Hello,
I have 2 functions (a and b)
a = function(n) { matrix (runif(n*2,0.0,1), n) }>
>
> b = function (m, matrix) {
> n=nrow (matrix)
> p=ceiling (n/m)
> lapply (1:p, function (l,n,m) {
> inf = ((l-1)*m)+1
> if (l<p) sup=((l-1)*m)+m
> else sup=n
> data.frame (matrix (lapply (inf: sup, function(i)
> c(i, Inf, matrix[i,1], matrix[i,2]) ), nrow=m ) )
> }, n=n, m=m
> )
> }
>
> >my.matrix = a(7)
> >my.matrix
> [,1] [,2]
> [1,] 0.708060983 0.3242221
> [2,] 0.356736311 0.1454096
> [3,] 0.402880340 0.4763676
> [4,] 0.795947223 0.4052168
> [5,] 0.001620093 0.2618591
> [6,] 0.192215589 0.6595275
> [7,] 0.539199304 0.5402015
>
> > b (m=6,matrix=my_matrix)
> [[1]]
> matrix.lapply.inf.sup..function.i..c.i..Inf..matrix.i..1...matrix.i..
> 1 1.0000000, Inf, 0.7080610, 0.3242221
> 2 2.0000000, Inf, 0.3567363, 0.1454096
> 3 3.0000000, Inf, 0.4028803, 0.4763676
> 4 4.0000000, Inf, 0.7959472, 0.4052168
> 5 5.000000000, Inf, 0.001620093, 0.261859077
> 6 6.0000000, Inf, 0.1922156, 0.6595275
>
> [[2]]
> matrix.lapply.inf.sup..function.i..c.i..Inf..matrix.i..1...matrix.i..
> 1 7.0000000, Inf, 0.5391993, 0.5402015
> 2 7.0000000, Inf, 0.5391993, 0.5402015
> 3 7.0000000, Inf, 0.5391993, 0.5402015
> 4 7.0000000, Inf, 0.5391993, 0.5402015
> 5 7.0000000, Inf, 0.5391993, 0.5402015
> 6 7.0000000, Inf, 0.5391993, 0.5402015
>
It seems like the second list is filled with repeated rows (from 2 to 6)
I would like the second list to stop in the last row of the my_matrix
So, I would like to have the following result:> b (m=6,matrix=my_matrix)
[[1]]
matrix.lapply.inf.sup..function.i..c.i..Inf..matrix.i..1...matrix.i..
1 1.0000000, Inf, 0.7080610, 0.3242221
2 2.0000000, Inf, 0.3567363, 0.1454096
3 3.0000000, Inf, 0.4028803, 0.4763676
4 4.0000000, Inf, 0.7959472, 0.4052168
5 5.000000000, Inf, 0.001620093, 0.261859077
6 6.0000000, Inf, 0.1922156, 0.6595275
[[2]]
matrix.lapply.inf.sup..function.i..c.i..Inf..matrix.i..1...matrix.i..
1 7.0000000, Inf, 0.5391993, 0.5402015
Can`t I do this with an apply function? Is there any more efficient way
that substituting the lapply by a for loop?
THanks,
Rui
[[alternative HTML version deleted]]
Hello,
When you don't know what's going on, break long instructions into
simpler ones.
R is good at doing a lot in one line, to debug don't do that.
b = function (m, mat) {
n=nrow (mat)
p=ceiling (n/m)
lapply (1:p, function (l,n,m) {
inf = ((l-1)*m)+1
if (l<p) sup=((l-1)*m)+m else sup=n
lst <- lapply( inf:sup, function(i)
c(i, Inf, mat[i, 1], mat[i, 2]) )
data.frame( matrix(lst, nrow=length(lst)) )
#data.frame( matrix(unlist(lst), nrow=length(lst)) )
}, n=n, m=m
)
}
It returns a list of data.frames with 1 column only.
If you want 4 column data.frames, uncomment the instruction above (and
comment out the previuos one).
Hope this helps,
Rui Barradas
Em 11-09-2012 19:44, Rui Esteves escreveu:> Hello,
>
> I have 2 functions (a and b)
>
> a = function(n) { matrix (runif(n*2,0.0,1), n) }
>>
>> b = function (m, matrix) {
>> n=nrow (matrix)
>> p=ceiling (n/m)
>> lapply (1:p, function (l,n,m) {
>> inf = ((l-1)*m)+1
>> if (l<p) sup=((l-1)*m)+m
>> else sup=n
>> data.frame (matrix (lapply (inf: sup,
function(i)
>> c(i, Inf, matrix[i,1], matrix[i,2]) ), nrow=m ) )
>> }, n=n, m=m
>> )
>> }
>>
>>> my.matrix = a(7)
>>> my.matrix
>> [,1] [,2]
>> [1,] 0.708060983 0.3242221
>> [2,] 0.356736311 0.1454096
>> [3,] 0.402880340 0.4763676
>> [4,] 0.795947223 0.4052168
>> [5,] 0.001620093 0.2618591
>> [6,] 0.192215589 0.6595275
>> [7,] 0.539199304 0.5402015
>>
>>> b (m=6,matrix=my_matrix)
>> [[1]]
>>
matrix.lapply.inf.sup..function.i..c.i..Inf..matrix.i..1...matrix.i..
>> 1 1.0000000, Inf, 0.7080610, 0.3242221
>> 2 2.0000000, Inf, 0.3567363, 0.1454096
>> 3 3.0000000, Inf, 0.4028803, 0.4763676
>> 4 4.0000000, Inf, 0.7959472, 0.4052168
>> 5 5.000000000, Inf, 0.001620093, 0.261859077
>> 6 6.0000000, Inf, 0.1922156, 0.6595275
>>
>> [[2]]
>>
matrix.lapply.inf.sup..function.i..c.i..Inf..matrix.i..1...matrix.i..
>> 1 7.0000000, Inf, 0.5391993, 0.5402015
>> 2 7.0000000, Inf, 0.5391993, 0.5402015
>> 3 7.0000000, Inf, 0.5391993, 0.5402015
>> 4 7.0000000, Inf, 0.5391993, 0.5402015
>> 5 7.0000000, Inf, 0.5391993, 0.5402015
>> 6 7.0000000, Inf, 0.5391993, 0.5402015
>>
> It seems like the second list is filled with repeated rows (from 2 to 6)
> I would like the second list to stop in the last row of the my_matrix
> So, I would like to have the following result:
>> b (m=6,matrix=my_matrix)
> [[1]]
> matrix.lapply.inf.sup..function.i..c.i..Inf..matrix.i..1...matrix.i..
> 1 1.0000000, Inf, 0.7080610, 0.3242221
> 2 2.0000000, Inf, 0.3567363, 0.1454096
> 3 3.0000000, Inf, 0.4028803, 0.4763676
> 4 4.0000000, Inf, 0.7959472, 0.4052168
> 5 5.000000000, Inf, 0.001620093, 0.261859077
> 6 6.0000000, Inf, 0.1922156, 0.6595275
>
> [[2]]
> matrix.lapply.inf.sup..function.i..c.i..Inf..matrix.i..1...matrix.i..
> 1 7.0000000, Inf, 0.5391993, 0.5402015
>
>
> Can`t I do this with an apply function? Is there any more efficient way
> that substituting the lapply by a for loop?
>
> THanks,
> Rui
>
> [[alternative HTML version deleted]]
>
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