aggregating columns based on commas

Question

I have the following dataframe and I'm trying to separate the commas and turn that particular name(s) into their own individual columns and specify if that particular column names exist (which are separated by commas) for that particular ID. (1 = Yes, 0 = No) Any help would be appreciated! Thanks!

ID<- c(1,2,3,4,5,6)
Details<- c("V1,V2", "V1,V3", "V1", "V2", "V3,V4", "V2,V3" )

data.frame <- data.frame(ID, Details, stringsAsFactors=FALSE)

DESIRED OUTPUT:

ID<-c(1,2,3,4,5,6)
V1<-c(1,1,1,0,0,0)
V2<-c(1,0,0,1,0,1)
V3<-c(0,1,0,0,1,1)
V4<-c(0,0,0,0,1,0)

data.frame1<-data.frame(ID, V1, V2, V3, V4, stringsAsFactors=FALSE)

Answer 1

A solution using the tidyverse package. dat is your example data frame. dat2 is the final data frame.

library(tidyverse)

dat2 <- dat %>%
  separate_rows(Details) %>%
  mutate(Value = 1L) %>%
  spread(Details, Value, fill = 0L)
dat2
#   ID V1 V2 V3 V4
# 1  1  1  1  0  0
# 2  2  1  0  1  0
# 3  3  1  0  0  0
# 4  4  0  1  0  0
# 5  5  0  0  1  1
# 6  6  0  1  1  0

Answer 2

One option with mtabulate from qdapTools

library(qdapTools)
cbind.data.frame(ID, # or data.frame$ID
                 mtabulate(strsplit(as.character(data.frame$Details), ",")))
# output
  ID V1 V2 V3 V4
1  1  1  1  0  0
2  2  1  0  1  0
3  3  1  0  0  0
4  4  0  1  0  0
5  5  0  0  1  1
6  6  0  1  1  0

Answer 3

Here is a base R solution. I have renamed your data.frames data1 and data2 .

data1 <- data.frame(ID, Details, stringsAsFactors=FALSE)
data2 <- data.frame(ID, V1, V2, V3, V4, stringsAsFactors=FALSE)        

nms <- unique(unlist(strsplit(data1$Details, ",")))
data3 <- cbind.data.frame(ID, sapply(nms, grepl, data1$Details))
data3[-1] <- lapply(data3[-1], as.integer)

Now compare data3 with your expected result data2 .

all.equal(data2, data3)
#[1] TRUE

Note, however, that

identical(data2, data3)
#[1] FALSE

This is because I have used as.integer and the values in data2 are of class "numeric" . If this makes a difference, you can change the lapply instruction above to use as.numeric .

Answer 4

using base R:

 xtabs(val~.,cbind.data.frame(ID=rep(ID,lengths(s<-strsplit(Details,","))),Details=unlist(s),val=1))
   Details
ID  V1 V2 V3 V4
  1  1  1  0  0
  2  1  0  1  0
  3  1  0  0  0
  4  0  1  0  0
  5  0  0  1  1
  6  0  1  1  0

Answer 5

The most straightforward way I see is to would be to build a data.frame for each of these vectors hidden in strings and bind them. purrr can help to make it quite compact. Note that column ID isn't needed, I'll work on Details directly.

library(purrr)
df <- map_dfr(strsplit(Details, ","),
              ~data.frame(t(setNames(rep(1, length(.x)), .x))))
df[is.na(df)] <- 0

#   V1 V2 V3 V4
# 1  1  1  0  0
# 2  1  0  1  0
# 3  1  0  0  0
# 4  0  1  0  0
# 5  0  0  1  1
# 6  0  1  1  0

You could also split and unlist to get distinct values, and then look them up in the original vector:

unique_v <- unique(unlist(strsplit(Details, ",")))
map_dfc(unique_v, ~as.numeric(grepl(.x, Details)))
# # A tibble: 6 x 4
#      V1    V2    V3    V4
#   <dbl> <dbl> <dbl> <dbl>
# 1     1     1     0     0
# 2     1     0     1     0
# 3     1     0     0     0
# 4     0     1     0     0
# 5     0     0     1     1
# 6     0     1     1     0

We could do some dirty string evaluation also if you know the number of columns:

m <- as.data.frame(matrix(0,ncol=4,nrow=6))
eval(parse(text=paste0("m[",ID,", c(",gsub("V","",Details),")] <- 1")))
#   V1 V2 V3 V4
# 1  1  1  0  0
# 2  1  0  1  0
# 3  1  0  0  0
# 4  0  1  0  0
# 5  0  0  1  1
# 6  0  1  1  0