R：data.frame列中的分割不平衡列表 [英] R: Split unbalanced list in data.frame column

问题描述

假设你有一个具有以下结构的数据框架：

Suppose you have a data frame with the following structure:

df <- data.frame(a=c(1,2,3,4), b=c("job1;job2", "job1a", "job4;job5;job6", "job9;job10;job11"))

其中列 b 是以分号分隔的列表（按行不平衡）。理想的数据框架将是：

where the column b is a semicolon-delimited list (unbalanced by row). The ideal data.frame would be:

id,job,jobNum
1,job1,1
1,job2,2
...
3,job6,3
4,job9,1
4,job10,2
4,job11,3

我有一个部分解决方案需要将近2小时（170K行）：

I have a partial solution that takes almost 2 hours (170K rows):

# Split the column by the semicolon.  Results in a list.
df$allJobs <- strsplit(df$b, ";", fixed=TRUE)

# Function to reshape column that is a list as a data.frame
simpleStack <- function(data){
    start <- as.data.frame.list(data)                       
    names(start) <-c("id", "job")
    return(start)
}
# pylr!
system.time(df2 <- ddply(df, .(id), simpleStack))

这似乎是一个大小问题，因为如果我运行

It appears to be a size issue, because if I run

system.time(df2 <- ddply(df[1:4000,c("id", "allJobs")], .(id), simpleStack))

只需要9秒钟。首先将数据转换成一组数据。（具有不同的功能）是快速的，但所需的`rbind'需要更长的时间。

it only takes 9 seconds. First converting to a set of data.frames with sapply (with a different function) is fast, but the required `rbind' takes even longer.

推荐答案

#Split by ; as before
allJobs <- strsplit(df$b, ";", fixed=TRUE)

#Replicate a by the number of jobs in each case
n <- sapply(allJobs, length)
id <- rep(df$a, times = n)

#Turn allJobs into a vector
job <- unlist(allJobs)

#Retrieve position of each job
jobNum <- unlist(lapply(n, seq_len))

#Combine into a data frame
df2 <- data.frame(id = id, job = job, jobNum = jobNum)

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