I am trying to write a function in R, for a simple time series regression (the result of this function is the output for more complicated ones). In the first part i define the variables and create some lags for the function, which are named ar_i depending on the used lag.
However in the second part i try to combine this lags in a matrix using a cbind function on the variables initially defined. As you can see the output is not the expected matrix, but the names of the lags themselves. I tried to solve this by using the noquote() and cat() function, but these don't seem to work.
Do you have any suggestions? Thanks in advance!!!
Pd: The code and the results are below.
trans <- dlpib
ar <- dlpib
linear <- 1:4
for (i in linear){
assign(paste("ar_",i,sep = ""), lag(ar,k=-i))
}
linear_dat <- cbind(paste("ar_",linear, collapse=',', sep = ""))
> linear_dat
[,1]
[1,] "ar_1,ar_2,ar_3,ar_4"
I think you could go about this more efficiently with sapply:
linear <- 1:4
linear_list <- lapply(linear, function(i) lag(ar, k=-i))
linear_dat <- do.call(cbind, linear_list)
colnames(linear_dat) <- paste0("ar_", linear)
Related
I have a dataset "res.sav" that I read in via haven. It contains 20 columns, called "Genes1_Acc4", "Genes2_Acc4" etc. I am trying to find a correlation coefficient between those and another column called "Condition". I want to separately list all coefficients.
I created two functions, cor.condition.cols and cor.func to do that. The first iterates through the filenames and works just fine. The second was supposed to give me my correlations which didn't work at all. I also created a new "cor.condition.Genes" which I would like to fill with the correlations, ideally as a matrix or dataframe.
I have tried to iterate through the columns with two functions. However, when I try to pass it, I get the error: "NAs introduced by conversion". This wouldn't be the end of the world (I tried also suppressWarning()). But the bigger problem I have that it seems like my function does not convert said columns into the numeric type I need for my cor() function. I receive the "y must be numeric" error when trying to run the cor() function. I tried to put several arguments within and without '' or "" without success.
When I ran str(cor.condition.cols) I only receive character strings, which makes me think that my function somehow messes up with the as.numeric function. Any suggestions of how else I could iter through these columns and transfer them?
Thanks guys :)
cor.condition.cols <- lapply(1:20, function(x){paste0("res$Genes", x, "_Acc4")})
#save acc_4 columns as numeric columns and calculate correlations
res <- (as.numeric("cor.condition.cols"))
cor.func <- function(x){
cor(res$Condition, x, use="complete.obs", method="pearson")
}
cor.condition.Genes <- cor.func(cor.condition.cols)
You can do:
cor.condition.cols <- paste0("Genes", 1:20, "_Acc4")
res2 <- as.numeric(as.matrix(res[cor.condition.cols]))
cor.condition.Genes <- cor(res2, res$Condition, use="complete.obs", method="pearson")
eventually the short variant:
cor.condition.cols <- paste0("Genes", 1:20, "_Acc4")
cor.condition.Genes <- cor(res[cor.condition.cols], res$Condition, use="complete.obs")
Here is an example with other data:
cor(iris[-(4:5)], iris[[4]])
I'm trying to understand the answer to this question using R and I'm struggling a lot.
The dataset for the R code can be found with this code
library(devtools)
install_github("genomicsclass/GSE5859Subset")
library(GSE5859Subset)
data(GSE5859Subset) ##this loads the three tables you need
Here is the question
Write a function that takes a vector of values e and a binary vector group coding two groups, and returns the p-value from a t-test: t.test( e[group==1], e[group==0])$p.value.
Now define g to code cases (1) and controls (0) like this g <- factor(sampleInfo$group)
Next use the function apply to run a t-test for each row of geneExpression and obtain the p-value. What is smallest p-value among all these t-tests?
The answer provided is
myttest <- function(e,group){
x <- e[group==1]
y <- e[group==0]
return( t.test(x,y)$p.value )
}
g <- factor(sampleInfo$group)
pvals <- apply(geneExpression,1,myttest, group=g)
min( pvals )
Which gives you the answer of 1.406803e-21.
What exactly is the input of the "e" argument of the myttest function when you run this? Is it possible to write this function as a formula like
t.test(DV ~ sampleInfo$group)
The t test is comparing the gene expression values of the 24 people (the values of which I believe are in the "geneExpression" matrix) by what group they were
in which you can find in sampleInfo's "group" column. I've run t tests so many times in R, but for some reason I can't wrap my mind around what's going on in this code.
You question seems to be about understanding the function apply().
For the technical description, see ?apply.
My quick explanation: the apply() line of code in your question applies the following function to each of the rows of geneExpression
myttest(e=x, group=g)
where x is a placeholder for each row.
To help make sense of it, a for loop version of that apply() line would look something like:
N <- nrows(geneExpression) #so we don't have to type this twice
pvals <- numeric(N) #empty vector to store results
# what 'apply' does (but it does it very quickly and with less typing from us)
for(i in 1:N) {
pvals[i] <- myttest(geneExpression[i,], group=g[i])
}
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 erroneous syntax below illustrates what I am trying to achieve:
crs_mat[,1:2] <- sapply(crs_mat[,1:2], function(x) {
# Replace first phrase
gsub("mpg","MPG",x),
# Replace second phrase
gsub("gear", "GeArr",x)
# Ideally, perform other changes
})
Naturally, the code is not syntactically correct and fails. To summarise, I would like to do the following:
Go through all the values in first two columns (Var1 and Var2) and perform simple replacements via gsub.
