Re: [Numpy-discussion] Tuple outer product?
Robert Kern wrote: On Fri, Sep 25, 2009 at 17:38, Mads Ipsen m...@comxnet.dk wrote: Yes, but it should also work for [2.1,3.2,4.5] combined with [4.6,-2.3,5.6] - forgot to tell that. In [5]: np.transpose(np.meshgrid([2.1,3.2,4.5], [4.6,-2.3,5.6])) Out[5]: array([[[ 2.1, 4.6], [ 2.1, -2.3], [ 2.1, 5.6]], [[ 3.2, 4.6], [ 3.2, -2.3], [ 3.2, 5.6]], [[ 4.5, 4.6], [ 4.5, -2.3], [ 4.5, 5.6]]]) Point taken :-) -- ++ | Mads Ipsen, Scientific developer | +--+-+ | QuantumWise A/S | phone: +45-29716388 | | Nørresøgade 27A | www:www.quantumwise.com | | DK-1370 Copenhagen, Denmark | email: m...@quantumwise.com | +--+-+ ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] unpacking bytes directly in numpy
On Sat, Sep 26, 2009 at 10:33 PM, Thomas Robitaille
thomas.robitai...@gmail.com wrote:
Hi,
To convert some bytes to e.g. a 32-bit int, I can do
bytes = f.read(4)
i = struct.unpack('i', bytes)[0]
and the convert it to np.int32 with
i = np.int32(i)
However, is there a more direct way of directly transforming bytes
into a np.int32 type without the intermediate 'struct.unpack' step?
Assuming you have an array of bytes, you could just use view:
# x is an array of bytes, whose length is a multiple of 4
x.view(np.int32)
cheers,
David
___
NumPy-Discussion mailing list
NumPy-Discussion@scipy.org
http://mail.scipy.org/mailman/listinfo/numpy-discussion
[Numpy-discussion] unpacking bytes directly in numpy
Hi,
To convert some bytes to e.g. a 32-bit int, I can do
bytes = f.read(4)
i = struct.unpack('i', bytes)[0]
and the convert it to np.int32 with
i = np.int32(i)
However, is there a more direct way of directly transforming bytes
into a np.int32 type without the intermediate 'struct.unpack' step?
Thanks for any help,
Tom
___
NumPy-Discussion mailing list
NumPy-Discussion@scipy.org
http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] np.any and np.all short-circuiting
Hello David, thank you. I followed your suggestion but I was unable to make it work. I surprisingly found that with numpy in a different folder, it worked. I am afraid it is due to the fact that the first one is not a linux filesystem and cannot deal with permission and ownership. This would make sense and agree with a recent post about nose having problems with certain permissions or ownerships. Problem solved. I checked out the svn version into a linux filesystem and it works. Thanks again to all those that offered help. Best, Luca ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
[Numpy-discussion] python reduce vs numpy reduce for outer product
I'm encountering behavior that I think makes sense, but I'm not sure if there's some numpy function I'm unaware of that might speed up this operation. I have a (potentially very long) sequence of vectors, but for examples' sake, I'll stick with three: [A,B,C] with lengths na,nb, and nc. To get the result I want, I first reshape them to (na,1,1) , (1,nb,1) and (1,1,nc) and do: reduce(np.multiply,[A,B,C]) and the result is what I want... The curious thing is that np.prod.reduce([A,B,C]) throws ValueError: setting an array element with a sequence. Presumably this is because np.prod.reduce is trying to operate elemnt-wise without broadcasting. But is there a way to make the ufunc broadcast faster than doing the python-level reduce? (I tried np.prod(broadcast_arrays([A,B,C]),axis=0), but that seemed slower, presumably because it needs to allocate the full array for all three instead of just once). Or, if there's a better way to just start with the first 3 1d vectorsand jump straight to the broadcast product (basically, an outer product over arbitrary number of dimensions...)? ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] python reduce vs numpy reduce for outer product
On Sat, Sep 26, 2009 at 17:17, Erik Tollerud erik.tolle...@gmail.com wrote: I'm encountering behavior that I think makes sense, but I'm not sure if there's some numpy function I'm unaware of that might speed up this operation. I have a (potentially very long) sequence of vectors, but for examples' sake, I'll stick with three: [A,B,C] with lengths na,nb, and nc. To get the result I want, I first reshape them to (na,1,1) , (1,nb,1) and (1,1,nc) and do: reduce(np.multiply,[A,B,C]) and the result is what I want... The curious thing is that np.prod.reduce([A,B,C]) I'm sure you mean np.multiply.reduce(). throws ValueError: setting an array element with a sequence. Presumably this is because np.prod.reduce is trying to operate elemnt-wise without