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How do I extract a single column from a data.frame as a data.frame?
(3 answers)
Closed 1 year ago.
I am simply extracting a single row from a data.frame. Consider for example
d=data.frame(a=1:3,b=1:3)
d[1,] # returns a data.frame
# a b
# 1 1 1
The output matched my expectation. The result was not as I expected though when dealing with a data.frame that contains a single column.
d=data.frame(a=1:3)
d[1,] # returns an integer
# [1] 1
Indeed, here, the extracted data is not a data.frame anymore but an integer! To me, it seems a little strange that the same function on the same data type wants to return different data types. One of the issue with this conversion is the loss of the column name.
To solve the issue, I did
extractRow = function(d,index)
{
if (ncol(d) > 1)
{
return(d[index,])
} else
{
d2 = as.data.frame(d[index,])
names(d2) = names(d)
return(d2)
}
}
d=data.frame(a=1:3,b=1:3)
extractRow(d,1)
# a b
# 1 1 1
d=data.frame(a=1:3)
extractRow(d,1)
# a
# 1 1
But it seems unnecessarily cumbersome. Is there a better solution?
Just subset with the drop = FALSE option:
extractRow = function(d, index) {
return(d[index, , drop=FALSE])
}
R tries to simplify data.frame cuts by default, the same thing happens with columns:
d[, "a"]
# [1] 1 2 3
Alternatives are:
d[1, , drop = FALSE]
tibble::tibble which has drop = FALSE by default
I can't tell you why that happens - it seems weird. One workaround would be to use slice from dplyr (although using a library seems unecessary for such a simple task).
library(dplyr)
slice(d, 1)
a
1 1
data.frames will simplify to vectors or scallars whith base subsetting [,].
If you want to avoid that, you can use tibbles instead:
> tibble(a=1:2)[1,]
# A tibble: 1 x 1
a
<int>
1 1
tibble(a=1:2)[1,] %>% class
[1] "tbl_df" "tbl" "data.frame"
Related
I want to write a function in R that does the following:
I have a table of cases, and some data. I want to find the correct row matching to each observation from the data. Example:
crit1 <- c(1,1,2)
crit2 <- c("yes","no","no")
Cases <- matrix(c(crit1,crit2),ncol=2,byrow=FALSE)
data1 <- c(1,2,1)
data2 <- c("no","no","yes")
data <- matrix(c(data1,data2),ncol=2,byrow=FALSE)
Now I want a function that returns for each row of my data, the matching row from Cases, the result would be the vector
c(2,3,1)
Are you sure you want to be using matrices for this?
Note that the numeric data in crit1 and data1 has been converted to string (matrices can only store one data type):
typeof(data[ , 1L])
# [1] character
In R, a data.frame is a much more natural choice for what you're after. data.table is (among many other things) a toolset for working with "enhanced" data.frames; See the Introduction.
I would create your data as:
Cases = data.table(crit1, crit2)
data = data.table(data1, data2)
We can get the matching row indices as asked by doing a keyed join (See the vignette on keys):
setkey(Cases) # key by all columns
Cases
# crit1 crit2
# 1: 1 no
# 2: 1 yes
# 3: 2 no
setkey(data)
data
# data1 data2
# 1: 1 no
# 2: 1 yes
# 3: 2 no
Cases[data, which=TRUE]
# [1] 1 2 3
This differs from 2,3,1 because the order of your data has changed, but note that the answer is still correct.
If you don't want to change the order of your data, it's slightly more complicated (but more readable if you're not used to data.table syntax):
Cases = data.table(crit1, crit2)
data = data.table(data1, data2)
Cases[data, on = setNames(names(data), names(Cases)), which=TRUE]
# [1] 2 3 1
The on= part creates the mapping between the columns of data and those of Cases.
We could write this in a bit more SQL-like fashion as:
Cases[data, on = .(crit1 == data1, crit2 == data2), which=TRUE]
# [1] 2 3 1
This is shorter and more readable for your sample data, but not as extensible if your data has many columns or if you don't know the column names in advance.
The prodlim package has a function for that:
library(prodlim)
row.match(data,Cases)
[1] 2 3 1
for a dataframe df, I need to find the unique values for some_col. Tried the following
length(unique(df["some_col"]))
but this is not giving the expected results. However length(unique(some_vector)) works on a vector and gives the expected results.
