Convert data.frame column to a vector?

Question

I have a dataframe such as:

a1 = c(1, 2, 3, 4, 5)
a2 = c(6, 7, 8, 9, 10)
a3 = c(11, 12, 13, 14, 15)
aframe = data.frame(a1, a2, a3)

I tried the following to convert one of the columns to a vector, but it doesn't work:

avector <- as.vector(aframe['a2'])
class(avector) 
[1] "data.frame"

This is the only solution I could come up with, but I'm assuming there has to be a better way to do this:

class(aframe['a2']) 
[1] "data.frame"
avector = c()
for(atmp in aframe['a2']) { avector <- atmp }
class(avector)
[1] "numeric"

Note: My vocabulary above may be off, so please correct me if so. I'm still learning the world of R. Additionally, any explanation of what's going on here is appreciated (i.e. relating to Python or some other language would help!)

As you're seeing in the answers, a close reading of ?'[.data.frame' will take you very far. — joran
– joran, Commented Aug 15, 2011 at 20:34

joran · Accepted Answer · 2011-08-15 20:19:35Z

269

I'm going to attempt to explain this without making any mistakes, but I'm betting this will attract a clarification or two in the comments.

A data frame is a list. When you subset a data frame using the name of a column and [, what you're getting is a sublist (or a sub data frame). If you want the actual atomic column, you could use [[, or somewhat confusingly (to me) you could do aframe[,2] which returns a vector, not a sublist.

So try running this sequence and maybe things will be clearer:

avector <- as.vector(aframe['a2'])
class(avector) 

avector <- aframe[['a2']]
class(avector)

avector <- aframe[,2]
class(avector)

answered Aug 15, 2011 at 20:19

joran

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5 Comments

Iterator Over a year ago

+1 This is useful. I had gotten used to using aframe[,"a2"] because of the ability to use this with both data frames and matrices & seem to get the same results - a vector.

hadley Over a year ago

[..., drop = F] will always return a data frame

jimjamslam Over a year ago

This is particularly good to know because the df$x syntax returns a vector. I used this syntax for a long time, but when I had to start using df['name'] or df[n] to retrieve columns, I hit problems when I tried to send them to functions that expected vectors. Using df[[n]] or df[['x']] cleared things right up.

bli Over a year ago

Why does as.vector seem to silently have no effect? Shouldn't this either return a vector or conspicuously fail?

Matt Over a year ago

aframe[['a2']] is very useful with sf objects because aframe[,"a2"] will return two columns because the geometry column is included.

Andrew Brēza · Accepted Answer · 2018-01-08 17:21:10Z

87

There's now an easy way to do this using dplyr.

dplyr::pull(aframe, a2)

answered Jan 8, 2018 at 17:21

Andrew Brēza

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Comments

James · Accepted Answer · 2011-08-15 20:20:05Z

39

You could use $ extraction:

class(aframe$a1)
[1] "numeric"

or the double square bracket:

class(aframe[["a1"]])
[1] "numeric"

answered Aug 15, 2011 at 20:20

James

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Comments

Dirk is no longer here · Accepted Answer · 2011-08-15 20:19:54Z

23

You do not need as.vector(), but you do need correct indexing: avector <- aframe[ , "a2"]

The one other thing to be aware of is the drop=FALSE option to [:

R> aframe <- data.frame(a1=c1:5, a2=6:10, a3=11:15)
R> aframe
  a1 a2 a3
1  1  6 11
2  2  7 12
3  3  8 13
4  4  9 14
5  5 10 15
R> avector <- aframe[, "a2"]
R> avector
[1]  6  7  8  9 10
R> avector <- aframe[, "a2", drop=FALSE]
R> avector
  a2
1  6
2  7
3  8
4  9
5 10
R>

answered Aug 15, 2011 at 20:19

Dirk is no longer here

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3 Comments

Iterator Over a year ago

+1: The reminder of drop=FALSE is useful - this helps me in cases where I may select N columns from a data.frame, in those cases where N=1.

Roman Luštrik Over a year ago

I use this when I can't foresee the number of columns selected and in case one column comes up, the result still gets passed as a data.frame with n columns. A vector may throw a monkey wrench into the functions down the line.

