similar to: A question about data frame

Hi all - I've been working on a lite ORM and database abstraction package for R. Formatting complex queries by hand has always been an error-prone hassle, so I've tried to do away with that as much as possible, instead, using R objects to represent elements of a database system (statements, clauses, operators, functions, fields, tables, etc.). R5 classes have made the development of this

Hello there, would you please look into my codes? Here I have following: > set.seed(100) > samp <- sample(c(1,-1,0), 20, replace=T); samp [1] 1 1 -1 1 -1 -1 0 -1 -1 1 -1 0 1 -1 0 0 1 -1 -1 0 Here I want to calculate the length of each unique number for above vector. How can I do that? Thanks in advance [[alternative HTML version deleted]]

Fellow R programmers, I'd like to submit SQLDF statements with R objects as column names. For example, I want to assign "X" to "var1" (var1<-"X") and then refer to "var1" in the SQLDF statement. SQLDF needs to understand that when I reference "var1", it should look for "X" in the dataframe. This is necessary because my SQLDF

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

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

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

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

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

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

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

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

Dear all, I am using grep but I did not understand the problem as I am doing something wrong.Please help me. I am using this code- sf=data.frame(sapply(df[],function(x) grep('\\.&\\,', df[,9]))) the thing is i have a data frame(df) like this- 10 135349467 g G 4 0 0 5 ,,,., 10 135349468 t T 2 0 0 5 ,,c., 10 135349469 g G 7 0 0 5 ,,a., 10 135349470 c C 8 0 0 5 ,,,., 10 135349471

Sorry for this dumb question. Suppose I have a named array ww defined as ww <- 1:5 names(ww) <- c("a", "b", "c", "d", "e") How can I extract the whole array of numbers without the names? ww[1:5] does not work while ww[[1]] can only extract one number at a time. Thanks, 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

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

I suspect that this is simple, but thanks in advance for any advice... I have a dataframe, t2: V1 V2 aaa 3 aaaa 4 aaaaaa 6 a 1 aa 2 V2 is the length of the string in V1 using nchar(as.character(t1$V1)) I'd like to create a third column, that contains a string of the length of V2, but containing an alternate text, e.g. V1 V2 V3

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

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

Hello, I wrote this code which works fine on a single observation: x<-100 y<-200 z<-125 aa<-150 if(x<z && y>z) {aa-z} result: 25 I am trying to apply this logic where x,y,z,and aa are arrays but with very little success. I have tried using loops and whiles but I always get errors of various types. I have consulted a few manuals but with limited success. My hopeful