Some preceding steps while the df is created
df <- read.csv(file, header=T)
typeof(df) #=> "list"
typeof(unique(df["some_col"])) #=> "list"
length(unique(df["some_col"])) #=> 1
Try with [[ instead of [. [ returns a list (a data.frame in fact), [[ returns a vector.
df <- data.frame( some_col = c(1,2,3,4),
another_col = c(4,5,6,7) )
length(unique(df[["some_col"]]))
#[1] 4
class( df[["some_col"]] )
[1] "numeric"
class( df["some_col"] )
[1] "data.frame"
You're getting a value of 1 because the list is of length 1 (1 column), even though that 1 element contains several values.
you need to use
length(unique(unlist(df[c("some_col")])))
When you call column by df[c("some_col")] or by df["some_col"] ; it pulls it as a list. Unlist will convert it into the vector and you can work easily with it. When you call column by df$some_col .. it pulls the data column as vector
I think you might just be missing a ,
Try
length(unique(df[,"some_col"]))
In response to comment :
df <- data.frame(cbind(A=c(1:10),B=rep(c("A","B"),5)))
df["B"]
Output :
B
1 A
2 B
3 A
4 B
5 A
6 B
7 A
8 B
9 A
10 B
and
length(unique(df[,"B"]))
Output:
[1] 1
Which is the same incorrect/undesirable output as the OP posted
HOWEVER With a comma ,
df[,"B"]
Output :
[1] A B A B A B A B A B
Levels: A B
and
length(unique(df[,"B"]))
Now gives you the correct/desired output by the OP. Which in this example is 2
[1] 2
The reason is that df["some_col"] calls a data.frame and length call to an object class data.frame counts the number of data.frames in that object which is 1, while df[,"some_col"] returns a vector and length call to a vector correctly returns the number of elements in that vector. So you see a comma (,) makes all the difference.
using tidyverse
df %>%
select("some_col") %>%
n_distinct()
The data.table package contains the convenient shorthand uniqueN. From the documentation
uniqueN is equivalent to length(unique(x)) when x is anatomic vector, and nrow(unique(x)) when x is a data.frame or data.table. The number of unique rows are computed directly without materialising the intermediate unique data.table and is therefore faster and memory efficient.
You can use it with a data frame:
df <- data.frame(some_col = c(1,2,3,4),
another_col = c(4,5,6,7) )
data.table::uniqueN(df[['some_col']])
[1] 4
or if you already have a data.table
dt <- setDT(df)
dt[,uniqueN(some_col)]
[1] 4
Here is another option:
df %>%
distinct(column_name) %>%
count()
or this without tidyverse:
count(distinct(df, column_name))
checking benchmarks in the web you will see that distinct() is fast.
Suppose that I have a vector x whose elements I want to use to extract columns from a matrix or data frame M.
If x[1] = "A", I cannot use M$x[1] to extract the column with header name A, because M$A is recognized while M$"A" is not. How can I remove the quotes so that M$x[1] is M$A rather than M$"A" in this instance?
Don't use $ in this case; use [ instead. Here's a minimal example (if I understand what you're trying to do).
mydf <- data.frame(A = 1:2, B = 3:4)
mydf
# A B
# 1 1 3
# 2 2 4
x <- c("A", "B")
x
# [1] "A" "B"
mydf[, x[1]] ## As a vector
# [1] 1 2
mydf[, x[1], drop = FALSE] ## As a single column `data.frame`
# A
# 1 1
# 2 2
I think you would find your answer in the R Inferno. Start around Circle 8: "Believing it does as intended", one of the "string not the name" sub-sections.... You might also find some explanation in the line The main difference is that $ does not allow computed indices, whereas [[ does. from the help page at ?Extract.
Note that this approach is taken because the question specified using the approach to extract columns from a matrix or data frame, in which case, the [row, column] mode of extraction is really the way to go anyway (and the $ approach would not work with a matrix).
Apologises for a semi 'double post'. I feel I should be able to crack this but I'm going round in circles. This is on a similar note to my previously well answered question:
Within ID, check for matches/differences
test <- data.frame(
ID=c(rep(1,3),rep(2,4),rep(3,2)),
DOD = c(rep("2000-03-01",3), rep("2002-05-01",4), rep("2006-09-01",2)),
DOV = c("2000-03-05","2000-06-05","2000-09-05",
"2004-03-05","2004-06-05","2004-09-05","2005-01-05",
"2006-10-03","2007-02-05")
)
What I want to do is tag the subject whose first vist (as at DOV) was less than 180 days from their diagnosis (DOD). I have the following from the plyr package.
ddply(test, "ID", function(x) ifelse( (as.numeric(x$DOV[1]) - as.numeric(x$DOD[1])) < 180,1,0))
Which gives:
ID V1
1 A 1
2 B 0
3 C 1
What I would like is a vector 1,1,1,0,0,0,0,1,1 so I can append it as a column to the data frame. Basically this ddply function is fine, it makes a 'lookup' table where I can see which IDs have a their first visit within 180 days of their diagnosis, which I could then take my original test and go through and make an indicator variable, but I should be able to do this is one step I'd have thought.
