Replicate each row of data.frame and specify the number of replications for each row?

I am programming in R and I got the following problem:

I have a data String jb, that is quite long. Heres a simple version of it:

jb:    a     b     frequency               jb.expanded: a    b   
       5     3        2                                 5    3
       5     7        1                                 5    3
       9     1        40                                5    7
       12    4        5                                 9    1
       12    5        13                                9    1
                                                        ...  ...

I want to replicate the rows and the frequency of the replication is the column frequency. That means, the first row is replicated two times, the second row is replicated 1 time and so on. I already solved that problem with the code

jb.expanded <- jb[rep(row.names(jb), jb$freqency), 1:2]

Now here is the problem:

Whenever any number in the frequency corner is greater than 10, the number of replicated columns is wrong. For example:

Frequency: 43 --> 14 columns
           40 --> 13 columns
           13 --> 11 columns
           14 --> 12 columns

Can you help me? I have no idea how to fix that, I also cannot find anything on the internet.

Thanks for your help!

Solution 1:

Update

Upon revisiting this question, I have a feeling that @Codoremifa was correct in their assumption that your "frequency" column might be a factor.

Here's an example if that were the case. It won't match your actual data since I don't know what other levels are in your dataset.

mydf$F2 <- factor(as.character(mydf$frequency))
## expandRows(mydf, "F2")
mydf[rep(rownames(mydf), mydf$F2), ]
#      a b frequency F2
# 1    5 3         2  2
# 1.1  5 3         2  2
# 1.2  5 3         2  2
# 2    5 7         1  1
# 3    9 1        40 40
# 3.1  9 1        40 40
# 3.2  9 1        40 40
# 3.3  9 1        40 40
# 4   12 4         5  5
# 4.1 12 4         5  5
# 4.2 12 4         5  5
# 4.3 12 4         5  5
# 4.4 12 4         5  5
# 5   12 5        13 13
# 5.1 12 5        13 13

Hmmm. That doesn't look like 61 rows to me. Why not? Because rep uses the numeric values underlying the factor, which is quite different in this case from the displayed value:

as.numeric(mydf$F2)
# [1] 3 1 4 5 2

To properly convert it, you would need:

as.numeric(as.character(mydf$F2))
# [1]  2  1 40  5 13

Original answer

A while ago I wrote a function that is a bit more of a generalization of @Simono101's answer. The function looks like this:

expandRows <- function(dataset, count, count.is.col = TRUE) {
  if (!isTRUE(count.is.col)) {
    if (length(count) == 1) {
      dataset[rep(rownames(dataset), each = count), ]
    } else {
      if (length(count) != nrow(dataset)) {
        stop("Expand vector does not match number of rows in data.frame")
      }
      dataset[rep(rownames(dataset), count), ]
    }
  } else {
    dataset[rep(rownames(dataset), dataset[[count]]), 
            setdiff(names(dataset), names(dataset[count]))]
  }
}

For your purposes, you could just use expandRows(mydf, "frequency")

head(expandRows(mydf, "frequency"))
#     a b
# 1   5 3
# 1.1 5 3
# 2   5 7
# 3   9 1
# 3.1 9 1
# 3.2 9 1

Other options are to repeat each row the same number of times:

expandRows(mydf, 2, count.is.col=FALSE)
#      a b frequency
# 1    5 3         2
# 1.1  5 3         2
# 2    5 7         1
# 2.1  5 7         1
# 3    9 1        40
# 3.1  9 1        40
# 4   12 4         5
# 4.1 12 4         5
# 5   12 5        13
# 5.1 12 5        13

Or to specify a vector of how many times to repeat each row.

expandRows(mydf, c(1, 2, 1, 0, 2), count.is.col=FALSE)
#      a b frequency
# 1    5 3         2
# 2    5 7         1
# 2.1  5 7         1
# 3    9 1        40
# 5   12 5        13
# 5.1 12 5        13

Note the required count.is.col = FALSE argument in those last two options.

Solution 2:

Nearly. You want to pass [ a vector of row indices, not row.names. Try this...

jb[ rep( seq_len( nrow(jb) ) , times = jb$frequency ) , ]

rep( seq_len( nrow(jb) ) , times = jb$frequency ) 
# [1] 1 1 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
# [39] 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5

Solution 3:

This might be more of a comment but seeing that all the other answers are suggesting new options - if you correct the spelling of jb$freqency when creating jb.expanded, and convert jb$frequency to an integer then the construction you mention in your question also works.

And why I suspect jb$frequency is a factor is because the incorrect frequencies are neatly ordered as 11,12,13,14.