Create an empty data.frame

Question

I'm trying to initialize a data.frame without any rows. Basically, I want to specify the data types for each column and name them, but not have any rows created as a result.

The best I've been able to do so far is something like:

df <- data.frame(Date=as.Date("01/01/2000", format="%m/%d/%Y"), 
                 File="", User="", stringsAsFactors=FALSE)
df <- df[-1,]

Which creates a data.frame with a single row containing all of the data types and column names I wanted, but also creates a useless row which then needs to be removed.

Is there a better way to do this?

Answer 1

Just initialize it with empty vectors:

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

---

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.

Answer 2

This question didn't specifically address my concerns (outlined here) but in case anyone wants to do this with a parameterized number of columns and no coercion:

> require(dplyr)
> dbNames <- c('a','b','c','d')
> emptyTableOut <- 
    data.frame(
        character(), 
        matrix(integer(), ncol = 3, nrow = 0), stringsAsFactors = FALSE
    ) %>% 
    setNames(nm = c(dbNames))
> glimpse(emptyTableOut)
Observations: 0
Variables: 4
$ a <chr> 
$ b <int> 
$ c <int> 
$ d <int>

As divibisan states on the linked question,

...the reason [coercion] occurs [when cbinding matrices and their constituent types] is that a matrix can only have a single data type. When you cbind 2 matrices, the result is still a matrix and so the variables are all coerced into a single type before converting to a data.frame

Answer 3

If you want to create an empty data.frame with dynamic names (colnames in a variable), this can help:

names <- c("v","u","w")
df <- data.frame()
for (k in names) df[[k]]<-as.numeric()

You can change the types as well if you need so. like:

names <- c("u", "v")
df <- data.frame()
df[[names[1]]] <- as.numeric()
df[[names[2]]] <- as.character()

Answer 4

Say your column names are dynamic, you can create an empty row-named matrix and transform it to a data frame.

nms <- sample(LETTERS,sample(1:10))
as.data.frame(t(matrix(nrow=length(nms),ncol=0,dimnames=list(nms))))

Answer 5

I created empty data frame using following code

df = data.frame(id = numeric(0), jobs = numeric(0));

and tried to bind some rows to populate the same as follows.

newrow = c(3, 4)
df <- rbind(df, newrow)

but it started giving incorrect column names as follows

  X3 X4
1  3  4

Solution to this is to convert newrow to type df as follows

newrow = data.frame(id=3, jobs=4)
df <- rbind(df, newrow)

now gives correct data frame when displayed with column names as follows

  id nobs
1  3   4 

Answer 6

If you want to declare such a data.frame with many columns, it'll probably be a pain to type all the column classes out by hand. Especially if you can make use of rep, this approach is easy and fast (about 15% faster than the other solution that can be generalized like this):

If your desired column classes are in a vector colClasses, you can do the following:

library(data.table)
setnames(setDF(lapply(colClasses, function(x) eval(call(x)))), col.names)

lapply will result in a list of desired length, each element of which is simply an empty typed vector like numeric() or integer().

setDF converts this list by reference to a data.frame.

setnames adds the desired names by reference.

Speed comparison:

classes <- c("character", "numeric", "factor",
             "integer", "logical","raw", "complex")

NN <- 300
colClasses <- sample(classes, NN, replace = TRUE)
col.names <- paste0("V", 1:NN)

setDF(lapply(colClasses, function(x) eval(call(x))))

library(microbenchmark)
microbenchmark(times = 1000,
               read = read.table(text = "", colClasses = colClasses,
                                 col.names = col.names),
               DT = setnames(setDF(lapply(colClasses, function(x)
                 eval(call(x)))), col.names))
# Unit: milliseconds
#  expr      min       lq     mean   median       uq      max neval cld
#  read 2.598226 2.707445 3.247340 2.747835 2.800134 22.46545  1000   b
#    DT 2.257448 2.357754 2.895453 2.401408 2.453778 17.20883  1000  a 

It's also faster than using structure in a similar way:

microbenchmark(times = 1000,
               DT = setnames(setDF(lapply(colClasses, function(x)
                 eval(call(x)))), col.names),
               struct = eval(parse(text=paste0(
                 "structure(list(", 
                 paste(paste0(col.names, "=", 
                              colClasses, "()"), collapse = ","),
                 "), class = \"data.frame\")"))))
#Unit: milliseconds
#   expr      min       lq     mean   median       uq       max neval cld
#     DT 2.068121 2.167180 2.821868 2.211214 2.268569 143.70901  1000  a 
# struct 2.613944 2.723053 3.177748 2.767746 2.831422  21.44862  1000   b