Re: [Numpy-discussion] Numpy-scalars vs Numpy 0-d arrays: copy or not copy?
On 10/20/06, Travis Oliphant [EMAIL PROTECTED] wrote: How about this. To get the i,j,k,l element a[i:i+1,j:j+1,k:k+1,l:l+1].squeeze() -Travis I think all this can be condensed in a method call or similar mehcanism, natively provided by ndarray type. Or should this be seen as a special use case? -- Lisandro Dalcín --- Centro Internacional de Métodos Computacionales en Ingeniería (CIMEC) Instituto de Desarrollo Tecnológico para la Industria Química (INTEC) Consejo Nacional de Investigaciones Científicas y Técnicas (CONICET) PTLC - Güemes 3450, (3000) Santa Fe, Argentina Tel/Fax: +54-(0)342-451.1594 - Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with pre-integrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.as-us.falkag.net/sel?cmd=lnkkid=120709bid=263057dat=121642 ___ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/numpy-discussion
[Numpy-discussion] Numpy-scalars vs Numpy 0-d arrays: copy or not copy?
Hi! I am confused with Numpy behavior with its scalar or 0-d arrays objects: numpy.__version__ '1.0rc2' a = numpy.array((1,2,3)) b = a[:2] b += 1 b array([2, 3]) a array([2, 3, 3]) type(b) type 'numpy.ndarray' To this point all is ok for me: subarrays share (by default) memory with their parent array. But: c = a[2] c += 1 c 4 a array([2, 3, 3]) type(c) type 'numpy.int32' id(c) 169457808 c += 1 id(c) 169737448 That's really confusing, because slices (from __getslice__ method) are not copies (they share memory), and items (single elements from __getitem__ ) are copies to one of the scalar objects provided by Numpy. I can understand that numpy.scalars do not provide inplace operations (like Python standard scalars, they are immutable), so I'd like to use 0-d Numpy.ndarrays. But: d = numpy.array(a[2],copy=False) d += 1 d array(4) a array([2, 3, 3]) type(d) type 'numpy.ndarray' d.shape () id(d) 169621280 d += 1 id(d) 169621280 This is not a solution because d is a copy since construction time... My question is: is there a way to get a single element of an array into a 0-d array which shares memory with its parent array? Thx for your help, Sebastien - Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with pre-integrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.as-us.falkag.net/sel?cmd=lnkkid=120709bid=263057dat=121642 ___ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/numpy-discussion
Re: [Numpy-discussion] Numpy-scalars vs Numpy 0-d arrays: copy or not copy?
A Divendres 20 Octubre 2006 11:42, Sebastien Bardeau va escriure: [snip] I can understand that numpy.scalars do not provide inplace operations (like Python standard scalars, they are immutable), so I'd like to use 0-d Numpy.ndarrays. But: d = numpy.array(a[2],copy=False) d += 1 d array(4) a array([2, 3, 3]) type(d) type 'numpy.ndarray' d.shape () id(d) 169621280 d += 1 id(d) 169621280 This is not a solution because d is a copy since construction time... My question is: is there a way to get a single element of an array into a 0-d array which shares memory with its parent array? One possible solution (there can be more) is using ndarray: In [47]: a=numpy.array([1,2,3], dtype=i4) In [48]: n=1# the position that you want to share In [49]: b=numpy.ndarray(buffer=a[n:n+1], shape=(), dtype=i4) In [50]: a Out[50]: array([1, 2, 3]) In [51]: b Out[51]: array(2) In [52]: b += 1 In [53]: b Out[53]: array(3) In [54]: a Out[54]: array([1, 3, 3]) Cheers, -- 0,0 Francesc Altet http://www.carabos.com/ V V Cárabos Coop. V. Enjoy Data - - Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with pre-integrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.as-us.falkag.net/sel?cmd=lnkkid=120709bid=263057dat=121642 ___ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/numpy-discussion
Re: [Numpy-discussion] Numpy-scalars vs Numpy 0-d arrays: copy or not copy?
On Fri, Oct 20, 2006 at 11:42:26AM +0200, Sebastien Bardeau wrote: a = numpy.array((1,2,3)) b = a[:2] Here you index by a slice. c = a[2] Whereas here you index by a scalar. So you want to do b = a[[2]] b += 1 or in the general case b = a[slice(2,3)] b += 1 Regards Stéfan - Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with pre-integrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.as-us.falkag.net/sel?cmd=lnkkid=120709bid=263057dat=121642 ___ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/numpy-discussion
Re: [Numpy-discussion] Numpy-scalars vs Numpy 0-d arrays: copy or not copy?
