将逗号分隔的条目转换为列

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.

我有一个包含多个列的数据集，其中一列是反应时间的列。这些反应时间以逗号分隔，表示不同试验的反应时间（同一参与者）。

For example: row 1 (i.e.: the data from participant 1) has the following under the column "reaction times"

例如：第1行（即：来自参与者1的数据）在“反应时间”栏下有以下内容

reaction_times
2000,1450,1800,2200

Hence these are the reaction times of participant 1 for trials 1,2,3,4.

因此，这些是参与者1对试验1,2,3,4的反应时间。

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.

我现在想要创建一个新的数据集，其中这些试验的反应时间都形成单独的列。这样我就可以计算每次试验的平均反应时间。

              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).

我尝试了“reshape2”-package中的“colsplit”，但这似乎并没有将我的数据拆分成新列（可能是因为我的数据都在1个单元格中）。

Any suggestions?

有什么建议么？

4 个解决方案

#1

I think you are looking for the strsplit() function;

我想你正在寻找strsplit（）函数;

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:

请注意，strsplit返回一个列表，在这种情况下只有一个元素。这是因为strsplit将向量作为输入。因此，您还可以将单个单元格字符的长向量放入函数中，并返回该向量的拆分列表。在一个更相关的例子中，这看起来像：

# 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:

最后，计算每人的平均值：

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

#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:

一个漂亮的，如果相当沉重的方式是将read.csv与textConnection结合使用。假设您的数据在数据框中，df：

x <- read.csv(textConnection(df[["reaction times"]]))

#3

Old question, but I came across it from another recent question (which seems unrelated).

老问题，但我从另一个最近的问题（似乎无关）中遇到了它。

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.

两个现有的答案都是合适的，但我想分享一个与我创建的名为“splitstackshape”的软件包相关的答案，该软件包很快且语法简单明了。

Here's some sample data:

这是一些示例数据：

set.seed(1)
dat = data.frame(
  reaction_time = apply(matrix(round(
    runif(24, 1, 2000)), 6, 4), 1, paste, collapse = ","))

This is the splitting:

这是分裂：

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:

并且，可选地，如果您需要使用rowMeans：

rowMeans(cSplit(dat, "reaction_time", ","))
# [1] 1139.00 1097.75 1453.50  840.75  889.25 1096.75

#4

Another option using dplyr and tidyr with Paul Hiemstra's example data is:

使用dplyr和tidyr与Paul Hiemstra的示例数据的另一个选项是：

# 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)

#1