slicing : http://docs.scipy.org/doc/numpy/reference/arrays.indexing.html On Sun, Nov 21, 2010 at 11:28 AM, John Salvatier <jsalvati@u.washington.edu> wrote:

yes use the symbol ':'

so you want

t[:,x,y]

2010/11/21 Ernest Adrogué <eadrogue@gmx.net>:

...

Hi,

Suppose an array of shape (N,2,2), that is N arrays of shape (2,2). I want to select an element (x,y) from each one of the subarrays, so I get a 1-dimensional array of length N. For instance:

In [228]: t=np.arange(8).reshape(2,2,2)

In [229]: t Out[229]: array([[[0, 1],        [2, 3]],

      [[4, 5],        [6, 7]]])

In [230]: x=[0,1]

In [231]: y=[1,1]

In [232]: t[[0,1],x,y] Out[232]: array([1, 7])

This way, I get the elements (0,1) and (1,1) which is what I wanted. The question is: is it possible to omit the [0,1] in the index?

Thanks in advance.

-- Ernest _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

On 11/21/10 11:37 AM, Ernest Adrogué wrote:

...

so you want

t[:,x,y]

I tried that, but it's not the same:

In [307]: t[[0,1],x,y] Out[307]: array([1, 7])

In [308]: t[:,x,y] Out[308]: array([[1, 3], [5, 7]])

what is your t? Here's my example, which I think matches what you asked for: In [1]: import numpy as np In [2]: a = np.arange(12) In [3]: a.shape = (3,2,2) In [4]: a Out[4]: array([[[ 0, 1], [ 2, 3]], [[ 4, 5], [ 6, 7]], [[ 8, 9], [10, 11]]]) In [5]: a[:,1,0] Out[5]: array([ 2, 6, 10]) -- Christopher Barker, Ph.D. Oceanographer Emergency Response Division NOAA/NOS/OR&R (206) 526-6959 voice 7600 Sand Point Way NE (206) 526-6329 fax Seattle, WA 98115 (206) 526-6317 main reception Chris.Barker@noaa.gov

22/11/10 @ 11:08 (-0800), thus spake Christopher Barker:

On 11/21/10 11:37 AM, Ernest Adrogué wrote:

...

...

so you want

t[:,x,y]

I tried that, but it's not the same:

In [307]: t[[0,1],x,y] Out[307]: array([1, 7])

In [308]: t[:,x,y] Out[308]: array([[1, 3], [5, 7]])

what is your t? Here's my example, which I think matches what you asked for:

In [1]: import numpy as np

In [2]: a = np.arange(12)

In [3]: a.shape = (3,2,2)

In [4]: a Out[4]: array([[[ 0, 1], [ 2, 3]],

[[ 4, 5], [ 6, 7]],

[[ 8, 9], [10, 11]]])

In [5]: a[:,1,0] Out[5]: array([ 2, 6, 10])

This works with scalar indices, but not with arrays. The problem is that I don't want always the same element from each subarray, but an arbitrary element, say the (1,0) from the first, the (0,0) from the second, and so on, so I have to use arrays. -- Ernest

22/11/10 @ 11:20 (-0800), thus spake John Salvatier:

I didn't realize the x's and y's were varying the first time around. There's probably a way to omit it, but I think the conceptually simplest way is probably what you had to begin with. Build an index by saying i = numpy.arange(0, t.shape[0])

then you can do t[i, x,y]

Exactly. I was just wondering if I can speed this up by omitting building the "arange array". This is inside a function that gets called a lot, so I suppose it would make a difference if I can get rid of it. -- Ernest

I think that the only speedup you will get is defining an index only once and reusing it. 2010/11/22 Ernest Adrogué <eadrogue@gmx.net>:

22/11/10 @ 14:04 (-0600), thus spake Robert Kern:

...

...

This way, I get the elements (0,1) and (1,1) which is what I wanted. The question is: is it possible to omit the [0,1] in the index?

No, but you can write generic code for it:

  t[np.arange(t.shape[0]), x, y]

Thank you. This is what I wanted to know.

-- Ernest _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion