Hi Berend:
Thanks for your reply.
dat<-data.frame(x1=1:3,x2=4:6,x3=7:9)
z<-c(0.1,10,100)
#I wanna 0.1*x1,10*x2,100*x3
Option2 is similar as "dat*rep(z,each=length(z))",and the latter is simpler
than Option2 in expression.
At 2012-12-31 00:31:42,"Berend Hasselman" <b...@xs4all.nl> wrote:
>
>On 30-12-2012, at 11:26, meng <laomen...@163.com> wrote:
>
>> hi all:
>> Here's a dataframe(dat) and a vector(z):
>>
>> dat:
>> x1 x2 x3
>> 0.2 1.2 2.5
>> 0.5 2 5
>> 0.8 3 6.2
>>
>>> z
>> [1] 10 100 100
>>
>> I wanna do the following:
>> 10*x1,100*x2,1000*x3
>>
>> My solution is using the loop for z and dat(since the length of z is the
>> same as ncol of dat),which is tedious.
>> I wanna an efficient solution to do it .
>
>
>Data:
>
>xdat <- data.frame(x1=c(.2,.5,.8),x2=c(1.2,2,3),x3=c(2.5,5,6.2))
>z <- c(10,100,1000)
>
>Assuming you want a dataframe as result you can do one of the following
>
>Option 1:
>------------
>as.data.frame(sapply(1:length(z),function(k)xdat[k]*z[k]))
>
>Option 2:
>------------
>as.data.frame(as.matrix(xdat)*rep(z,each=length(z)))
>
>Option 3:
>------------
>as.data.frame(t(t(as.matrix(xdat))*z))
>
>Option 4:
>------------
>as.data.frame(as.matrix(xdat)%*%diag(z))
>
>
>The suggested solution
>
>as.matrix(xdat)*z
>
>results in
>
> x1 x2 x3
>[1,] 2 12 25
>[2,] 50 200 500
>[3,] 800 3000 6200
>
>
>and that doesn't seem to be what you want.
>
>Berend
>
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