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It is straight forward to obtain unique values of a column using unique. However, I am looking to do the same but for multiple columns in a dataframe and store them in a list, all using base R. Importantly, it is not combinations I need but simply unique values for each individual column. I currently have the below:
# dummy data
df = data.frame(a = LETTERS[1:4]
,b = 1:4)
# for loop
cols = names(df)
unique_values_by_col = list()
for (i in cols)
{
x = unique(i)
unique_values_by_col[[i]] = x
}
The problem comes when displaying unique_values_by_col as it shows as empty. I believe the problem is i is being passed to the loop as a text not a variable.
Any help would be greatly appreciated. Thank you.
Why not avoid the for loop altogether using lapply:
lapply(df, unique)
Resulting in:
> $a
> [1] A B C D
> Levels: A B C D
> $b
> [1] 1 2 3 4
Or you have also apply that is specifically done to be run on column or line:
apply(df,2,unique)
result:
> apply(df,2,unique)
a b
[1,] "A" "1"
[2,] "B" "2"
[3,] "C" "3"
[4,] "D" "4"
thought if you want a list lapply return you a list so may be better
Your for loop is almost right, just needs one fix to work:
# for loop
cols = names(df)
unique_values_by_col = list()
for (i in cols) {
x = unique(df[[i]])
unique_values_by_col[[i]] = x
}
unique_values_by_col
# $a
# [1] A B C D
# Levels: A B C D
#
# $b
# [1] 1 2 3 4
i is just a character, the name of a column within df so unique(i) doesn't make sense.
Anyhow, the most standard way for this task is lapply() as shown by demirev.
Could this be what you're trying to do?
Map(unique,df)
Result:
$a
[1] A B C D
Levels: A B C D
$b
[1] 1 2 3 4
I have data frame with some numerical variables and some categorical factor variables. The order of levels for those factors is not the way I want them to be.
numbers <- 1:4
letters <- factor(c("a", "b", "c", "d"))
df <- data.frame(numbers, letters)
df
# numbers letters
# 1 1 a
# 2 2 b
# 3 3 c
# 4 4 d
If I change the order of the levels, the letters no longer are with their corresponding numbers (my data is total nonsense from this point on).
levels(df$letters) <- c("d", "c", "b", "a")
df
# numbers letters
# 1 1 d
# 2 2 c
# 3 3 b
# 4 4 a
I simply want to change the level order, so when plotting, the bars are shown in the desired order - which may differ from default alphabetical order.
Use the levels argument of factor:
df <- data.frame(f = 1:4, g = letters[1:4])
df
# f g
# 1 1 a
# 2 2 b
# 3 3 c
# 4 4 d
levels(df$g)
# [1] "a" "b" "c" "d"
df$g <- factor(df$g, levels = letters[4:1])
# levels(df$g)
# [1] "d" "c" "b" "a"
df
# f g
# 1 1 a
# 2 2 b
# 3 3 c
# 4 4 d
some more, just for the record
## reorder is a base function
df$letters <- reorder(df$letters, new.order=letters[4:1])
library(gdata)
df$letters <- reorder.factor(df$letters, letters[4:1])
You may also find useful Relevel and combine_factor.
Since this question was last active Hadley has released his new forcats package for manipulating factors and I'm finding it outrageously useful. Examples from the OP's data frame:
levels(df$letters)
# [1] "a" "b" "c" "d"
To reverse levels:
library(forcats)
fct_rev(df$letters) %>% levels
# [1] "d" "c" "b" "a"
To add more levels:
fct_expand(df$letters, "e") %>% levels
# [1] "a" "b" "c" "d" "e"
And many more useful fct_xxx() functions.
so what you want, in R lexicon, is to change only the labels for a given factor variable (ie, leave the data as well as the factor levels, unchanged).
df$letters = factor(df$letters, labels=c("d", "c", "b", "a"))
given that you want to change only the datapoint-to-label mapping and not the data or the factor schema (how the datapoints are binned into individual bins or factor values, it might help to know how the mapping is originally set when you initially create the factor.
