And I forget one thing: computing the 1-D index from the row and column is also straightforward.
# --- import rpy2.robjects as ro row_i = 1 col_i = 2 idx = (row_i-1) + m.dim[0] * (col_i - 1) m = ro.r.matrix(ro.IntVector(range(9)), nrow=3, ncol=3) m[idx] = 33 print(m) # --- L. Laurent Gautier wrote: > The code snippet I gave was incorrect: > > # --- > import rpy2.robjects as ro > import rpy2.rlike.container as rlc > > m = ro.r.matrix(range(9), nrow=3, ncol=3) > print(m) > > idx = rlc.TaggedList([ro.IntVector([1, ]), > ro.IntVector([2, ])]) > m2 = m.assign(idx, 33) > print(m2) > > idx = rlc.TaggedList([ro.IntVector([1, 2, 3]), > ro.IntVector([1, 2, 3])]) > m3 = m.assign(idx, 33) > print(m3) > > idx = rlc.TaggedList([ro.r.cbind(ro.IntVector([1, 2, 3]), > ro.IntVector([1, 2, 3])), > ]) > m4 = m.assign(idx, 33) > print(m4) > # --- > > That complication is due to the fact that the R function "[<-" is called > behind the hood. > It has the advantage that specific signatures to "[<-" will be handled > gracefully, the disadvantage is the verbosity. > The other potential issue is the copying (I have ideas for that). > > > You could also consider going the numpy route: > > import numpy > > # using the 'm' from the previous code snippet > nm = numpy.asarray(m) > nm.shape = (3,3) > nm.strides = nm.strides[::-1] > > print(nm) > > # zero-offset (while R has one-offset) > nm[1-1][2-1] = 33 > > print(nm) > > > > > > L. > > > > > > Sancar Adali wrote: >> If I modify the code in the following way >> m = ro.r['matrix'](range(4), nrow=2, ncol=2) >> >> idx = ro.IntVector([1,4]) >> m2 = m.assign(idx, 33) >> >> print(m2) >> >> I get the following matrix >> >> [,1] [,2] >> [1,] 33 2 >> [2,] 1 33 >> >> I think the indices are like vector indices here, a single index, In >> this case 1 and 4 is assigned to 33 which are [1,1] and [2,2] >> respectively . Is there a way to access the matrix elements using pair >> of indices? >> >> On Sat, Feb 21, 2009 at 2:46 PM, Laurent Gautier <lgaut...@gmail.com> >> wrote: >>> Sancar Adali wrote: >>>> I'm trying to access and update a particular element of a matrix to >>>> update >>>> it >>>> >>>> for example in R, >>>> >>>> mat=matrix(0,nrow=2,ncol=2) >>>> mat[0,0]= mat[0,0]+1 >>> In R, vector indexing starts at one; zero are silently ignored. >>> >>>> How do I do this in rpy2 >>>> do I have to use the low-level interface or can I use high-level >>>> interface >>>> >>> Using the high-level interface is possible. >>> >>> One way is: >>> >>> import rpy2.robjects as ro >>> >>> m = ro.r['matrix'](range(4), nrow=2, ncol=2) >>> >>> idx = ro.IntVector([1,1]) >>> m2 = m.assign(idx, 33) >>> >>> print(m2) >>> >>> >>> Not as elegant as one could wish... the 2.1 series for rpy2 will >>> hopefully >>> have something nicer. >>> >>> >>> L. >>> >>> PS: using the rpy2-numpy compatibility layer is an another way. >>> >> >> >> > ------------------------------------------------------------------------------ Open Source Business Conference (OSBC), March 24-25, 2009, San Francisco, CA -OSBC tackles the biggest issue in open source: Open Sourcing the Enterprise -Strategies to boost innovation and cut costs with open source participation -Receive a $600 discount off the registration fee with the source code: SFAD http://p.sf.net/sfu/XcvMzF8H _______________________________________________ rpy-list mailing list rpy-list@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/rpy-list