Ideally, I would like to avoid defining a separate function, as discussed in the linked post and keep everything within the sapply syntax
I don't want a nested loop
I had a look at the broadly similar subject discussed here and here but, if possible, I would like to avoid making use of plyr. I'm also interested in replacing the column values not in creating new columns and I would like to avoid specifying any column names. While working with my existing data frame it is more convenient for me to use column numbers.
Edit
Following very useful comments, what I'm trying to achieve can be summarised in the solution below:
fun.clean.columns <- function(x, str_width = 15) {
# Make character
x <- as.character(x)
# Replace various phrases
x <- gsub("perc85","something else", x)
x <- gsub("again", x)
x <- gsub("more","even more", x)
x <- gsub("abc","ohmg", x)
# Clean spaces
x <- trimws(x)
# Wrap strings
x <- str_wrap(x, width = str_width)
# Return object
return(x)
}
mean_data[,1:2] <- sapply(mean_data[,1:2], fun.clean.columns)
I don't need this function in my global.env so I can run rm after this but even nicer solution would involve squeezing this within the apply syntax.
We can use mgsub from library(qdap) to replace multiple patterns. Here, I am looping the first and second column using lapply and assign the results back to the crs_mat[,1:2]. Note that I am using lapply instead of sapply as lapply keeps the structure intact
library(qdap)
crs_mat[,1:2] <- lapply(crs_mat[,1:2], mgsub,
pattern=c('mpg', 'gear'), replacement=c('MPG', 'GeArr'))
Here is a start of a solution for you, I think you're capable of extending it yourself. There's probably more elegant approaches available, but I don't see them atm.
crs_mat[,1:2] <- sapply(crs_mat[,1:2], function(x) {
# Replace first phrase
step1 <- gsub("mpg","MPG",x)
# Replace second phrase. Note that this operates on a modified dataframe.
step2 <- gsub("gear", "GeArr",step1)
# Ideally, perform other changes
return(step2)
#or one nested line, not practical if more needs to be done
#return(gsub("gear", "GeArr",gsub("mpg","MPG",x)))
})
Ran a bunch of regressions and now I am trying to collect their p values and put them into a vector.
x=summary(reg2)$coefficients[4,4] #p value from the first regression, p-val is in row 4, col 4
for (i in 3:1000){
currentreg=summary(paste("reg",i,sep=""))
assign(x,c(x,currentreg$coefficients[4,4]))
}
I also tried eval(parse(currentreg)) and eval(parse(summary(paste("reg",i,sep="")))) with no luck. I always have this problem with telling R "Hey don't treat this as a string, treat it as a variable" and vice versa.
While it would be better to store the objects in a list and loop over that, you're asking for get:
currentreg <- summary(get(paste("reg", i, sep="")))
If you had a list of objects, models <- list(reg2, reg3, reg4, ...). You can then loop over this list with sapply to achieve the desired result (looping, collecting the results into a vector):
x <- sapply(models, function(z) { summary(z)$coeficients[4,4] })
You can use
sapply(mget(ls(pattern = "^reg\\d+$")), function(x) summary(x)$coefficients[4,4])
to create a vector with all p-values.
I am trying to use the interp1 function in R for linearly interpolating a matrix without using a for loop. So far I have tried:
bthD <- c(0,2,3,4,5) # original depth vector
bthA <- c(4000,3500,3200,3000,2800) # original array of area
Temp <- c(4.5,4.2,4.2,4,5,5,4.5,4.2,4.2,4)
Temp <- matrix(Temp,2) # matrix for temperature measurements
# -- interpolating bathymetry data --
depthTemp <- c(0.5,1,2,3,4)
layerZ <- seq(depthTemp[1],depthTemp[5],0.1)
library(signal)
layerA <- interp1(bthD,bthA,layerZ);
# -- interpolate= matrix --
layerT <- list()
for (i in 1:2){
t <- Temp[i,]
layerT[[i]] <- interp1(depthTemp,t,layerZ)
}
layerT <- do.call(rbind,layerT)
So, here I have used interp1 on each row of the matrix in a for loop. I would like to know how I could do this without using a for loop. I can do this in matlab by transposing the matrix as follows:
layerT = interp1(depthTemp,Temp',layerZ)'; % matlab code
but when I attempt to do this in R
layerT <- interp1(depthTemp,t(Temp),layerZ)
it does not return a matrix of interpolated results, but a numeric array. How can I ensure that R returns a matrix of the interpolated values?
There is nothing wrong with your approach; I probably would avoid the intermediate t <-
If you want to feel R-ish, try
apply(Temp,1,function(t) interp1(depthTemp,t,layerZ))
You may have to add a t(ranspose) in front of all if you really need it that way.
Since this is a 3d-field, per-row interpolation might not be optimal. My favorite is interp.loess in package tgp, but for regular spacings other options might by available. The method does not work for you mini-example (which is fine for the question), but required a larger grid.