Convert comma separated string to numeric columns - r

Question

I have a dataset with several columns, one of which is a column for reaction times. These reaction times are comma separated to denote the reaction times (of the same participant) for the different trials.
For example: row 1 (i.e.: the data from participant 1) has the following under the column "reaction times"
reaction_times
2000,1450,1800,2200
Hence these are the reaction times of participant 1 for trials 1,2,3,4.
I now want to create a new data set in which the reaction times for these trials all form individual columns. This way I can calculate the mean reaction time for each trial.
trial 1 trial 2 trial 3 trial 4
participant 1: 2000 1450 1800 2200
I tried the colsplit from the reshape2 package but that doesn't seem to split my data into new columns (perhaps because my data is all in 1 cell).
Any suggestions?

Answer 1

I think you are looking for the strsplit() function;
a = "2000,1450,1800,2200"
strsplit(a, ",")
[[1]]
[1] "2000" "1450" "1800" "2200"
Notice that strsplit returns a list, in this case with only one element. This is because strsplit takes vectors as input. Therefore, you can also put a long vector of your single cell characters into the function and get back a splitted list of that vector. In a more relevant example this look like:
# Create some example data
dat = data.frame(reaction_time =
apply(matrix(round(runif(100, 1, 2000)),
25, 4), 1, paste, collapse = ","),
stringsAsFactors=FALSE)
splitdat = do.call("rbind", strsplit(dat$reaction_time, ","))
splitdat = data.frame(apply(splitdat, 2, as.numeric))
names(splitdat) = paste("trial", 1:4, sep = "")
head(splitdat)
trial1 trial2 trial3 trial4
1 597 1071 1430 997
2 614 322 1242 1140
3 1522 1679 51 1120
4 225 1988 1938 1068
5 621 623 1174 55
6 1918 1828 136 1816
and finally, to calculate the mean per person:
apply(splitdat, 1, mean)
[1] 1187.50 361.25 963.75 1017.00 916.25 1409.50 730.00 1310.75 1133.75
[10] 851.25 914.75 881.25 889.00 1014.75 676.75 850.50 805.00 1460.00
[19] 901.00 1443.50 507.25 691.50 1090.00 833.25 669.25

Answer 2

A nifty, if rather heavy-handed, way is to use read.csv in conjunction with textConnection. Assuming your data is in a data frame, df:
x <- read.csv(textConnection(df[["reaction times"]]))

Answer 3

Old question, but I came across it from another recent question (which seems unrelated).
Both existing answers are appropriate, but I wanted to share an answer related to a package I have created called "splitstackshape" that is fast and has straightforward syntax.
Here's some sample data:
set.seed(1)
dat = data.frame(
reaction_time = apply(matrix(round(
runif(24, 1, 2000)), 6, 4), 1, paste, collapse = ","))
This is the splitting:
library(splitstackshape)
cSplit(dat, "reaction_time", ",")
# reaction_time_1 reaction_time_2 reaction_time_3 reaction_time_4
# 1: 532 1889 1374 761
# 2: 745 1322 769 1555
# 3: 1146 1259 1540 1869
# 4: 1817 125 996 425
# 5: 404 413 1436 1304
# 6: 1797 354 1984 252
And, optionally, if you need to take the rowMeans:
rowMeans(cSplit(dat, "reaction_time", ","))
# [1] 1139.00 1097.75 1453.50 840.75 889.25 1096.75

Answer 4

Another option using dplyr and tidyr with Paul Hiemstra's example data is:
# create example data
data = data.frame(reaction_time =
apply(matrix(round(runif(100, 1, 2000)),
25, 4), 1, paste, collapse = ","),
stringsAsFactors=FALSE)
head(data)
# clean data
data2 <- data %>% mutate(split_reaction_time = str_split(as.character(reaction_time), ",")) %>% unnest(split_reaction_time)
data2$col_names <- c("trial1", "trial2", "trial3", "trial4")
data2 <- data2 %>% spread(key = col_names, value = split_reaction_time) %>% select(-reaction_time)
head(data2)