change data.frame() behavior-CodePudding

# example data
df <- structure(c(-1, 0, 0, -1, -1, 2), .Dim = 2:3)

# how it looks like
df

r$> df
     [,1] [,2] [,3]
[1,]   -1    0   -1
[2,]    0   -1    2

# construct a data.frame using two rows of df
data.frame(df[1:2, ])

r$> data.frame(df[1:2,])
  X1 X2 X3
1 -1  0 -1
2  0 -1  2

# construct a data.frame using only one row of df
data.frame(df[1, ])

r$> data.frame(df[1, ])
  df.1...
1      -1
2       0
3      -1

data.frame(df[1:2,]) has the same structure as df, but data.frame(df[1, ]) has not.

Is it possible to let data.frame(df[1, ]) create a data frame with only one row, but not one column?

CodePudding user response：

Let us call the input m instead of df because it is a matrix and not a data.frame.

When indexing a matrix there is an optional drop= argument which defaults to TRUE which means that the result of a scalar index will be to drop the dimensions and so convert the result to an ordinary vector. drop has no effect if the index has length > 1. In that case we always get a matrix. Stated in another way by default or if drop = TRUE then we get a vector if the index is scalar and a matrix otherwise. If drop = FALSE then we always get a matrix result. Thus, specify that drop=FALSE.

m <- structure(c(-1, 0, 0, -1, -1, 2), .Dim = 2:3)

m[1, ] # vector
## [1] -1  0 -1

m[1,, drop = TRUE] # same
## [1] -1  0 -1

m[1,, drop = FALSE] # matrix
##      [,1] [,2] [,3]
## [1,]   -1    0   -1

m[1:2,, drop = FALSE] # matrix
##      [,1] [,2] [,3]
## [1,]   -1    0   -1
## [2,]    0   -1    2

Now when we apply data.frame we get a single row.

data.frame( m[1,, drop = FALSE] )
##   X1 X2 X3
## 1 -1  0 -1

data.frame(m[1:2,, drop = FALSE])
##   X1 X2 X3
## 1 -1  0 -1
## 2  0 -1  2

If we already have a vector so that it is too late to use drop = FALSE then we can use rbind to turn it into a matrix.

data.frame(rbind(m[1, ]))
##   X1 X2 X3
## 1 -1  0 -1

data.frame(rbind(m[1:2, ]))
##   X1 X2 X3
## 1 -1  0 -1
## 2  0 -1  2

CodePudding user response：

A possible solution:

df <- structure(c(-1, 0, 0, -1, -1, 2), .Dim = 2:3)

data.frame(t(df[1, ]))

#>   X1 X2 X3
#> 1 -1  0 -1

CodePudding user response：

The issue is that selecting just one row from a matrix returns a vector, not a matrix:

str(df[1:2, ])
#>  num [1:2, 1:3] -1 0 0 -1 -1 2
str(df[1, ])
#>  num [1:3] -1 0 -1

You would have to turn it into a matrix again, to get your expected behavior:

df[1, ] |> 
  matrix(nrow = 1) |> 
  data.frame()
#>   X1 X2 X3
#> 1 -1  0 -1

^{Created on 2022-02-05 by the reprex package (v2.0.1)}