Re: A question about numpy
On 14-04-2015 23:49, Rob Gaddi wrote: On Tue, 14 Apr 2015 23:41:56 +0100, Paulo da Silva wrote: Supposing I have 2 vectors v1 and v2 and a value (constant) k. I want to build a vector r with all values of v1 greater than k and the others from v2. You're looking for numpy.where() . That's it! Thank you. -- https://mail.python.org/mailman/listinfo/python-list
A question about numpy
I am new to numpy ... Supposing I have 2 vectors v1 and v2 and a value (constant) k. I want to build a vector r with all values of v1 greater than k and the others from v2. I found 2 ways, but not sure if they are the best solution: 1. r1=v1.copy() r2=v2.copy() r1[r1k]=0 r2[r2=k]=0 r=r1+r2 2. r=v1*(v1=k)+v2*(v2k) Are there any better sols.? Thanks for help/comments. -- https://mail.python.org/mailman/listinfo/python-list
Re: A question about numpy
On Tue, 14 Apr 2015 23:41:56 +0100, Paulo da Silva wrote: Supposing I have 2 vectors v1 and v2 and a value (constant) k. I want to build a vector r with all values of v1 greater than k and the others from v2. You're looking for numpy.where() . -- Rob Gaddi, Highland Technology -- www.highlandtechnology.com Email address domain is currently out of order. See above to fix. -- https://mail.python.org/mailman/listinfo/python-list
Re: question about numpy, subclassing, and a DeprecationWarning
On 6/27/12 10:02 PM, Jason Swails wrote: Hello, I'm running into an unexpected issue in a program I'm writing, and I was hoping someone could provide some clarification for me. I'm trying to subclass numpy.ndarray (basically create a class to handle a 3D grid). When I instantiate a numpy.ndarray, everything works as expected. When I call numpy.ndarray's constructor directly within my subclass, I get a deprecation warning about object.__init__ not taking arguments. Presumably this means that ndarray's __init__ is somehow (for some reason?) calling object's __init__... This is some sample code: import numpy as np class derived(np.ndarray): ... def __init__(self, stuff): ... np.ndarray.__init__(self, stuff) ... l = derived((2,3)) __main__:3: DeprecationWarning: object.__init__() takes no parameters l derived([[ 8.87744455e+159, 6.42896975e-109, 5.56218818e+180], [ 1.79996515e+219, 2.41625066e+198, 5.15855295e+307]]) Am I doing something blatantly stupid? Is there a better way of going about this? I suppose I could create a normal class and just put the grid points in a ndarray as an attribute to the class, but I would rather subclass ndarray directly (not sure I have a good reason for it, though). Suggestions on what I should do? numpy.ndarray does not have its own __init__(), just a __new__(). It's __init__() is the same as object.__init__(), which takes no arguments. [~] |3 np.ndarray.__init__ is object.__init__ True There is no need to call np.ndarray.__init__() explicitly. http://docs.scipy.org/doc/numpy/user/basics.subclassing.html#a-brief-python-primer-on-new-and-init You will also want to ask numpy questions on the numpy mailing list. http://www.scipy.org/Mailing_Lists Personally, I recommend not subclassing ndarray at all. It rarely works out well. -- 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: a question about numpy
On 2009-09-09 20:46 PM, rechard wrote: Robert Kern wrote: On 2009-09-08 20:45 PM, hi_roger wrote: hello, i want to ask a question about numpy. i know how to select a submatrix using the slice object in numpy. But how can i select a submatrix A[i1,i2,i3;j1,j2,j3] (elements in A on line i1,i2,i3 and column j1,j2,j3 , and i1,i2,i3,j1,j2,j3 are all arbitrary numbers ) The submatrix must share data memory with original matrix. Any one help? thank you Sturla is almost correct. What you really want is this: i = [[i1], [i2], [i3]] j = [[j1, j2, j3]] B = A[i, j] http://docs.scipy.org/doc/numpy/reference/arrays.indexing.html#integer If you have more numpy questions, please ask on the numpy mailing list. http://www.scipy.org/Mailing_Lists wow..., thank you all :) but it seems it is impossible to make the submatrix share data with the original matrix in pure numpy code ? That is correct. -- 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: a question about numpy