I'd also like to use base if possible. I had a method with 'by', but again it only gave one result per ID and was also a list. Have been trying with aggregate but getting things like 'by has to be a list', then 'it's not the same length' and using the formula method of input I'm stumped 'cbind(DOV,DOD) ~ ID'...
Appreciate the input, keen to learn!
After wrapping as.Date around the creation of those date columns, this returns the desired marking vector assuming the df named 'test' is sorted by ID (and done in base):
# could put an ordering operation here if needed
0 + unlist( # to make vector from list and coerce logical to integer
lapply(split(test, test$ID), # to apply fn with ID
function(x) rep( # to extend a listwise value across all ID's
min(x$DOV-x$DOD) <180, # compare the minimum of a set of intervals
NROW(x)) ) )
11 12 13 21 22 23 24 31 32 # the labels
1 1 1 0 0 0 0 1 1 # the values
I have added to data.frame function stringsAsFactors=FALSE:
test <- data.frame(ID=c(rep(1,3),rep(2,4),rep(3,2)),
DOD = c(rep("2000-03-01",3), rep("2002-05-01",4), rep("2006-09-01",2)),
DOV = c("2000-03-05","2000-06-05","2000-09-05","2004-03-05",
"2004-06-05","2004-09-05","2005-01-05","2006-10-03","2007-02-05")
, stringsAsFactors=FALSE)
CODE
test$V1 <- ifelse(c(FALSE, diff(test$ID) == 0), 0,
1*(as.numeric(as.Date(test$DOV)-as.Date(test$DOD))<180))
test$V1 <- ave(test$V1,test$ID,FUN=max)
How do I add a column in the middle of an R data frame? I want to see if I have a column named "LastName" and then add it as the third column if it does not already exist.
One approach is to just add the column to the end of the data frame, and then use subsetting to move it into the desired position:
d$LastName <- c("Flim", "Flom", "Flam")
bar <- d[c("x", "y", "Lastname", "fac")]
1) Testing for existence: Use %in% on the colnames, e.g.
> example(data.frame) # to get 'd'
> "fac" %in% colnames(d)
[1] TRUE
> "bar" %in% colnames(d)
[1] FALSE
2) You essentially have to create a new data.frame from the first half of the old, your new column, and the second half:
> bar <- data.frame(d[1:3,1:2], LastName=c("Flim", "Flom", "Flam"), fac=d[1:3,3])
> bar
x y LastName fac
1 1 1 Flim C
2 1 2 Flom A
3 1 3 Flam A
>
Of the many silly little helper functions I've written, this gets used every time I load R. It just makes a list of the column names and indices but I use it constantly.
##creates an object from a data.frame listing the column names and location
namesind=function(df){
temp1=names(df)
temp2=seq(1,length(temp1))
temp3=data.frame(temp1,temp2)
names(temp3)=c("VAR","COL")
return(temp3)
rm(temp1,temp2,temp3)
}
ni <- namesind
Use ni to see your column numbers. (ni is just an alias for namesind, I never use namesind but thought it was a better name originally) Then if you want insert your column in say, position 12, and your data.frame is named bob with 20 columns, it would be
bob2 <- data.frame(bob[,1:11],newcolumn, bob[,12:20]
though I liked the add at the end and rearrange answer from Hadley as well.
Dirk Eddelbuettel's answer works, but you don't need to indicate row numbers or specify entries in the lastname column. This code should do it for a data frame named df:
if(!("LastName" %in% names(df))){
df <- cbind(df[1:2],LastName=NA,df[3:length(df)])
}
(this defaults LastName to NA, but you could just as easily use "LastName='Smith'")
or using cbind:
> example(data.frame) # to get 'd'
> bar <- cbind(d[1:3,1:2],LastName=c("Flim", "Flom", "Flam"),fac=d[1:3,3])
> bar
x y LastName fac
1 1 1 Flim A
2 1 2 Flom B
3 1 3 Flam B
I always thought something like append() [though unfortunate the name is] should be a generic function
## redefine append() as generic function
append.default <- append
append <- `body<-`(args(append),value=quote(UseMethod("append")))
append.data.frame <- function(x,values,after=length(x))
`row.names<-`(data.frame(append.default(x,values,after)),
row.names(x))
## apply the function
d <- (if( !"LastName" %in% names(d) )
append(d,values=list(LastName=c("Flim","Flom","Flam")),after=2) else d)