the rules are simple:
labels are mapped to levels by index value (ie, the value
at levels[2] is given the label, label[2]);
factor levels can be set explicitly by passing them in via the the
levels argument; or
if no value is supplied for the levels argument, the default
value is used which is the result calling unique on the data vector
passed in (for the data argument);
labels can be set explicitly via the labels argument; or
if no value is supplied for the labels argument, the default value is
used which is just the levels vector
Dealing with factors in R is quite peculiar job, I must admit... While reordering the factor levels, you're not reordering underlying numerical values. Here's a little demonstration:
> numbers = 1:4
> letters = factor(letters[1:4])
> dtf <- data.frame(numbers, letters)
> dtf
numbers letters
1 1 a
2 2 b
3 3 c
4 4 d
> sapply(dtf, class)
numbers letters
"integer" "factor"
Now, if you convert this factor to numeric, you'll get:
# return underlying numerical values
1> with(dtf, as.numeric(letters))
[1] 1 2 3 4
# change levels
1> levels(dtf$letters) <- letters[4:1]
1> dtf
numbers letters
1 1 d
2 2 c
3 3 b
4 4 a
# return numerical values once again
1> with(dtf, as.numeric(letters))
[1] 1 2 3 4
As you can see... by changing levels, you change levels only (who would tell, eh?), not the numerical values! But, when you use factor function as #Jonathan Chang suggested, something different happens: you change numerical values themselves.
You're getting error once again 'cause you do levels and then try to relevel it with factor. Don't do it!!! Do not use levels or you'll mess things up (unless you know exactly what you're doing).
One lil' suggestion: avoid naming your objects with an identical name as R's objects (df is density function for F distribution, letters gives lowercase alphabet letters). In this particular case, your code would not be faulty, but sometimes it can be... but this can create confusion, and we don't want that, do we?!? =)
Instead, use something like this (I'll go from the beginning once again):
> dtf <- data.frame(f = 1:4, g = factor(letters[1:4]))
> dtf
f g
1 1 a
2 2 b
3 3 c
4 4 d
> with(dtf, as.numeric(g))
[1] 1 2 3 4
> dtf$g <- factor(dtf$g, levels = letters[4:1])
> dtf
f g
1 1 a
2 2 b
3 3 c
4 4 d
> with(dtf, as.numeric(g))
[1] 4 3 2 1
Note that you can also name you data.frame with df and letters instead of g, and the result will be OK. Actually, this code is identical with the one you posted, only the names are changed. This part factor(dtf$letter, levels = letters[4:1]) wouldn't throw an error, but it can be confounding!
Read the ?factor manual thoroughly! What's the difference between factor(g, levels = letters[4:1]) and factor(g, labels = letters[4:1])? What's similar in levels(g) <- letters[4:1] and g <- factor(g, labels = letters[4:1])?
You can put ggplot syntax, so we can help you more on this one!
Cheers!!!
Edit:
ggplot2 actually requires to change both levels and values? Hm... I'll dig this one out...
I wish to add another case where the levels could be strings carrying numbers alongwith some special characters : like below example
df <- data.frame(x = c("15-25", "0-4", "5-10", "11-14", "100+"))
The default levels of x is :
df$x
# [1] 15-25 0-4 5-10 11-14 100+
# Levels: 0-4 100+ 11-14 15-25 5-10
Here if we want to reorder the factor levels according to the numeric value, without explicitly writing out the levels, what we could do is
library(gtools)
df$x <- factor(df$x, levels = mixedsort(df$x))
df$x
# [1] 15-25 0-4 5-10 11-14 100+
# Levels: 0-4 5-10 11-14 15-25 100+
as.numeric(df$x)
# [1] 4 1 2 3 5
I hope this can be considered as useful information for future readers.
Here's my function to reorder factors of a given dataframe:
reorderFactors <- function(df, column = "my_column_name",
desired_level_order = c("fac1", "fac2", "fac3")) {
x = df[[column]]
lvls_src = levels(x)
idxs_target <- vector(mode="numeric", length=0)
for (target in desired_level_order) {
idxs_target <- c(idxs_target, which(lvls_src == target))
}
x_new <- factor(x,levels(x)[idxs_target])
df[[column]] <- x_new
return (df)
}
Usage: reorderFactors(df, "my_col", desired_level_order = c("how","I","want"))
I would simply use the levels argument:
levels(df$letters) <- levels(df$letters)[c(4:1)]
To add yet another approach that is quite useful as it frees us from remembering functions from differents packages. The levels of a factor are just attributes, so one can do the following:
numbers <- 1:4
letters <- factor(c("a", "b", "c", "d"))
df <- data.frame(numbers, letters)
# Original attributes
> attributes(df$letters)
$levels
[1] "a" "b" "c" "d"
$class
[1] "factor"
# Modify attributes
attr(df$letters,"levels") <- c("d", "c", "b", "a")
> df$letters
[1] d c b a
Levels: d c b a
# New attributes
> attributes(df$letters)
$levels
[1] "d" "c" "b" "a"
$class
[1] "factor"
I don't find the help page for the replace function from the base package to be very helpful. Worst part, it has no examples which could help understand how it works.
Could you please explain how to use it? An example or two would be great.
If you look at the function (by typing it's name at the console) you will see that it is just a simple functionalized version of the [<- function which is described at ?"[". [ is a rather basic function to R so you would be well-advised to look at that page for further details. Especially important is learning that the index argument (the second argument in replace can be logical, numeric or character classed values. Recycling will occur when there are differing lengths of the second and third arguments:
You should "read" the function call as" "within the first argument, use the second argument as an index for placing the values of the third argument into the first":
> replace( 1:20, 10:15, 1:2)
[1] 1 2 3 4 5 6 7 8 9 1 2 1 2 1 2 16 17 18 19 20
Character indexing for a named vector:
> replace(c(a=1, b=2, c=3, d=4), "b", 10)
a b c d
1 10 3 4
Logical indexing:
> replace(x <- c(a=1, b=2, c=3, d=4), x>2, 10)
a b c d
1 2 10 10
You can also use logical tests
x <- data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2))
x
x$a <- replace(x$a, is.na(x$a), 0)
x
x$b <- replace(x$b, x$b==2, 333)
Here's two simple examples
> x <- letters[1:4]
> replace(x, 3, 'Z') #replacing 'c' by 'Z'
[1] "a" "b" "Z" "d"
>
> y <- 1:10
> replace(y, c(4,5), c(20,30)) # replacing 4th and 5th elements by 20 and 30
[1] 1 2 3 20 30 6 7 8 9 10
Be aware that the third parameter (value) in the examples given above: the value is a constant (e.g. 'Z' or c(20,30)).
Defining the third parameter using values from the data frame itself can lead to confusion.
E.g. with a simple data frame such as this (using dplyr::data_frame):
tmp <- data_frame(a=1:10, b=sample(LETTERS[24:26], 10, replace=T))
This will create somthing like this:
a b
(int) (chr)
1 1 X
2 2 Y
3 3 Y
4 4 X
5 5 Z
..etc
Now suppose you want wanted to do, was to multiply the values in column 'a' by 2, but only where column 'b' is "X". My immediate thought would be something like this:
with(tmp, replace(a, b=="X", a*2))
That will not provide the desired outcome, however. The a*2 will defined as a fixed vector rather than a reference to the 'a' column. The vector 'a*2' will thus be
[1] 2 4 6 8 10 12 14 16 18 20
at the start of the 'replace' operation. Thus, the first row where 'b' equals "X", the value in 'a' will be placed by 2. The second time, it will be replaced by 4, etc ... it will not be replaced by two-times-the-value-of-a in that particular row.
Here's an example where I found the replace( ) function helpful for giving me insight. The problem required a long integer vector be changed into a character vector and with its integers replaced by given character values.
## figuring out replace( )
(test <- c(rep(1,3),rep(2,2),rep(3,1)))
which looks like
[1] 1 1 1 2 2 3
and I want to replace every 1 with an A and 2 with a B and 3 with a C
letts <- c("A","B","C")
so in my own secret little "dirty-verse" I used a loop
for(i in 1:3)
{test <- replace(test,test==i,letts[i])}
which did what I wanted
test
[1] "A" "A" "A" "B" "B" "C"
In the first sentence I purposefully left out that the real objective was to make the big vector of integers a factor vector and assign the integer values (levels) some names (labels).