On 2009-09-08 20:45 PM, hi_roger wrote: hello, i want to ask a question about numpy. i know how to select a submatrix using the slice object in numpy. But how can i select a submatrix A[i1,i2,i3;j1,j2,j3] (elements in A on line i1,i2,i3 and column j1,j2,j3 , and i1,i2,i3,j1,j2,j3 are all arbitrary numbers ) The submatrix must share data memory with original matrix. Any one help? thank you Sturla is almost correct. What you really want is this: i = [[i1], [i2], [i3]] j = [[j1, j2, j3]] B = A[i, j] http://docs.scipy.org/doc/numpy/reference/arrays.indexing.html#integer If you have more numpy questions, please ask on the numpy mailing list. http://www.scipy.org/Mailing_Lists -- 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: a question about numpy
On 2009-09-08 22:03 PM, sturlamolden wrote: On 9 Sep, 03:45, hi_rogerrechardc...@gmail.com wrote: i know how to select a submatrix using the slice object in numpy. But how can i select a submatrix A[i1,i2,i3;j1,j2,j3] (elements in A on line i1,i2,i3 and column j1,j2,j3 , and i1,i2,i3,j1,j2,j3 are all arbitrary numbers ) You just pass an array of ints for each dimension. If you want a 3x3 submatrix, you must pass in two 3x3 arrays of indices. Oops! I just responded to the OP saying that you were almost correct; but I was just thinking of your latter 1D example rather than this correct statement. My apologies. -- 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: a question about numpy
Robert Kern wrote: On 2009-09-08 20:45 PM, hi_roger wrote: hello, i want to ask a question about numpy. i know how to select a submatrix using the slice object in numpy. But how can i select a submatrix A[i1,i2,i3;j1,j2,j3] (elements in A on line i1,i2,i3 and column j1,j2,j3 , and i1,i2,i3,j1,j2,j3 are all arbitrary numbers ) The submatrix must share data memory with original matrix. Any one help? thank you Sturla is almost correct. What you really want is this: i = [[i1], [i2], [i3]] j = [[j1, j2, j3]] B = A[i, j] http://docs.scipy.org/doc/numpy/reference/arrays.indexing.html#integer If you have more numpy questions, please ask on the numpy mailing list. http://www.scipy.org/Mailing_Lists wow..., thank you all :) but it seems it is impossible to make the submatrix share data with the original matrix in pure numpy code ? -- http://mail.python.org/mailman/listinfo/python-list
a question about numpy
hello, i want to ask a question about numpy. i know how to select a submatrix using the slice object in numpy. But how can i select a submatrix A[i1,i2,i3;j1,j2,j3] (elements in A on line i1,i2,i3 and column j1,j2,j3 , and i1,i2,i3,j1,j2,j3 are all arbitrary numbers ) The submatrix must share data memory with original matrix. Any one help? thank you -- http://mail.python.org/mailman/listinfo/python-list
Re: a question about numpy
On 9 Sep, 03:45, hi_roger rechardc...@gmail.com wrote: hello, i want to ask a question about numpy. i know how to select a submatrix using the slice object in numpy. But how can i select a submatrix A[i1,i2,i3;j1,j2,j3] (elements in A on line i1,i2,i3 and column j1,j2,j3 , and i1,i2,i3,j1,j2,j3 are all arbitrary numbers ) The submatrix must share data memory with original matrix. So the only way to do this is to make an ndarray subclass that overloads __getitem__, __setitem__, and __iter__, and takes care of the mapping into A. Thus you get a double indirection. -- http://mail.python.org/mailman/listinfo/python-list
Re: a question about numpy
On 9 Sep, 03:45, hi_roger rechardc...@gmail.com wrote:
i know how to select a submatrix using the slice object in numpy. But
how can i select a submatrix
A[i1,i2,i3;j1,j2,j3] (elements in A on line i1,i2,i3 and column
j1,j2,j3 , and i1,i2,i3,j1,j2,j3 are all arbitrary numbers )
You just pass an array of ints for each dimension. If you want a 3x3
submatrix, you must pass in two 3x3 arrays of indices.
The submatrix must share data memory with original matrix.
