sapply

问题 There's a workaround to allow access the index inside a s/lapply e.g. x <- list(a=11,b=12,c=13) lapply(seq_along(x), function(y, n, i) { paste(n[[i]], y[[i]]) }, y=x, n=names(x)) Is there any function like s/lapply (or like purrr::map() ) which allows access to the index in the simplest way possible, which I guess would be to simply supply its desired name to the initial function call and nothing more; map_with_index <- function(.x, .f, index) { # Same as purrr::map() # ..but whatever string

问题 I have the following sample code that uses sapply which takes long to process (since executed many times): samples = sapply(rowIndices, function(idx){ sample(vectorToDrawFrom, 1, TRUE, weights[idx, ]) }) The issue is that I have to draw from the weights which are in the matrix, dependent on the indices in rowIndices . Does somebody have a better idea in mind to draw from the rows of the matrix? Reproducable example: rowIndices = floor(runif(1000, 1, 100)) vectorToDrawFrom = runif(5000, 0.0, 2

问题 I have a named list of data and a custom function I want to apply to the data: #Some example data d.list <- list(a = c(1,2,3), b = c(4,5,6), c = c(7,8,9)) #A simple function to process some data, write an output graph, and return an output myfun <- function(data, f.name) { y <- list() y[1] <- data[1] + data[2] + data[3] y[2] <- data[3] - data[1]/ data[2] y[3] <- data[2] * data[3] - data[1] svg(filename = f.name, width = 7, height = 5, pointsize = 12) plot.new() plot(data, y) dev.off() return

问题 I have 3 columns. First column has unique ID, second and third columns have string data and some NA data. I need to extract info from column 2 and put it in separate columns and do the same thing for column 3. I am building a function as follows, using for loops. I need to split the columns after the third letter. [For example in the V1 column below, I need to break AAAbbb as AAA and bbb and put them in separate columns. I know I can use substr to do this. I am new to R, please help. UID * V1

问题 I have 3 columns. First column has unique ID, second and third columns have string data and some NA data. I need to extract info from column 2 and put it in separate columns and do the same thing for column 3. I am building a function as follows, using for loops. I need to split the columns after the third letter. [For example in the V1 column below, I need to break AAAbbb as AAA and bbb and put them in separate columns. I know I can use substr to do this. I am new to R, please help. UID * V1

问题 I guess I just don't see it, but all the similar thing I found on the Net, in the Mailinglist archives or the FAQ could not really elucidate my issue. The closest I have found was this: apply strsplit rowwise I have a df, with two character columns and one numerical column. Filled like this: df=data.frame(name1=c("A","B","C","D"), name2=c("B","A","D","C"), nums=c(1,1,4,4), stringsAsFactors=F) Now I would like to find the unique rows in this, however, only based on the two name columns. And

问题 I'm trying to replicate solution on applying multiple functions in sapply posted on R-Bloggers but I can't get it to work in the desired manner. I'm working with a simple data set, similar to the one generated below: require(datasets) crs_mat <- cor(mtcars) # Triangle function get_upper_tri <- function(cormat){ cormat[lower.tri(cormat)] <- NA return(cormat) } require(reshape2) crs_mat <- melt(get_upper_tri(crs_mat)) I would like to replace some text values across columns Var1 and Var2 . The

问题 It might not be very clear from the title but what I wish to do is: I have a dataframe df with, say, 200 columns and the first 80 columns are response variables (y1, y2, y3, ...) and the rest of 120 are predictors (x1, x2, x3, ...). I wish to compute a linear model for each pair – lm(yi ~ xi, data = df) . Many problems and solutions I have looked through online have a either a fixed response vs many predictors or the other way around, using lapply() and its related functions. Could anyone who

问题 It might not be very clear from the title but what I wish to do is: I have a dataframe df with, say, 200 columns and the first 80 columns are response variables (y1, y2, y3, ...) and the rest of 120 are predictors (x1, x2, x3, ...). I wish to compute a linear model for each pair – lm(yi ~ xi, data = df) . Many problems and solutions I have looked through online have a either a fixed response vs many predictors or the other way around, using lapply() and its related functions. Could anyone who

来源： https://stackoverflow.com/questions/51314582/rename-columns-of-dataframe-based-on-names-of-list-in-r