How to convert multiple columns to individual rows in R

Question

I have a data frame in R that has many rows (over 3000) with F0 (fundamental frequency) tracks of an utterance in it. The rows have the following information in them: speaker ID, group #, repetition #, accent type, sex, and then 50 columns of F0 points. The data looks like this:

Speaker Sex Group Repetition Accent    Word         1         2         3        4
    105   M     1          1      N AILMENT 102.31030 102.31030 102.31030 102.31127 
    105   M     1          1      N COLLEGE 111.80641 111.80313 111.68612 111.36020
    105   M     1          1      N  FATHER 124.06655 124.06655 124.06655 124.06655 

But instead of only going to X4, it has 50 points per row, so I have a 3562x56 data frame. I want to change it so each column of data in the F0 track (so after word, from 1:50) gets its own column, with the associated column number as another row. I want to keep all of the information in the first six columns with each data point as well, so it would look like this:

Speaker Sex Group Repetition Accent    Word       Num        F0
    105   M     1          1      N AILMENT         1 102.31030
    105   M     1          1      N AILMENT         2 102.31030
    105   M     1          1      N AILMENT         3 102.31030
    105   M     1          1      N AILMENT         4 102.31127
    ...
    105   M     1          1      N COLLEGE         1 111.80641 
    105   M     1          1      N COLLEGE         1 111.80313 
    105   M     1          1      N COLLEGE         1 111.68612 
    105   M     1          1      N COLLEGE         1 111.36020 
    ...

The code I tried to use, while tedious, is as follows:

x = 1
for (i in 1:dim(normrangef0)[1]) {
     for (j in 1:50) {
             norm.all$Speaker[x] <- normrangef0$Speaker[i]
             norm.all$Sex[x] <- normrangef0$Sex[i]
             norm.all$Group[x] <- normrangef0$Group[i]
             norm.all$Repetition[x] <- normrangef0$Repetition[i]
             norm.all$Word[x] <- normrangef0$Word[i]
             norm.all$Accent[x] <- normrangef0$Accent[i]
             norm.all$Time[x] <- j
             norm.all$F0[x] <- normrangef0[i,j+6]
             x = x+1    
    }
}

However, when I do this with norm.all as a NULL object (just defined by norm.all = c() ), I end up with a list of over 200k items, many of which are NAs. When I define norm.all as a data frame (either an empty one or one of all 0s, in the 178100x8 data frame, I get an error:

Error in $<-.data.frame(*tmp*, "Speaker", value = 105L) : replacement has 1 row, data has 0

Is my code just totally off? Is there another way to do this?

Answer 1

Use melt from "reshape2"

library(reshape2)
melt(mydf, id.vars=c("Speaker", "Sex", "Group", "Repetition", "Accent", "Word"))
#    Speaker Sex Group Repetition Accent    Word variable    value
# 1      105   M     1          1      N AILMENT        1 102.3103
# 2      105   M     1          1      N COLLEGE        1 111.8064
# 3      105   M     1          1      N  FATHER        1 124.0666
# 4      105   M     1          1      N AILMENT        2 102.3103
# 5      105   M     1          1      N COLLEGE        2 111.8031
# 6      105   M     1          1      N  FATHER        2 124.0666
# 7      105   M     1          1      N AILMENT        3 102.3103
# 8      105   M     1          1      N COLLEGE        3 111.6861
# 9      105   M     1          1      N  FATHER        3 124.0666
# 10     105   M     1          1      N AILMENT        4 102.3113
# 11     105   M     1          1      N COLLEGE        4 111.3602
# 12     105   M     1          1      N  FATHER        4 124.0666

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In base R, you can also use stack to stack the columns named 1 through 4, and cbind that with the first group of columns. Alternatively, unlist will also do this.

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You may also want to look into the "data.table" package to get a bit of a speed boost.