# solutions: R problem set for 140.778 # # Part A # load data expr <- read.table("expr.csv",sep=",",header=T,na.strings=c(".","-99")) biol <- read.table("biol.csv",sep=",",header=T,na.strings=c(".","-99")) # matrix, list or data.frame? is.matrix(expr);is.list(expr);is.data.frame(expr) is.matrix(biol);is.list(biol);is.data.frame(biol) # columns = numeric or factor (or both)? sapply(biol,mode) sapply(biol,is.factor) sapply(biol,is.numeric) sapply(expr,mode) sapply(expr,is.factor) sapply(expr,is.numeric) # find rows in biol with at least one NA biol[apply(biol,1,function(a) any(is.na(a))),] # find means of each column in expr sapply(expr,mean,na.rm=T) # make sample number the row names dimnames(expr)[[1]] <- as.character(expr\$sample) expr <- expr[,-1] sapply(expr,mean,na.rm=T) sapply(expr,sd,na.rm=T) sapply(expr,range,na.rm=T) # correlation matrix round(cor(expr,use="complete.obs"),2) round(cor(expr,use="pairwise.complete.obs"),2) # subtract mean of first two columns from each of the other columns expr <- expr[,-(1:2)] - apply(expr[,1:2],1,mean,na.rm=T) # scatterplot matrix pairs(expr) # fix up the biol data.frame biol\$f3 <- factor(substring(biol\$sample,1,1)) dimnames(biol)[[1]] <- substring(biol\$sample,2) biol <- biol[,-1] # check that the rownames for biol and expr are the same, just in different orders all(sort(dimnames(biol)[[1]]) == sort(dimnames(expr)[[1]])) # sort the rows in biol and expr by their rownames biol <- biol[sort(dimnames(biol)[[1]]),] expr <- expr[sort(dimnames(expr)[[1]]),] # find mean & SD of each column in expr apply(expr,2,function(a,b) tapply(a,b,function(x) c(mean(x,na.rm=T), sd(x,na.rm=T))),biol\$f1) # t.test comparing groups defined by biol\$f2 for each column of expr library(ctest) apply(expr,2,function(a,b) { x <- split(a,b) t.test(x[[1]],x[[2]])\$p.value }, biol\$f2)