Chema_arguez Over a year ago

In my case, I had to extract a vector from filtering rows, and it was the only option that has worked: aframe[aframe$a1 == 1, 'a2', drop =TRUE]. In this case it doesn´t make sense but it could help in other situations

Vaibhav Sharma · Accepted Answer · 2018-10-05 03:48:22Z

19

You can try something like this-

as.vector(unlist(aframe$a2))

answered Oct 5, 2018 at 3:48

Vaibhav Sharma

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2 Comments

p-robot Over a year ago

This is good if you want to compare two columns using identical.

shwan Over a year ago

This is also helpful if you don't know the column name ahead of time...i.e. as.vector(unlist(aframe[,1]))

joel.wilson · Accepted Answer · 2016-09-14 07:26:29Z

14

Another advantage of using the '[[' operator is that it works both with data.frame and data.table. So if the function has to be made running for both data.frame and data.table, and you want to extract a column from it as a vector then

data[["column_name"]]

is best.

answered Sep 14, 2016 at 7:26

joel.wilson

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Comments

Code42 · Accepted Answer · 2019-10-19 12:57:40Z

11

as.vector(unlist(aframe['a2']))

answered Oct 19, 2019 at 12:57

Code42

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shubham ranjan · Accepted Answer · 2017-07-02 13:52:26Z

6

a1 = c(1, 2, 3, 4, 5)
a2 = c(6, 7, 8, 9, 10)
a3 = c(11, 12, 13, 14, 15)
aframe = data.frame(a1, a2, a3)
avector <- as.vector(aframe['a2'])

avector<-unlist(avector)
#this will return a vector of type "integer"

answered Jul 2, 2017 at 13:52

shubham ranjan

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Comments

Ari B. Friedman · Accepted Answer · 2011-08-15 20:20:31Z

5

If you just use the extract operator it will work. By default, [] sets option drop=TRUE, which is what you want here. See ?'[' for more details.

>  a1 = c(1, 2, 3, 4, 5)
>  a2 = c(6, 7, 8, 9, 10)
>  a3 = c(11, 12, 13, 14, 15)
>  aframe = data.frame(a1, a2, a3)
> aframe[,'a2']
[1]  6  7  8  9 10
> class(aframe[,'a2'])
[1] "numeric"

answered Aug 15, 2011 at 20:20

Ari B. Friedman

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Comments

0range · Accepted Answer · 2020-04-10 02:15:01Z

3

We can also convert data.frame columns generically to a simple vector. as.vector is not enough as it retains the data.frame class and structure, so we also have to pull out the first (and only) element:

df_column_object <- aframe[,2]
simple_column <- df_column_object[[1]]

All the solutions suggested so far require hardcoding column titles. This makes them non-generic (imagine applying this to function arguments).

Alternatively, you could, of course read the column names from the column first and then insert them in the code in the other solutions.

answered Apr 10, 2020 at 2:15

0range

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Comments

Adrian DSouza · Accepted Answer · 2018-12-14 14:25:36Z

2

I use lists to filter dataframes by whether or not they have a value %in% a list.

I had been manually creating lists by exporting a 1 column dataframe to Excel where I would add " ", around each element, before pasting into R: list <- c("el1", "el2", ...) which was usually followed by FilteredData <- subset(Data, Column %in% list).

After searching stackoverflow and not finding an intuitive way to convert a 1 column dataframe into a list, I am now posting my first ever stackoverflow contribution:

# assuming you have a 1 column dataframe called "df"
list <- c()
for(i in 1:nrow(df)){
  list <- append(list, df[i,1])
}
View(list)
# This list is not a dataframe, it is a list of values
# You can filter a dataframe using "subset([Data], [Column] %in% list")

edited Dec 14, 2018 at 14:25

answered Dec 3, 2018 at 18:17

Adrian DSouza

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Comments

Quinten · Accepted Answer · 2022-08-27 09:15:38Z

2

Another option is using as.matrix with as.vector. This can be done for one column but is also possible if you want to convert all columns to one vector. Here is a reproducible example with first converting one column to a vector and second convert complete dataframe to one vector:

a1 = c(1, 2, 3, 4, 5)
a2 = c(6, 7, 8, 9, 10)
a3 = c(11, 12, 13, 14, 15)
aframe = data.frame(a1, a2, a3)

# Convert one column to vector
avector <- as.vector(as.matrix(aframe[,"a2"]))
class(avector)
#> [1] "numeric"
avector
#> [1]  6  7  8  9 10

# Convert all columns to one vector
avector <- as.vector(as.matrix(aframe))
class(avector)
#> [1] "numeric"
avector
#>  [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15

^{Created on 2022-08-27 with reprex v2.0.2}

answered Aug 27, 2022 at 9:15

Quinten

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1 Comment

Momchill Mar 24 at 21:05

This was especially useful in my case, where i couldn't expressly name the column, but was rather indexing it in loops. Interestingly enough it also works with as.vector(as.matrix(df[i,-1])). Most versatile answer!

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