broadcasting. No. np.multiply.reduce() is trying to coerce its argument into an array. You have given it a list with three arrays that do not have the compatible shapes. In [1]: a = arange(5).reshape([5,1,1]) In [2]: b = arange(6).reshape([1,6,1]) In [4]: c = arange(7).reshape([1,1,7]) In [5]: array([a,b,c]) --- ValueErrorTraceback (most recent call last) /Users/rkern/Downloads/ipython console in module() ValueError: setting an array element with a sequence. But is there a way to make the ufunc broadcast faster than doing the python-level reduce? (I tried np.prod(broadcast_arrays([A,B,C]),axis=0), but that seemed slower, presumably because it needs to allocate the full array for all three instead of just once). Basically yes because it is computing np.array(np.broadcast_arrays([A,B,C])). Or, if there's a better way to just start with the first 3 1d vectorsand jump straight to the broadcast product (basically, an outer product over arbitrary number of dimensions...)? Well, numpy doesn't support arbitrary numbers of dimensions, nor will your memory. You won't be able to do more than a handful of dimensions practically. Exactly what are you trying to do? Specifics, please, not toy examples. -- Robert Kern I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth. -- Umberto Eco ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] python reduce vs numpy reduce for outer product
I'm sure you mean np.multiply.reduce(). Yes, sorry - typo. Or, if there's a better way to just start with the first 3 1d vectorsand jump straight to the broadcast product (basically, an outer product over arbitrary number of dimensions...)? Well, numpy doesn't support arbitrary numbers of dimensions, nor will your memory. You won't be able to do more than a handful of dimensions practically. Exactly what are you trying to do? Specifics, please, not toy examples. Well, I'm not sure how to get too much more specific than what I just described. I am computing moments of n-d input arrays given a particular axis ... I want to take a sequence of 1D arrays, and get an output has as many dimensions as the input sequence's length, with each dimension's size matching the corresponding vector. Symbolically, A[i,j,k,...] = v0[i]*v1[j]*v2[k]*... A is then multiplied by the input n-d array (same shape as A), and that is the output. And yes, practically, this will only work until I run out of memory, but the reduce method works for the n=1,2, 3, and 4 cases, and potentially in the future it will be needed for with higher (up to maybe 8) dimensions that are small enough that they won't overwhelm the memory. So it seems like a bad idea to write custom versions for each potential dimensionality. ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] python reduce vs numpy reduce for outer product
On Sat, Sep 26, 2009 at 18:17, Erik Tollerud erik.tolle...@gmail.com wrote: I'm sure you mean np.multiply.reduce(). Yes, sorry - typo. Or, if there's a better way to just start with the first 3 1d vectorsand jump straight to the broadcast product (basically, an outer product over arbitrary number of dimensions...)? Well, numpy doesn't support arbitrary numbers of dimensions, nor will your memory. You won't be able to do more than a handful of dimensions practically. Exactly what are you trying to do? Specifics, please, not toy examples. Well, I'm not sure how to get too much more specific than what I just described. I am computing moments of n-d input arrays given a particular axis ... I want to take a sequence of 1D arrays, and get an output has as many dimensions as the input sequence's length, with each dimension's size matching the corresponding vector. Symbolically, A[i,j,k,...] = v0[i]*v1[j]*v2[k]*... A is then multiplied by the input n-d array (same shape as A), and that is the output. And yes, practically, this will only work until I run out of memory, but the reduce method works for the n=1,2, 3, and 4 cases, and potentially in the future it will be needed for with higher (up to maybe 8) dimensions that are small enough that they won't overwhelm the memory. So it seems like a bad idea to write custom versions for each potential dimensionality. Okay, that's the key fact I needed. When you said that you would have a long list of vectors, I was worried that you wanted a dimension for each of them. You probably aren't going to be able to beat reduce(np.multiply, ...). -- Robert Kern I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth. -- Umberto Eco ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
