Create an empty data.frame

问题:

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?

回答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.

回答2:

If you already have an existent data frame, let\'s say df that has the columns you want, then you can just create an empty data frame by removing all the rows:

empty_df = df[FALSE,]

Notice that df still contains the data, but empty_df doesn\'t.

I found this question looking for how to create a new instance with empty rows, so I think it might be helpful for some people.

回答3:

You can do it without specifying column types

df = data.frame(matrix(vector(), 0, 3,
                dimnames=list(c(), c(\"Date\", \"File\", \"User\"))),
                stringsAsFactors=F)

回答4:

You could use read.table with an empty string for the input text as follows:

colClasses = c(\"Date\", \"character\", \"character\")
col.names = c(\"Date\", \"File\", \"User\")

df <- read.table(text = \"\",
                 colClasses = colClasses,
                 col.names = col.names)

Alternatively specifying the col.names as a string:

df <- read.csv(text=\"Date,File,User\", colClasses = colClasses)

Thanks to Richard Scriven for the improvement

回答5:

The most efficient way to do this is to use structure to create a list that has the class \"data.frame\":

structure(list(Date = as.Date(character()), File = character(), User = character()), 
          class = \"data.frame\")
# [1] Date File User
# <0 rows> (or 0-length row.names)

To put this into perspective compared to the presently accepted answer, here\'s a simple benchmark:

s <- function() structure(list(Date = as.Date(character()), 
                               File = character(), 
                               User = character()), 
                          class = \"data.frame\")
d <- function() data.frame(Date = as.Date(character()),
                           File = character(), 
                           User = character(), 
                           stringsAsFactors = FALSE) 
library(\"microbenchmark\")
microbenchmark(s(), d())
# Unit: microseconds
#  expr     min       lq     mean   median      uq      max neval
#   s()  58.503  66.5860  90.7682  82.1735 101.803  469.560   100
#   d() 370.644 382.5755 523.3397 420.1025 604.654 1565.711   100

回答6:

If you are looking for shortness :

read.csv(text=\"col1,col2\")

so you don\'t need to specify the column names separately. You get the default column type logical until you fill the data frame.

回答7:

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 

回答8:

Just declare

table = data.frame()

when you try to rbind the first line it will create the columns

回答9:

If you don\'t mind not specifying data types explicitly, you can do it this way:

headers<-c(\"Date\",\"File\",\"User\")
df <- as.data.frame(matrix(,ncol=3,nrow=0))
names(df)<-headers

#then bind incoming data frame with col types to set data types
df<-rbind(df, new_df)

回答10:

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()

回答11:

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

回答12:

To create an empty data frame, pass in the number of rows and columns needed into the following function:

create_empty_table <- function(num_rows, num_cols) {
    frame <- data.frame(matrix(NA, nrow = num_rows, ncol = num_cols))
    return(frame)
}

To create an empty frame while specifying the class of each column, simply pass a vector of the desired data types into the following function:

create_empty_table <- function(num_rows, num_cols, type_vec) {
  frame <- data.frame(matrix(NA, nrow = num_rows, ncol = num_cols))
  for(i in 1:ncol(frame)) {
    print(type_vec[i])
    if(type_vec[i] == \'numeric\') {frame[,i] <- as.numeric(df[,i])}
    if(type_vec[i] == \'character\') {frame[,i] <- as.character(df[,i])}
    if(type_vec[i] == \'logical\') {frame[,i] <- as.logical(df[,i])}
    if(type_vec[i] == \'factor\') {frame[,i] <- as.factor(df[,i])}
  }
  return(frame)
}

Use as follows:

df <- create_empty_table(3, 3, c(\'character\',\'logical\',\'numeric\'))

Which gives:

   X1  X2 X3
1 <NA> NA NA
2 <NA> NA NA
3 <NA> NA NA

To confirm your choices, run the following:

lapply(df, class)

#output
$X1
[1] \"character\"

$X2
[1] \"logical\"

$X3
[1] \"numeric\"

回答13:

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))))

回答14:

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