Add column based on header cell; then remove header cell - r

Question

I have interesting data that is not uniform. A group of items are listed under the category name, but it is all in the same column. I need to add a column with the row corresponding to the item's category that it belongs to (then remove the category heading). The only way to distinguish a new category is determining whether the value under the year is empty.... My dputs should explain my issue more clearly.
Before:
structure(list(X = structure(c(13L, 1L, 19L, 16L, 5L, 17L, 11L,
8L, 2L, 10L, 4L, 6L, 18L, 15L, 21L, 12L, 14L, 9L, 3L, 20L, 7L
), .Label = c("-Burgers", "-Cameras", "-Shirts", "+Laptops",
"+Salads", "+TVs", "Caps", "Cell", "Clothes:", "Desktops", "Electronics",
"Flowers", "Food", "Garden Nomes", "Grills", "Hotdogs", "Nachoes",
"Outdoors:", "Pizza", "Shorts", "Swimming Gear"), class = "factor"),
X2000 = c(NA, 104L, 159L, 184L, 189L, 182L, NA, 49L, 28L,
46L, 34L, 43L, NA, 129L, 190L, 189L, 119L, NA, 45L, 80L,
80L), X2001 = c(NA, 147L, 192L, 164L, 174L, 196L, NA, 40L,
34L, 43L, 35L, 22L, NA, 114L, 130L, 120L, 145L, NA, 56L,
35L, 54L), X2002 = c(NA, 163L, 172L, 138L, 146L, 190L, NA,
38L, 40L, 21L, 22L, 33L, NA, 186L, 172L, 139L, 119L, NA,
88L, 78L, 91L), X2003 = c(NA, 125L, 152L, 182L, 148L, 125L,
NA, 36L, 44L, 34L, 27L, 50L, NA, 119L, 115L, 188L, 166L,
NA, 91L, 77L, 77L), X2004 = c(NA, 116L, 111L, 120L, 153L,
199L, NA, 49L, 48L, 43L, 37L, 32L, NA, 159L, 116L, 143L,
153L, NA, 18L, 53L, 51L)), .Names = c("X", "X2000", "X2001",
"X2002", "X2003", "X2004"), class = "data.frame", row.names = c(NA,
-21L))
After:
structure(list(X = structure(c(1L, 15L, 13L, 5L, 14L, 8L, 2L,
9L, 4L, 6L, 12L, 17L, 10L, 11L, 3L, 16L, 7L), .Label = c("-Burgers",
"-Cameras", "-Shirts", "+Laptops", "+Salads", "+TVs", "Caps",
"Cell", "Desktops", "Flowers", "Garden Nomes", "Grills", "Hotdogs",
"Nachoes", "Pizza", "Shorts", "Swimming Gear"), class = "factor"),
X.1 = structure(c(3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L,
4L, 4L, 4L, 4L, 1L, 1L, 1L), .Label = c("Clothes:", "Electronics",
"Food", "Outdoors:"), class = "factor"), X2000 = c(104L,
159L, 184L, 189L, 182L, 49L, 28L, 46L, 34L, 43L, 129L, 190L,
189L, 119L, 45L, 80L, 80L), X2001 = c(147L, 192L, 164L, 174L,
196L, 40L, 34L, 43L, 35L, 22L, 114L, 130L, 120L, 145L, 56L,
35L, 54L), X2002 = c(163L, 172L, 138L, 146L, 190L, 38L, 40L,
21L, 22L, 33L, 186L, 172L, 139L, 119L, 88L, 78L, 91L), X2003 = c(125L,
152L, 182L, 148L, 125L, 36L, 44L, 34L, 27L, 50L, 119L, 115L,
188L, 166L, 91L, 77L, 77L), X2004 = c(116L, 111L, 120L, 153L,
199L, 49L, 48L, 43L, 37L, 32L, 159L, 116L, 143L, 153L, 18L,
53L, 51L)), .Names = c("X", "X.1", "X2000", "X2001", "X2002",
"X2003", "X2004"), class = "data.frame", row.names = c(NA, -17L
))
The items are arbitrarily have + or - signs...I need that to remain the same. Also, some category headers have : while others do not.

Answer 1

We create an index based on the 'NA' values in columns other than the 1st ('indx'). We split the dataset using the 'indx', remove the first row i.e. NA values from columns 2nd to the last, cbind with the 1st row, 1st column value, rearrange the columns and rbind.
indx <- cumsum(!rowSums(!is.na(df1[-1])))
res <- do.call(rbind,lapply(split(df1, indx), function(x)
cbind(x, X.1= x[1,1])[-1,c(1,7,2:6)]))
row.names(res) <- NULL
all.equal(res, out, check.attributes=FALSE)
#[1] TRUE
where 'out' is the dput output of the expected result
Update
If the columns have '' instead of NA,
indx <- cumsum(!rowSums(df1[-1]!=''))
and do the rest as above. Having said that, when we have '' in a numeric column, the class will be either factor or character based on whether you specify stringsAsFactors=FALSE or =TRUE in the read.table/read.csv. So, keeping the '' as such will get the output also a factor/character class. I would convert the columns to their correct class first which will also coerce the '' to NA, i.e.
df1[-1] <- lapply(df1[-1], function(x) as.numeric(as.character(x)))
The as.character is only needed if the columns are factor class.
Once, we have done the conversion, the first approach should work fine as well.