Francesc Altet wrote: A Divendres 20 Octubre 2006 11:42, Sebastien Bardeau va escriure: [snip] I can understand that numpy.scalars do not provide inplace operations (like Python standard scalars, they are immutable), so I'd like to use 0-d Numpy.ndarrays. But: d = numpy.array(a[2],copy=False) d += 1 d array(4) a array([2, 3, 3]) type(d) type 'numpy.ndarray' d.shape () id(d) 169621280 d += 1 id(d) 169621280 This is not a solution because d is a copy since construction time... My question is: is there a way to get a single element of an array into a 0-d array which shares memory with its parent array? One possible solution (there can be more) is using ndarray: [SNIP] Here's a slightly more concise version of the same idea: b = a[n:n+1].reshape([]) -tim - Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with pre-integrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.as-us.falkag.net/sel?cmd=lnkkid=120709bid=263057dat=121642 ___ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/numpy-discussion
Re: [Numpy-discussion] Numpy-scalars vs Numpy 0-d arrays: copy or not copy?
Ooops sorry there was two mistakes with the 'hasslice' flag. This seems now to work for me. def __getitem__(self,index): # Index may be either an int or a tuple # Index length: if type(index) == int: # A single element through first dimension ilen = 1 index = (index,)# A tuple else: ilen = len(index) # Array rank: arank = len(self.shape) # Check if there is a slice: hasslice = False for i in index: if type(i) == slice: hasslice = True # Array is already a 0-d array: if arank == 0 and index == (0,): return self elif arank == 0: raise IndexError, 0-d array has only one element at index 0. # This will return a single element as a 0-d array: elif arank == ilen and not hasslice: # This ugly thing returns a numpy 0-D array AND NOT a numpy scalar! # (Numpy scalars do not share their data with the parent array) newindex = list(index) newindex[0] = slice(index[0],index[0]+1,None) newindex = tuple(newindex) return self[newindex].reshape(()) # This will return a n-D subarray (n=1): else: return self[index] Sebastien Bardeau wrote: One possible solution (there can be more) is using ndarray: In [47]: a=numpy.array([1,2,3], dtype=i4) In [48]: n=1# the position that you want to share In [49]: b=numpy.ndarray(buffer=a[n:n+1], shape=(), dtype=i4) Ok thanks. Actually that was also the solution I found. But this is much more complicated when arrays are N dimensional with N1, and above all if user asks for a slice in one or more dimension. Here is how I redefine the __getitem__ method for my arrays. Remember that the goal is to return a 0-d array rather than a numpy.scalar when I extract a single element out of a N-dimensional (N=1) array: def __getitem__(self,index): # Index may be either an int or a tuple # Index length: if type(index) == int: # A single element through first dimension ilen = 1 index = (index,)# A tuple else: ilen = len(index) # Array rank: arank = len(self.shape) # Check if there is a slice: for i in index: if type(i) == slice: hasslice = True else: hasslice = False # Array is already a 0-d array: if arank == 0 and index == (0,): return self[()] elif arank == 0: raise IndexError, 0-d array has only one element at index 0. # This will return a single element as a 0-d array: elif arank == ilen and hasslice: # This ugly thing returns a numpy 0-D array AND NOT a numpy scalar! # (Numpy scalars do not share their data with the parent array) newindex = list(index) newindex[0] = slice(index[0],index[0]+1,None) newindex = tuple(newindex) return self[newindex].reshape(()) # This will return a n-D subarray (n=1): else: return self[index] Well... I do not think this is very nice. Someone has another idea? My question in my first post was: is there a way to get a single element of an array into a 0-d array which shares memory with its parent array? Sebastien - Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with pre-integrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.as-us.falkag.net/sel?cmd=lnkkid=120709bid=263057dat=121642 ___ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/numpy-discussion -- - Sebastien Bardeau L3AB - CNRS UMR 5804 2 rue de l'observatoire BP 89 F - 33270 Floirac Tel: (+33) 5 57 77 61 46 - - Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with pre-integrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.as-us.falkag.net/sel?cmd=lnkkid=120709bid=263057dat=121642 ___ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/numpy-discussion
