Re: simpler way to get a particular banded matrix?
- To: mathgroup at smc.vnet.net
- Subject: [mg98107] Re: simpler way to get a particular banded matrix?
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Mon, 30 Mar 2009 04:44:26 -0500 (EST)
On 3/29/09 at 2:48 AM, bitbucket at comcast.net (rip pelletier) wrote: >The following command constructs a matrix (typically denoted M0, it >arises in filter banks in general, in Burrus, Gopinath, & Guo "intro >to wavelets and wavelet transforms: a primer" in particular). >I'm sure there's a way to do this without typing out all 5 bands. >Here's what i did: >SparseArray[Band[{1,1}]->H,{6,6}]+ >SparseArray[Band[{2,3}]->H,{6,6}]+ >SparseArray[Band[{3,5}]->H,{6,6}]+ >SparseArray[Band[{-2,-3},{1,1},{-1,-1}]->rH,{6,6}]+ >SparseArray[Band[{-3,-5},{1,1},{-1,-1}]->rH,{6,6}] >where >H={h0,h1,h2,h3,h4,h5}; rH=Reverse[H]; >I get what I want: >h0 0 0 0 0 0 >h2 h1 h0 0 0 0 >h4 h3 h2 h1 h0 0 >0 h5 h4 h3 h2 h1 >0 0 0 h5 h4 h3 >0 0 0 0 0 h5 >Any easier ways to get this? You can include a list of bands rather than summing separate arrays with a single band specified. That is: In[17]:= s = SparseArray[Band[{1, 1}] -> H, {6, 6}] + SparseArray[Band[{2, 3}] -> H, {6, 6}] + SparseArray[Band[{3, 5}] -> H, {6, 6}] + SparseArray[Band[{-2, -3}, {1, 1}, {-1, -1}] -> rH, {6, 6}] + SparseArray[Band[{-3, -5}, {1, 1}, {-1, -1}] -> rH, {6, 6}]; In[18]:= s == SparseArray[{Band[{1, 1}] -> H, Band[{2, 3}] -> H[[;; 4]], Band[{3, 5}] -> H[[;; 2]], Band[{2, 1}] -> H[[3 ;;]], Band[{3, 1}] -> H[[5 ;;]]}] Out[18]= True Note, I've also omitted specifying the array dimensions since this is fixed by the main diagonal and I have specified the off main diagonal bands to use just the part of H needed which eliminates the warning messages your code generates. However, I still specified each of the bands. So, this may not satisfy your criteria for easier.