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.
>>
>
>
>
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