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I have two arrays A and B with dimensions of (L, M, N, P) and (L, M, N), and I want to do for (i in 1:L) { for (j in 1:M) { for (k in 1:N) { if (abs(B[i, j, k]) > 10e-5) C[i, j, k,] <- A[i, j, k,]/B[i, j, k] else C[i, j, k,] <- 0 } } } How can I get C more efficiently than looping? Thanks, Gang

I have a fairly large data set with six variables set up like the following dummy: # Create fake data df <- data.frame(code = c(rep(1001, 8), rep(1002, 8)), year = rep(c(rep(2011, 4), rep(2012, 4)), 2), fund = rep(c("10E", "10E", "10E", "27E"), 4), func = rep(c("110000", "122000", "214000", "158000"), 4), obj = rep("100", 16), amount = round(rnorm(16, 50000, 10000))) What I w...

dear all, here's a couple of questions that puzzled me in these last hours: ##### issue 1 qnorm(1-10e-100)!=qnorm(10e-100) qnorm(1-1e-10) == -qnorm(1e-10) # turns on to be FALSE. Ok I'm not a computer scientist but, # but I had a look at the R inferno so I write: all.equal(qnorm(1-1e-10) , -qnorm(1e-10)) # which turns TRUE, as one would expect, but all.equal(qnorm(1-1e-100) , -qnorm(1e-100...

Hi, I have a function that generates a set of data but I am having problems determining the parameters using the nls fitting procedure. #### "MH"<-function(field,diameter,mu=10e-7,sig=0.1,Ms=100,chi=0){ #variables mu, sig, chi, Ms #input: field and diameter #all in CGS rho <- 5 kb <- 1.38e-16 t <- 300 length.d<-length(diameter) length.H<-length(field) M<-double(length.H) for (i in 1:length.H){ S1<-0 S2<-0 H <- field[i] for (j in 1:length.d){...

Hy, I have dowloaded the R-version for Windows. I would want to plot different pictures on the same device, but I am not able to do it (i.e I would like make 2 qqplot on the same graphic). Can you help me? Thanks in advance Davide reply to: d.tarfanelli at libero.it -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

...d can I avoid this? func <- function(y, a, rate, sad){ f3 <- function(z){ f1 <- function(y,a,n){ dpois(y,a*n) } f2 <- function(n,rate){ dexp(n,rate) } f <- function(n){ f1(y,a,n)*f2(n,rate) } r <- 0 r1 <- 1 x1 <- 0 dx <- 20 while(r1 > 10e-500){ r1 <- integrate(f,x1,x1+dx)$value r <- r + r1 x1 <- x1 + dx } r + integrate(f,x1,Inf)$valu } sapply(y,f3) } func(200,0.1,0.1,sad=Exp) Thanks in advance.

...=1) [1] 0.3678793 > dnbinom(1,size=3e+15,mu=1) [1] 0.3678793 - very close to Poisson. But then I increase the size further, and it goes off somewhat: > dnbinom(1,size=5e+15,mu=1) [1] 0.3658024 > dnbinom(1,size=7e+15,mu=1) [1] 0.3572676 ...until suddenly it returns 1: > dnbinom(1,size=10e+15,mu=1) [1] 1 This turned out to be a big and hard to track down issue for me. The bug confused the optimizer of the likelihood function, which happened to move too far on the size dimension and began to discover "very good" parameters. I fixed the problem by adding a logical check, whi...

Dear all, I would like to (i) produce boxplot graphs with axis in logarithm in base 10 and (ii) showing the values on the axis in 10^exponent format rather than 10E+exponent. To illustrate with an example, I have some widely spread data that I chart plot using boxplot() [figure on the left]; the log="y" option of boxplot() I obtained the natural logarithm conversion of the data and the unfriendly notation baseE+exponent [figure on the centre]; i...

...es are correlated or not. (pearson correlation) cor.test(x,y) give a p=5.87.... Because the x, y is not normal distributed (qqplot indicate that) I also perform (spearman rank correlation) cor.test(x,y,method="spearman") give a very significant result p<10e-4 I don't know how to explain this. Will this result tell us that x, y are correlated but not a linear one, and we can't use the coefficient estimated by spearman rank correlation because its interpretation is not quite clear. [[alternative HTML version deleted]]

I tried to fit data with the following function: fit<-nls(y~ Is*(1-exp(-l*x))+Iph,start=list(Is=-2e-5,l=2.3,Iph=-0.3 ),control=list(maxiter=500,minFactor=1/10000,tol=10e-05),trace=TRUE) But I get only a singular Gradient warning... the data can by found attached(there are two sampels of data col 1/2 and 3/4). I tried to fix it by chanching the start parameters but that didn't solve the problem. Would it be a possibiliti to use the selfstart Model? How? Thanks...

