variable
variable

Reputation: 1023

Apply function to every column in data frame operating on names

I'm trying to write something that will take a data frame and plot every numeric column, with simple plots for non-normal distributions and control charts for normal ones:

library(plyr)
library(qcc)
library(ggplot2)

#generate data frame
data <- data.frame(seq_len(10),LETTERS[seq_len(10)],rnorm(10,5,3),rep(1,10),rep(2,10),rnorm(10,3,1),runif(10))

##checks heterogeneity
has_range <- function(data) { if(all( abs(data - mean(data)) == 0)) FALSE else TRUE}

##test for normality
normtest <- function(data) {if(has_range(data) == FALSE) FALSE else {
  if(shapiro.test(data)$p.value < 0.05) FALSE else TRUE}}

##Control charts for Normal data, simple plots otherwise
drawplot<-function(data, ref=NULL) {
  Sys.sleep(.1)
  print(names(data))
  if(normtest(data) == FALSE) {
    plot(x=ref, y=data, ylab=names(data))
  } else {
    qcc(data,type="xbar.one", labels=ref, ylab=names(data))  
  }
}

## Apply drawplot to all numeric columns in data frame
colwise(drawplot, is.numeric, ref=data[[2]])(data)

The problem is that every apply family function seems to strip column names and I can't use the column names to label the plots:

print(names(data))

Gives NULL results.

Also there's a seemingly unrelated error that is cropping up:

Error: length(rows) == 1 is not TRUE

Upvotes: 2

Views: 1239

Answers (1)

mnel
mnel

Reputation: 115382

You need to create a function that uses the names, otherwise the names won't be accessible within the function.

You can't pass x = NULL to plot, so I've rewritten a bit of your function (qcc was kicking up a fuss with an atomic vector for x as well)

Something like

drawplot<-function(n, data, ref=NULL) {
  Sys.sleep(.1)
  print(n)


  if(normtest(data[[n]]) == FALSE) {
    if(is.null(ref)){ref <- seq_along(data[[n]])}
    plot(x=ref, y=data[[n]], ylab=n)
  } else {
    qcc(data[,n, drop=FALSE], type="xbar.one", labels=ref,ylab = n)  
  }
}

lapply(names(Filter(is.numeric,dd)), drawplot, data = dd)

Note that this function would work with position indexing as well (but the labels wouldn't be as pretty)

Upvotes: 2

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