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I have a data frame presenting different variables in column and for different species name in rows as follow. In the column a have only one independent variables(Proteome size) and all the others are independent variables. I want to use the PIC package to correlate all these dependent variables separately to my independent variables, a way to generate the R coefficient and p-value.

data <- read.table("t1.txt", row.names = 1, header = T,sep='')

data 
path1 path2 path3 path4 Proteome_size
ahli 4.0 -3.2 0.4 -1.0 -0.2
alayoni 3.8 3.4 -1.8 2.2 -0.3
alfaroi 3.5 2.6 0.8 -2.4 -0.5
aliniger 4.0 0.1 -1.7 2.4 0.0
allisoni 4.4 2.0 -3.7 0.5 -0.1
allogus 4.0 -2.8 0.6 -1.0 0.6
altitudinalis 3.8 2.9 -6.1 2.3 1.2
alumina 3.6 0.7 1.5 -2.7 -1.3
alutaceus 3.6 1.2 -0.6 -1.7 0.3
angusticeps 3.8 4.6 -2.0 1.2 1.7

I have this code following code that can correlate any one of the ‘path*’ column to proteome size:

data <- read.csv("data.csv", row.names = 1
tree <- read.tree("anolis.phy")
Proteome <- data[, "Proteome_size"]
awe <- data[, "path1"]
names(Proteome_size) <- names(var1) <- rownames(data)
hPic <- pic(proteome_size, tree)
aPic <- pic(gene1, tree)
picModel <- lm(hPic ~ aPic - 1)
summary(picModel)

My question is that, how can I implement this code to do that for all the columns simultaneously and store the statistics info as (R and p_value) for each variable somewhere ?

Thanks

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  • $\begingroup$ Welcome to the site. Have you tried to do a loop for each column to correlate them with the Proteome_size? $\endgroup$
    – llrs
    Commented Nov 8, 2018 at 16:01
  • $\begingroup$ Thanks for your quick response. I am a beginner in R. To do that need to create a vector before as : gene.names <- colnames(data[, "proteome_size"]). unfortunately it failed and the vector is NULL with length 0. I am not so good in for loop too $\endgroup$ Commented Nov 8, 2018 at 21:33
  • $\begingroup$ Sorry, I don't actually understand what the variable awe is doing. Also I don't see which variable will change through the loop. It's quite easy to write a for loop in R if you can tell which variable needs to be investigated! $\endgroup$
    – gabt
    Commented Nov 9, 2018 at 9:04
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    $\begingroup$ Your code gene.names <- colnames(data[, "proteome_size"]) produces a NULL vector since R is case sensitive and your data frame does not have a "proteome_size" column. $\endgroup$
    – haci
    Commented Nov 12, 2018 at 18:06

2 Answers 2

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One option is to use one of the apply family functions. If you enclose your code above (which generates errors by the way) in a function, you can then "apply" this function on the rows or columns of your data with something like apply(X, MARGIN, FUN, …), the margin being 1 for rows or 2 for columns (as in your case).

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Like haci mentioned, you can use the apply family of functions to loop over all your columns.

apply(data,2, function(path_name){ summary(lm(data$Proteome_size ~ path_name -1))$adj.r.squared })    

this gives you the adjusted r square value for your linear model, and

apply(data,2, function(path_name){ summary(lm(data$Proteome_size ~ path_name -1))$coefficients[,4] })    

would give you the p-values.

What happens here is apply takes your matrix and chops it up into pieces(a piece is a column here because I gave the option 2) and each pieces is fed into the function as 'path_name' and the output for all the pieces is given back to you.

You might get a warning about getting a perfect fit because I am correlating 'Proteome_size' against itself.

Also,

awe <- data[, "path1"]
awe <- data$path1

are the same when dealing with columns and the second can be more convenient because you can search for autocomplete by hitting tab after typing the $

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