...if it uses a register tied to a register def in that instruction. Is that right? Why would we ever expect the def slot for early-clobber VNInfo to point to the same instruction as the early-clobber slot originally queried? In other words, why would we ever expect this: Idx = 10r EC = 10e VNI->def = 10B (or 10r or whatever) I would expect VNI->def to point to a completely different instruction or a PHI-def. Otherwise we have a use-before-def, right? My problem boils down to trying to find the defining instruction for a virtual register use. Given an instruction using regis...

Could some kindly tell me if I am supposed to be getting the same test statistic value with var.equal=TRUE and var.equal=FALSE in t.test ? set.seed(1066) x1 <- rnorm(50) x2 <- rnorm(50) t.test(x1, x2, var.equal=FALSE)$statistic # 0.5989774 t.test(x1, x2, var.equal=TRUE)$statistic # 0.5989774 ??? Here are my own calculations that shows that perhaps the result when var.equal=TRUE is

...t;>>> well. The 32 bit version puts out a histogram that has an expected, >>>> almost symmetric unimodal distribution. The 64 bit version >>>> created a >>>> bimodal distribution with one large mode near 0 and another smaller >>>> mode near 10E+37. Postcript output attached. >>>> >>> >>> >>> >>> -- >>> Anirban Mukherjee | Assistant Professor, Marketing | LKCSB, SMU >>> 5062 School of Business, 50 Stamford Road, Singapore 178899 | >>> +65-6828-1932 >>> >...

Would like to know how much data can R process - number of rows and columns? [[alternative HTML version deleted]]

...0 getdents(4, /* 6 entries */, 4096) = 144 open("/dev/bus/usb/002/013", O_RDWR) = -1 EACCES (Permission denied) open("/dev/bus/usb/002/013", O_RDONLY) = 5 ioctl(5, USBDEVFS_CONNECTINFO, 0x7ffff1226ac0) = -1 EPERM (Operation not permitted) read(5, "\22\1\20\1\0\0\0\10e\6aQ\2\0\1\2\0\1"..., 18) = 18 read(5, "\t\2\"\0\1\1\3\200"..., 8) = 8 read(5, "2\t\4\0\0\1\3\0\0\4\t!\0\1\0\1\"\33\0\7\5\201\3\10\0 "..., 26) = 26 close(5) = 0 open("/dev/bus/usb/002/004", O_RDWR) = -1 EACCES (Permis...

Is there a proper function for transforming a character to a number instead of using i=match('c',letters) # 3 Thanks. Josef Eschgf??ller -- Josef Eschgf??ller Dipartimento Matematico Universita' di Ferrara http://felix.unife.it

...0.096 (max possible= 0.911 )Likelihood ratio test= 23.33 on 3 df, p=3.444e-05Wald test = 21.76 on 3 df, p=7.312e-05Score (logrank) test = 24.46 on 3 df, p=2.001e-05 I think, this means that the two variables I tested are of interest to build the model with a p-value of about 10e-5. This also mean that having the Herceptincat =1 is significantly different from having a Herceptincat = 0. Moreover, having a nodecat_all iof 2 is significantly different from having a nodecat equals to 0.However, this does not tell me much about the variables alone. summary(coxph(Surv(Time_to...

Dear R users, First of all, thanks for the incredibly fast answers and help of Rolf, Marc and Robert. Yes, I noticed that it was a lot of permutacions, but my intention was to make this process automatic and take only 5.000 - 10.000 permutations. Therefore, I wanted only to take that "interesting permutations" with "some information" [inter-block permutations]. The

Dear R-users Please excuse me if this topic has been covered before, but I was unable to find anything relevant by searching I am currently doing a comparison of two biological variables that have a highly significant linear relationship. I know that the p-value of linear regression is not so interesting in itself, but this particular value does raise a question. How does R calculate

Dear R-Helpers, I have a dataframe (g10df) formatted like this: GENE PVAL 1 KCTD12 4.06904e-22 2 UNC93A 9.91852e-22 3 CDKN3 1.24695e-21 4 CLEC2B 4.71759e-21 5 DAB2 1.12062e-20 The rows are ranked in ascending order by PVAL, and I need to end up with the same relative order. There are duplicate entries for genes in the first column with corresponding