使用列名称中的时间值重塑R中的数据

Question

I have a data frame which looks like this (simplified):

     data1.time1 data1.time2 data2.time1 data2.time2 data3.time1 group
 1          1.53        2.01        6.49        5.22        3.46    A
 ...
 24         2.12        3.14        4.96        4.89        3.81    C

where there are actually dataK.timeT for K in 1..27 and T in some (but maybe not all) of 1..8.

I would like to rearrange the data into K data frames so that I can plot, for each K, the summary data (for now let's say mean and mean ± standard deviation) for each of the three groups A, B, and C. That is, I want 27 graphs with three lines per graph, and also marks for the deviations.

Once I rearrange the data it should be easy enough to collapse by group, compute summary statistics, etc. But I'm not really sure how to get the data into this form. I looked at the reshape package, which suggests melting it into a key-value store format and rearranging from there, but it doesn't seem to support the columns containing the T values as I have here.

Is there a good way to do this? I'm quite willing to use something other than R to do this, since I can just import the results into R after transforming.

Answer 1

After creating fake data with a structure similar to yours, we convert from wide to long format, making a "tidy" data frame that is ready for plotting with ggplot2 .

library(reshape2)
library(ggplot2)
library(dplyr)

Create fake data

set.seed(194)
dat = data.frame(replicate(27*8, cumsum(rnorm(24*3))))

names(dat) = paste0(rep(paste0("data",1:27), each=8), ".", rep(paste0("time",1:8), 27))

dat$group = rep(LETTERS[1:3], each=24)

Remove some columns so that number of time points will be different for different data sources:

dat = dat[ , -c(2,4,9,43,56,78,100:103,115:116,134:136,202,205)]

Reshape from wide to long format

datl = melt(dat, id.var="group")

Split data source and time point into separate columns:

datl$source = gsub("(.*)\\..*","\\1", datl$variable)
datl$time = as.numeric(gsub(".*time(.*)","\\1", datl$variable))

# Order data frame names by number (rather than alphabetically)
datl$source = factor(datl$source, levels=paste0("data",1:length(unique(datl$source))))

Plot the data using ggplot2

# Helper function for plotting standard deviation
sdFnc = function(x) {
  vals = c(mean(x) - sd(x), mean(x) + sd(x))
  names(vals) = c("ymin", "ymax")
  vals
}

pd = position_dodge(0.7)

ggplot(datl, aes(time, value, group=group, color=group)) + 
  stat_summary(fun.y=mean, geom="line", position=pd) +
  stat_summary(fun.data=sdFnc, geom="errorbar", width=0.4, position=pd) +
  stat_summary(fun.y=mean, geom="point", position=pd) +
  facet_wrap(~source, ncol=3) +
  theme_bw()

Original (unnecessarily complicated) reshaping code. (Note, this code will no longer work with the updated (fake) data set, because the number of time columns is no longer uniform):

# Convert data source from wide to long
datl = data.frame()
for (i in seq(1,27*8,8)) {

  tmp.dat = dat[, c(i:(i+7),grep("group",names(dat)))]
  tmp.dat$source = gsub("(.*)\\..*", "\\1", names(tmp.dat)[1])
  names(tmp.dat)[1:8] = 1:8

  #datl = rbind(datl, tmp.dat)
  datl = bind_rows(datl, tmp.dat)  # Updated based on comment
}

datl$source = factor(datl$source, levels=paste0("data",1:27))

# Convert time from wide to long
datl = melt(datl, id.var = c("source","group"), variable.name="time")

Answer 2

Could do something like this with dplyr:

for(i in 1:K){ ## for 1:27
  my.data.ind <- paste0("data",i,"|group") ## "datai|group"
  one.month <- select(data, contains(my.data.ind) %>% ## grab cols that have these
                  group_by(group) %>% ## group by your group
                  summarise_each(funs(mean), funs(sd)) ## find mean for each col within each group
}

That should leave you with a 3xT data frame that has the average value of each group over time T