That is the tricky part. An ndarray must be indexable using an array
of strides. That is in C:
void *get_element_ptr( PyArrayObject *a, int indices[])
{
char *out = a-data;
int d;
for (d=0; d a-nd; d++)
out += indices[d] * a-strides[d];
return (void *)out;
}
If you slice with an array or list of ints in Python, the resulting
array cannot be indexed like this. Therefore NumPy is forced to make a
copy. So if I do
import numpy as np
a = np.zeros((10,10))
b = a[[1,3,5],[6,7,8]]
I get this:
b.flags['OWNDATA']
True
But:
c = a[::2,::2]
c.flags['OWNDATA']
False
This is because C can still be indexed with strides as shown above,
and no copy is made.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Newbie question about numpy
On Thu, 24 Aug 2006 17:23:49 GMT, Dennis Lee Bieber [EMAIL PROTECTED] wrote: On Thu, 24 Aug 2006 16:38:45 +0100, Paul Johnston [EMAIL PROTECTED] declaimed the following in comp.lang.python: I know its a long time since my degree but that's not matrix multiplication is it ? Define matrix multiplication... What you see appears to be multiplication of corresponding elements. Were you expecting a dot product, or a cross product, or something else? That as explained in http://people.hofstra.edu/faculty/stefan_Waner/RealWorld/tutorialsf1/frames3_2.html As I say its been a long time :-) Thanks to everyone for the help. Paul -- http://mail.python.org/mailman/listinfo/python-list
Newbie question about numpy
Hi I'm new to python and have just been taking a look at what it has to offer. I noted the lack of matrices so installed numpy I know the documentation is chargable so wanted a quick play to see if I should buy it However _ from numpy import * a = array([[1,2,3],[4,5,6],[1,2,3]]) b = array([[1,3,6],[2,5,1],[1,1,1]]) print 'a = \n',a,\n print 'b = \n',b,\n print 'a has shape ', a.shape print 'b has shape ', b.shape, \n print a * b is \n, a * b _ Gives me _ a = [[1 2 3] [4 5 6] [1 2 3]] b = [[1 3 6] [2 5 1] [1 1 1]] a has shape (3, 3) b has shape (3, 3) a * b is [[ 1 6 18] [ 8 25 6] [ 1 2 3]] _ I know its a long time since my degree but that's not matrix multiplication is it ? TIA Paul -- http://mail.python.org/mailman/listinfo/python-list
Re: Newbie question about numpy
At Thursday 24/8/2006 12:38, Paul Johnston wrote: from numpy import * a = array([[1,2,3],[4,5,6],[1,2,3]]) b = array([[1,3,6],[2,5,1],[1,1,1]]) print a * b is \n, a * b I know its a long time since my degree but that's not matrix multiplication is it ? No, it's plain element-by-element multiplication. You want matrixmultiply: http://numpy.scipy.org/numpydoc/numpy-9.html#pgfId-36542 Gabriel Genellina Softlab SRL __ Preguntá. Respondé. Descubrí. Todo lo que querías saber, y lo que ni imaginabas, está en Yahoo! Respuestas (Beta). ¡Probalo ya! http://www.yahoo.com.ar/respuestas -- http://mail.python.org/mailman/listinfo/python-list
Re: Newbie question about numpy
Paul Johnston wrote: Hi I'm new to python and have just been taking a look at what it has to offer. I noted the lack of matrices so installed numpy You will want to ask numpy questions on the numpy list. http://www.scipy.org/Mailing_Lists numpy arrays are not matrices; they are arrays. All of the arithmetic operations on them are done element-wise. The dot() function will do matrix multiplication. There is a matrix class (with the constructor numpy.mat(some_array)) that derives from arrays and overrides the * operator to do matrix multiplication if that is what you want. I prefer using dot() on regular arrays, myself. -- 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Newbie question about numpy
Paul Johnston wrote: (snip) I noted the lack of matrices so installed numpy (snip) _ from numpy import * a = array([[1,2,3],[4,5,6],[1,2,3]]) b = array([[1,3,6],[2,5,1],[1,1,1]]) (snip) print a * b is \n, a * b _ (snip) a * b is [[ 1 6 18] [ 8 25 6] [ 1 2 3]] _ I know its a long time since my degree but that's not matrix multiplication is it ? You consider that a and b are matrices, but for the python interpreter they are arrays so a*b returns the multiplication of 2 arrays. For matrices multiplication, you could get a hint by typing the following in the interpreter : import numpy dir(numpy) help(numpy.matrixmultiply)#type q to exit which could make you want to try the following code : from numpy import * a = array([[1,2,3],[4,5,6],[1,2,3]]) b = array([[1,3,6],[2,5,1],[1,1,1]]) print matrixmultiply(a,b) ... output : ... array([[ 8, 16, 11], [20, 43, 35], [ 8, 16, 11]]) ... HIH, avell -- http://mail.python.org/mailman/listinfo/python-list
Re: Newbie question about numpy
Avell Diroll wrote: For matrices multiplication, you could get a hint by typing the following in the interpreter : import numpy dir(numpy) help(numpy.matrixmultiply)#type q to exit Note that the name matrixmultiply() has been deprecated in favor of dot() for many, many years now even in Numeric, numpy's predecessor. It has finally been removed in recent versions of numpy. -- 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 -- http://mail.python.org/mailman/listinfo/python-list
