R Define Dimensions of Empty Data Frame

R define dimensions of empty data frame

Just create a data frame of empty vectors:

collect1 <- data.frame(id = character(0), max1 = numeric(0), max2 = numeric(0))

But if you know how many rows you're going to have in advance, you should just create the data frame with that many rows to start with.

Create empty data frame with 200 rows and no columns

This can work (if the call to 2 functions is not considered 2 commands):

data.frame(matrix(, nrow=200, ncol=0))
#data frame with 0 columns and 200 rows

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Edit: Another option is data.frame()[1:200, ]:

data.frame()[1:200, ]
# data frame with 0 columns and 200 rows

Create an empty data.frame

Just initialize it with empty vectors:

df <- data.frame(Date=as.Date(character()),
                 File=character(), 
                 User=character(), 
                 stringsAsFactors=FALSE) 

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Here's an other example with different column types :

df <- data.frame(Doubles=double(),
                 Ints=integer(),
                 Factors=factor(),
                 Logicals=logical(),
                 Characters=character(),
                 stringsAsFactors=FALSE)

str(df)
> str(df)
'data.frame':   0 obs. of  5 variables:
 $ Doubles   : num 
 $ Ints      : int 
 $ Factors   : Factor w/ 0 levels: 
 $ Logicals  : logi 
 $ Characters: chr 

N.B. :

Initializing a data.frame with an empty column of the wrong type does not prevent further additions of rows having columns of different types.

This method is just a bit safer in the sense that you'll have the correct column types from the beginning, hence if your code relies on some column type checking, it will work even with a data.frame with zero rows.

Create empty data frame with column names by assigning a string vector?

How about:

df <- data.frame(matrix(ncol = 3, nrow = 0))
x <- c("name", "age", "gender")
colnames(df) <- x

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To do all these operations in one-liner:

setNames(data.frame(matrix(ncol = 3, nrow = 0)), c("name", "age", "gender"))

#[1] name   age    gender
#<0 rows> (or 0-length row.names)

Or

data.frame(matrix(ncol=3,nrow=0, dimnames=list(NULL, c("name", "age", "gender"))))

How to create an empty dataframe with specified columns AND number of rows in R?

If I understand the question correctly, the following hack will do what the OP wants. It creates the number of rows by setting the row.names attribute directly. And if a dataframe has row names, it must have the corresponding rows.

empty_df <- old_df[0, ]
attr(empty_df, 'row.names') <- 1:nrow(new_df)

str(empty_df)
#'data.frame':  300 obs. of  5 variables:
# $ Sepal.Length: num 
# $ Sepal.Width : num 
# $ Petal.Length: num 
# $ Petal.Width : num 
# $ Species     : Factor w/ 3 levels "setosa","versicolor",..:

The dataframe empty_df now has 300 rows.

Data creation code.

The test data creation code uses the built-in dataset iris.

set.seed(1234)

old_df <- iris
new_df <- rbind(iris, iris)
new_df <- new_df[, sample(ncol(new_df))]

how do you create an empty data frame using a function in R?

I think this does what you want:

do.call(data.frame,setNames(lapply(xv,function(e) vector(typeof(e))),xv));
## [1] Users        Transactions Workload
## <0 rows> (or 0-length row.names)

Edit: Thanks to @Zheyuan for making me think a little more about my solution. Since the input (as specified by the OP) is an atomic vector, it cannot contain heterogeneous data types, so my lapply() call that generates a separate zero-length vector according to each input element's type offers no benefit. It would offer a benefit if xv was a list, which can contain heterogeneous data types, but since xv is also being used to set the names of the resulting data.frame, it would be very questionable if it contained non-character elements. So my solution is actually not as sensible as I thought it was.

Here's a more sensible solution using the do.call(data.frame,...) pattern, which replaces the lapply() call with rep():

do.call(data.frame,setNames(rep(list(character()),length(xv)),xv));
## [1] Users        Transactions Workload
## <0 rows> (or 0-length row.names)

working with empty data frames in R

Here are 4 ways to grow your data.frame:

col1 <- letters[1:3] # [1] "a" "b" "c"
col2 <- letters[4:6] # [1] "d" "e" "f"

1- Start by assigning the first column

df1 <- data.frame(col1,stringsAsFactors = FALSE)
df1$col2 <- col2

2- Grow a list first, convert afterwards

l2 <- list()
l2$col1 <- col1
l2$col2 <- col2
df2 <- data.frame(l2,stringsAsFactors = FALSE)

3- Define the data.frame with columns initiated with the right length:

df3 <- data.frame(col1 = character(3), col2 = character(3))
df3$col1 <- col1
df3$col2 <- col2

4- Set rownames when you define it so it has 0 column and n rows

df4 <- data.frame(row.names = 1:3)
df4$col1 <- col1
df4$col2 <- col2

Check that it's all equivalent:

identical(df1,df2) # [1] TRUE
identical(df1,df3) # [1] TRUE
identical(df1,df4) # [1] TRUE

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