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
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.Tagged
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,] 133
I think the indices are like vector indices here, a single index, In
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-
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
How do I do this in rpy2
do I have to use the low-level interface or can I use high-level interface
--
Sancar Adali
Johns Hopkins University
Graduate
