similar to: Extract values from a named array

I want to pass a predefined string ww ("fa*fb+fc") to function lme so that I can run > lme(y ~ fa*fb+fc, random = ~1|subj, model) I've tried something like > lme(y ~ paste(ww), random = ~1|subj, model) and > lme(y ~ sprintf(ww), random = ~1|subj, model) but both give me the following error: Error in model.frame(formula, rownames, variables, varnames, extras,

I can run one-sample t-test on an array, for example a matrix myData1, with the following apply(myData1, 2, t.test) Is there a similar fashion using apply() or something else to run 2-sample t-test with datasets from two groups, myData1 and myData2, without looping? TIA, Gang

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

Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace = TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE))) x R1 R2 1 C B 2 B B 3 C E 4 E C 5 E B 6 D E 7 E E 8 D F 9 C D 10 A E Notice that the factor levels between

A very simple question. With a data frame like this: > n = c(2, 3, 5) > s = c("aa", "bb", "cc") > df = data.frame(n, s) I want df$s[1] or df[1,2], but how can I get rid of the extra line in the output about the factor levels: > df$s[1] [1] aa Levels: aa bb cc Thanks, Gang

Anybody knows what functions can be used to calculate variance/covariance with complex numbers? var and cov don't seem to work: > a 1 V1 0.00810014+0.00169366i V2 0.00813054+0.00158251i V3 0.00805489+0.00163295i V4 0.00809141+0.00159533i V5 0.00813976+0.00161850i > var(a) 1 1 1.141556e-09 Warning message: In var(a) : imaginary parts discarded in

Untested, but I expect that setting the levels to be the same across the two factors levels(tmp$R1) <- levels(tmp$R2) <- LETTERS[1:6] and proceeding as before should be fine. Best, Ista On Jul 6, 2017 6:54 PM, "Gang Chen" <gangchen6 at gmail.com> wrote: Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp

Suppose I have a dataframe 'd' defined as L3 <- LETTERS[1:3] d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace = TRUE)) (d <- d0[d0$fac %in% c('A', 'B'),]) x y fac 2 1 2 B 3 1 3 A 4 1 4 A 5 1 5 A 6 1 6 B 8 1 8 A Even though factor 'fac' in 'd' only has 2 levels, but it seems to bear the birthmark

Suppose I create an R program called myTest.R with only one line like the following: type <- as.integer(readline("input type (1: type1; 2: type2)? ")) Then I'd like to run myTest.R in batch mode by constructing an input file called answers.R with the following: source("myTest.R") 1 When I ran the following at the terminal: R CMD BATCH answer.R output.Rout it failed

Suppose that we have the following dataframe: set.seed(1) (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace = TRUE), R2 = sample(LETTERS[1:5], 10, replace = TRUE))) x R1 R2 1 1 B B 2 2 B A 3 3 C D 4 4 E B 5 5 B D 6 6 E C 7 7 E D 8 8 D E 9 9 D B 10 10 A D I want to do the following: if the difference between the level index of factor

In a classical meta analysis model y_i = X_i * beta_i + e_i, data {y_i} are assumed to be independent effect sizes. However, I'm encountering the following two scenarios: (1) Each source has multiple effect sizes, thus {y_i} are not fully independent with each other. (2) Each source has multiple effect sizes, each of the effect size from a source can be categorized as one of a factor levels

I know how to convert a simple dataframe from wide to long format with one varying factor. However, for a dataset with two factors like the following, Subj T1_Cond1 T1_Cond2 T2_Cond1 T2_Cond2 1 0.125869 4.108232 1.099392 5.556614 2 1.427940 2.170026 0.120748 1.176353 How to elegantly convert to a long form as Subj Time Cond Value 1 1 1 0.125869 1

How about foo <- with(list(r1 = tmp$R1, r2 = tmp$R2, swapme = (as.numeric(tmp$R1) - as.numeric(tmp$R2)) %% 2 != 0), { tmp[swapme, "R1"] <- r2[swapme] tmp[swapme, "R2"] <- r1[swapme] tmp }) Best, Ista On Thu, Jul 6, 2017 at 4:06 PM, Gang Chen <gangchen6 at gmail.com> wrote: > Suppose that we have the following

Suppose I have a 4-D array X with dimensions (dx, dy, dz, dp). I want to collapse the first 3 dimensions of X to make a 2-D array Y with dimensions (dx*dy*dz, dp). Instead of awkward looping, what is a good way to do this? Is there a similar function like reshape in Matlab? Thanks, Gang

Hi, I want to store some number of outputs from running a bunch of analyses such as lm() into an array. I know how to do this with a one-dimensional array (vector) by creating myArray <- vector(mode='list', length=10) and storing each lm() result into a component of myArray. My question is, how can do this for a multiple dimensional array? It seems array() does not have such a

I have some data 'myData' in wide form (attached at the end), and would like to convert it to long form. I wish to have five variables in the result: 1) Subj: factor 2) Group: between-subjects factor (2 levels: s / w) 3) Reference: within-subject factor (2 levels: Me / She) 4) F: within-subject factor (2 levels: F1 / F2) 5) J: within-subject factor (2 levels: J1 / J2) As this is the

With the following example using contr.sum for both factors, > dd <- data.frame(a = gl(3,4), b = gl(4,1,12)) # balanced 2-way > model.matrix(~ a * b, dd, contrasts = list(a="contr.sum", b="contr.sum")) (Intercept) a1 a2 b1 b2 b3 a1:b1 a2:b1 a1:b2 a2:b2 a1:b3 a2:b3 1 1 1 0 1 0 0 1 0 0 0 0 0 2 1 1 0 0 1 0

Hi experts, I just started to learn R today, and tried to work with an add-on package sem. I have a version of 2.3.1 on MacOS X 10.4.6 with sem put under /Library/Frameworks/R.framework/Versions/2.3/Resources/library However when I typed library(sem) the following error showed up: Error in library(sem) : 'sem' is not a valid package -- installed < 2.0.0? Why is this? Thank

Why does R CMD INSTALL work for some packages (e.g., lme4) but not others (e.g., nlme)? Thanks, Gang

I want to run a R program, prog.R, interactively. My question is, is there a way I can start prog.R on the shell terminal when invoking R, instead of using source() inside R? TIA, Gang