I believe the solutions proposed ignore the recursive nature of the
original problem and hence produce wrong solutions.
P <- c(5, 7, 6.01, 6.01, 7)
m <- rep(6, 5)
S0 <- as.numeric(P>(m*1.005))
Then the original loop from Worik gives
S <- S0
for(i in 2:length(S)){
if(S[i]==0 && S[i-1] == 1){
Use 'diff' to determine where the changes are:
> S <- sample(0:1,30,TRUE)
> S
[1] 0 0 1 1 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 1 1 0 1 0 0 0 0 0 1 0
> which(diff(S) == -1)
[1] 4 9 13 15 18 21 23 29
>
then use the indices for the other processing.
On Thu, Apr 7, 2011 at 10:30 PM, Worik R wrote:
> Fr
Hi,
this response uses the previous responses with an example:
#Assume you have 100 observations
n=100
#Simulate a time series of prices
error=rnorm(n,0,3)
raw.price=rpois(n,100)
lag.price=c(rpois(1,100),raw.price[1:99])
price=lag.price+error
#Say you want the moving average based on this
#and
Kenn,
I find your solution more elegant.
2011/4/8 Kenn Konstabel :
> 2011/4/8 Juan Carlos Borrás :
>> #Use the indexes of S in a sapply function.
>>
>> N <- 10
>> S <- sample(c(0,1), size=N, replace=TRUE)
>> v1 <- sapply(c(1:N-1), function(i) S[i]&&S[i+1])
>
> You can achieve the same v1 using
>
>
2011/4/8 Juan Carlos Borrás :
> #Use the indexes of S in a sapply function.
>
> N <- 10
> S <- sample(c(0,1), size=N, replace=TRUE)
> v1 <- sapply(c(1:N-1), function(i) S[i]&&S[i+1])
You can achieve the same v1 using
v1.2 <- S[2:N-1] & S[2:N]
.. or if you insist on having NA as the first elemen
#Use the indexes of S in a sapply function.
N <- 10
S <- sample(c(0,1), size=N, replace=TRUE)
v1 <- sapply(c(1:N-1), function(i) S[i]&&S[i+1])
# Then
v2 <- (P > m)
# And I guess you can fill up the rest. Beware of the boundary
condition (the NA in v1)
Cheers,
jcb!
___
http:/
Thank you so much to Jim and Mark for their advices. Now I solved the
problem I had using a new approach.
Best,
Jorge
On Mon, Apr 21, 2008 at 12:53 PM, jim holtman <[EMAIL PROTECTED]> wrote:
> Will this do it for you:
>
> # Seed and data frames X, Y and Z
> set.seed(123)
> X=matrix(rnorm(300),
Will this do it for you:
# Seed and data frames X, Y and Z
set.seed(123)
X=matrix(rnorm(300),ncol=5)
Y=matrix(rpois(300,10),ncol=5)
Z=matrix(rexp(300,1),ncol=5)
index <- seq(1, by=3, length=nrow(X))
FINAL <- matrix(ncol=5, nrow=3*nrow(X))
FINAL[index,] <- X
FINAL[index + 1,] <- Y
FINAL[index + 2,
I knew it was simple. Thanks very much.
-tom
jim holtman <[EMAIL PROTECTED]> wrote:
> ?table to count the factors
>
> > x
> [1] "a" "b" "c" "d" "e"
> > paste(head(x, -1), tail(x, -1), sep='')
> [1] "ab" "bc" "cd" "de"
>
>
> On 10/15/07, Tom Sgouros <[EMAIL PROTECTED]> wrote:
> >
> > Hi
?table to count the factors
> x
[1] "a" "b" "c" "d" "e"
> paste(head(x, -1), tail(x, -1), sep='')
[1] "ab" "bc" "cd" "de"
On 10/15/07, Tom Sgouros <[EMAIL PROTECTED]> wrote:
>
> Hi All:
>
> I feel like there must be a slick R-native no-loop way to get the counts
> for the entries in a factor,
10 matches
Mail list logo
