How to add rows to empty data frames with header in R? [duplicate]

Answer 1

Colnames <- names(compData)
compData <- rbind(compData, c(5, 443))
names(compData) <- Colnames

Answer 2

If you have data of the same type*, you can do the following:

1. Convert actual data frame to a matrix.
   as.matrix(compData)
2. Add the new row to the end.
   rbind(as.matrix(compData), c(5,443))
3. Convert the matrix back to a data frame.
   as.data.frame(rbind(as.matrix(compData), c(5,443)))

In short:
compData <- as.data.frame(rbind(as.matrix(compData), c(5,443)))

*If you have data of the same type, you might want to keep them in a matrix.

Answer 3

You can assign to the dataframe by index:

compData <- data.frame(A= numeric(0), B= numeric(0))
compData
compData[1, ] <- c(5, 443)
compData

Which gives:

> compData <- data.frame(A= numeric(0), B= numeric(0))
> compData
[1] A B
<0 rows> (or 0-length row.names)
> compData[1, ] <- c(5, 443)
> compData
  A   B
1 5 443

Answer 4

Adding to a zero-row data.frame will act differently to adding to an data.frame that already contains rows

From ?rbind

The rbind data frame method first drops all zero-column and zero-row arguments. (If that leaves none, it returns the first argument with columns otherwise a zero-column zero-row data frame.) It then takes the classes of the columns from the first data frame, and matches columns by name (rather than by position). Factors have their levels expanded as necessary (in the order of the levels of the levelsets of the factors encountered) and the result is an ordered factor if and only if all the components were ordered factors. (The last point differs from S-PLUS.) Old-style categories (integer vectors with levels) are promoted to factors.

You have a number of options --

the most straightforward

 compData[1, ] <- c(5, 443)

more complicated

Or you could coerce c(5,433) to a list or data.frame

rbind(compData,setNames(as.list(c(5,443)), names(compData)))

or

rbind(compData,do.call(data.frame,setNames(as.list(c(5,443)), names(compData))))

But in this case you might as well do

do.call(data.frame,setNames(as.list(c(5,443)), names(compData)))

data.table option

You could use the data.table function rbindlist which does less checking and thus preserves the names of the first data.frame

library(data.table)
rbindlist(list(compData, as.list(c(5,443))

Answer 5

I just got a simpler way to do it... as follows

compData <- data.frame(A= numeric(0), B= numeric(0))
compData
compData[nrow(compData)+1, ] <- c(5, 443)
compData

Answer 6

You can use the function structure with the .Names argument:

compData <- structure(rbind(compData,c(5,443)), .Names = names(compData))

#  A   B
#1 5 443