How to convert a >2 column dataframe to 2 columns - r

Question

I am having problems transforming my data.
I have a dataframe, which tells which transitions were made, and how many times this sequence of transitions has occured. The different columns are corresponding with the situation in period 10, 11 and 12 (and there are more in my data). I want to summarize this data, and want to know how many times people went from A to C, A to D, but also C to G, etc. So basically I want to aggregate this data based on each column with the second column. My ultimate goal is to turn this data into a sankey diagram.
To illustrate:
df<-data.frame(s10=unlist(strsplit("AAAABBBBBC","")),
s11=unlist(strsplit("CCDDEEFFFF","")),
s12=unlist(strsplit("GHIGJKMNNN","")),
freq=c(10,20,30,40,50,60,70, 40, 20, 20))
s10 s11 s12 freq
1 A C G 10
2 A C H 20
3 A D I 30
4 A D G 40
5 B E J 50
6 B E K 60
7 B F M 70
8 B F N 40
9 B F N 20
10 C F N 20
And I aim to get this result:
colA colB freq
1 A C 30
2 A D 70
3 B E 110
4 B F 130
5 C F 20
6 C G 10
7 C H 20
8 D G 40
9 D I 30
10 E J 50
11 E K 60
12 F M 70
13 F N 80
I got this result by first aggregating the sum of frequencies for s10 and s11, and for s11 and s12, and then changing the column names and binding them together. It works for, but I intend to do this with more columns, so I'm sure there is a more efficient way to do this. See the code I used below:
bl1 <- df %>%
group_by(s10, s11) %>%
summarise(freq = sum(freq)) %>%
as.data.frame()
bl2 <- df %>%
group_by(s11, s12) %>%
summarise(freq = sum(freq)) %>%
as.data.frame()
colnames(bl1) <- c('colA', 'colB','freq' )
colnames(bl2) <- c('colA', 'colB','freq' )
rbind(bl1, bl2)
Any help is much appreciated!

Answer 1

You can rbind the selected columns of the data.frames together and then use aggregate. The only trick is to rename the columns so that they match. For this, I use setNames.
aggregate(freq ~ colA + colB,
data=rbind(setNames(df[c("s10", "s11", "freq")], c("colA", "colB", "freq")),
setNames(df[c("s11", "s12", "freq")], c("colA", "colB", "freq"))),
FUN=sum)
this returns the desired result.
colA colB freq
1 A C 30
2 A D 70
3 B E 110
4 B F 130
5 C F 20
6 C G 10
7 D G 40
8 C H 20
9 D I 30
10 E J 50
11 E K 60
12 F M 70
13 F N 80

Answer 2

OK. I gave it a go and had some fun with benchmarking. An alternate aproach (that I used) was to use aggregate() itself. See fun1 for implementation. I have made it to fit this particular example, and of course it will need tweaking to work with data frames of other widths
Edit: I have removed dataframe creation from functions and added Benchmarking output1
require(dplyr); require(microbenchmark)
df<-data.frame(s10=unlist(strsplit("AAAABBBBBC","")),
s11=unlist(strsplit("CCDDEEFFFF","")),
s12=unlist(strsplit("GHIGJKMNNN","")),
freq=c(10,20,30,40,50,60,70, 40, 20, 20))
fun0<- function(){
bl1 <- df %>%
group_by(s10, s11) %>%
summarise(freq = sum(freq)) %>%
as.data.frame()
bl2 <- df %>%
group_by(s11, s12) %>%
summarise(freq = sum(freq)) %>%
as.data.frame()
colnames(bl1) <- c('colA', 'colB','freq' )
colnames(bl2) <- c('colA', 'colB','freq' )
return(rbind(bl1, bl2))
}
fun1<- function(){
a<- apply(df[,c(1,2)], 1, function(x)paste(x[1],x[2], sep="",collapse = "" ))
b<- apply(df[,c(2,3)], 1, function(x)paste(x[1],x[2], sep="",collapse = "" ))
z<-data.frame(x=c(a,b),f=rep(df$freq,2))
return( aggregate( f~x , data=z, FUN=sum) )
}
fun0()
fun1()
#benchmarking
MB_res <- microbenchmark( fun0=fun0(), fun1=fun1() , times=1000)
MB_res
Results were:
Unit: milliseconds
expr min lq mean median uq max neval
fun0 2.218889 2.587820 2.773454 2.676921 2.785586 6.020277 1000
fun1 1.472971 1.737751 1.908966 1.842152 1.910118 8.915407 1000