Am 05.02.2012 15:54, schrieb Alaios:
Dear all
I am using lapply (actually mclapply that share the same syntax).
I want to call the same function that takes as input a vector. My initial data
structure is a matrix that I want to cut it to multiple vectors (one vector for
every row of the matrix
after running your code, you will notice
> that this is actually the correlation between
> DataArray_1["A2","B1","D1",] and DataArray_2["A1",C1","D1",] and so on.
>
> The code gives the correct result only in case where elements
> cor
1, D1 and A1, C2, D1.
So the "1:length(c)" writes only the correlation between the B and C out
of the whole correlation array.
That's also why the sequence in the second apply function is changed.
Regards Christian.
Poersching schrieb:
> Hey,
> I think I have a solutio
s just wondering if there are any other alternative way-outs
> to solving my problem. That's why I tried with apply functions
> (sapply)assuming that this might work out faster even fractionally as
> compared to for-loop.
>
> Cheers,
> Sauvik
>
> On Mon, Jul 27, 200
cor values greater 0.5 or smaller -0.5, like:
findIndex(Correl,Correl[Correl>0.5])
findIndex(Correl,Correl[Correl<(-0.5)])
I have changed the code of the function findIndex in line which
contents: el[j]<-which(is.element(data,element[j]))
Rigards,
Christian
>
> On Sun, Jul 26, 2009
stvienna wiener schrieb:
> Hi all,
>
> I am plotting a financial time series, but I need a more detailed X-Axis.
>
> Example:
> x <- zoo(rnorm(1:6000), as.Date("1992-11-11")+c(1:6000))
> plot(x)
>
> The X-Axis is labeled "1995", "2000" and "2005".
> I would need either "1995", "1997", etc. or maybe
Sauvik De schrieb:
> Hi Gabor:
> Many thanks for your prompt reply!
> The code is fine. But I need it in more general form as I had mentioned that
> I need to input any 0 to find its dimension-names.
>
> Actually, I was using "sapply" to calculate correlation and this idea was
> required in the mid
Mary A. Marion schrieb:
> Hello,
>
> I am plotting two distributions and want to draw a vertical line at
> the critical point 149.
> How can I stop it from going further up than the norm(140,15) curve?
>
> x<-seq(75,225,0.1)
> plot(x,dnorm(x,mean=140, sd=15), type='l', col='navy')
> abline(v = 149,
lanc...@fns.uniba.sk schrieb:
> Good day,
>
> I'm trying to get more time series in one plot. As there are bigger
> differences in values of variables I need logaritmic y axis.
>
> The code I use is the following:
>
> nvz_3_data <- read.csv('/home/tomas/R_outputs/nvz_3.csv')
> date <- (nvz_3_data$d
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