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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:
Just declare
table = data.frame()
when you try to rbind
the first line it will create the columns
回答7:
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8:
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9:
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10:
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11:
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12:
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13:
By Using data.table
we can specify data types for each column.
library(data.table)
data=data.table(a=numeric(), b=numeric(), c=numeric())
回答14:
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15:
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
回答16:
You can also extract the metadata (column names and types) from a dataframe (e.g. if you are controlling a BUG which is only triggered with certain inputs and need a empty dummy dataframe):
colums_and_types <- sapply(df, class)
# prints: "c('col1', 'col2')"
print(dput(as.character(names(colums_and_types))))
# prints: "c('integer', 'factor')"
dput(as.character(as.vector(colums_and_types)))
And then use the read.table
to create the empty dataframe
read.table(text = "",
colClasses = c('integer', 'factor'),
col.names = c('col1', 'col2'))