R:在data.frame列中拆分不平衡列表

时间:2011-01-18 14:00:56

标签: r dataframe plyr

假设您有一个具有以下结构的数据框:

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data.frame将是:

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近2个小时(170K行):

# 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))

这似乎是一个尺寸问题,因为如果我运行

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

只需9秒钟。首先使用sapply(使用不同的函数)转换为一组data.frames很快,但所需的`rbind'需要更长的时间。

2 个答案:

答案 0 :(得分:12)

#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)

答案 1 :(得分:6)

我的“splitstacksahpe”软件包中的

cSplit旨在处理这种数据操作。

这是针对这个问题的行动:

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

# install.packages("splitstackshape")
library(splitstackshape)
cSplit(df, "b", ";", "long", makeEqual = FALSE)
#    a b_new
# 1: 1  job1
# 2: 1  job2
# 3: 2 job1a
# 4: 3  job4
# 5: 3  job5
# 6: 3  job6
# 7: 4  job9
# 8: 4 job10
# 9: 4 job11

您也可以在“dplyr”中使用strsplit,然后使用“tidyr”中的unnest进行跟进,如下所示:

library(dplyr)
library(tidyr)
df %>% 
  mutate(b = strsplit(as.character(b), ";", fixed = TRUE)) %>% 
  unnest(b)
#   a     b
# 1 1  job1
# 2 1  job2
# 3 2 job1a
# 4 3  job4
# 5 3  job5
# 6 3  job6
# 7 4  job9
# 8 4 job10
# 9 4 job11