So another way of doing the replace( ) application here would be
(test <- factor(test,labels=letts))
[1] A A A B B C
Levels: A B C
I am using matching operators to grab values that appear in a matrix from a separate data frame. However, the resulting matrix has the values in the order they appear in the data frame, not in the original matrix. Is there any way to preserve the order of the original matrix using the matching operator?
Here is a quick example:
vec=c("b","a","c"); vec
df=data.frame(row.names=letters[1:5],values=1:5); df
df[rownames(df) %in% vec,1]
This produces > [1] 1 2 3 which is the order "a" "b" "c" appears in the data frame. However, I would like to generate >[1] 2 1 3 which is the order they appear in the original vector.
Thanks!
Use match.
df[match(vec, rownames(df)), ]
# [1] 2 1 3
Be aware that if you have duplicate values in either vec or rownames(df), match may not behave as expected.
Edit:
I just realized that row name indexing will solve your issue a bit more simply and elegantly:
df[vec, ]
# [1] 2 1 3
Use match (and get rid of the NA values for elements in either vector for those that don't match in the other):
Filter(function(x) !is.na(x), match(rownames(df), vec))
Since row name indexing also works on vectors, we can take this one step further and define:
'%ino%' <- function(x, table) {
xSeq <- seq(along = x)
names(xSeq) <- x
Out <- xSeq[as.character(table)]
Out[!is.na(Out)]
}
We now have the desired result:
df[rownames(df) %ino% vec, 1]
[1] 2 1 3
Inside the function, names() does an auto convert to character and table is changed with as.character(), so this also works correctly when the inputs to %ino% are numbers:
LETTERS[1:26 %in% 4:1]
[1] "A" "B" "C" "D"
LETTERS[1:26 %ino% 4:1]
[1] "D" "C" "B" "A"
Following %in%, missing values are removed:
LETTERS[1:26 %in% 3:-5]
[1] "A" "B" "C"
LETTERS[1:26 %ino% 3:-5]
[1] "C" "B" "A"
With %in% the logical sequence is repeated along the dimension of the object being subsetted, this is not the case with %ino%:
data.frame(letters, LETTERS)[1:5 %in% 3:-5,]
letters LETTERS
1 a A
2 b B
3 c C
6 f F
7 g G
8 h H
11 k K
12 l L
13 m M
16 p P
17 q Q
18 r R
21 u U
22 v V
23 w W
26 z Z
data.frame(letters, LETTERS)[1:5 %ino% 3:-5,]
letters LETTERS
3 c C
2 b B
1 a A
I have a list and I want to remove a single element from it. How can I do this?
I've tried looking up what I think the obvious names for this function would be in the reference manual and I haven't found anything appropriate.
If you don't want to modify the list in-place (e.g. for passing the list with an element removed to a function), you can use indexing: negative indices mean "don't include this element".
x <- list("a", "b", "c", "d", "e"); # example list
x[-2]; # without 2nd element
x[-c(2, 3)]; # without 2nd and 3rd
Also, logical index vectors are useful:
x[x != "b"]; # without elements that are "b"
This works with dataframes, too:
df <- data.frame(number = 1:5, name = letters[1:5])
df[df$name != "b", ]; # rows without "b"
df[df$number %% 2 == 1, ] # rows with odd numbers only
I don't know R at all, but a bit of creative googling led me here: http://tolstoy.newcastle.edu.au/R/help/05/04/1919.html
The key quote from there:
I do not find explicit documentation for R on how to remove elements from lists, but trial and error tells me
myList[[5]] <- NULL
will remove the 5th element and then "close up" the hole caused by deletion of that element. That suffles the index values, So I have to be careful in dropping elements. I must work from the back of the list to the front.
A response to that post later in the thread states:
For deleting an element of a list, see R FAQ 7.1
And the relevant section of the R FAQ says:
... Do not set x[i] or x[[i]] to NULL, because this will remove the corresponding component from the list.
Which seems to tell you (in a somewhat backwards way) how to remove an element.
I would like to add that if it's a named list you can simply use within.
l <- list(a = 1, b = 2)
> within(l, rm(a))
$b
[1] 2
So you can overwrite the original list
l <- within(l, rm(a))
to remove element named a from list l.
Here is how the remove the last element of a list in R:
x <- list("a", "b", "c", "d", "e")
x[length(x)] <- NULL
If x might be a vector then you would need to create a new object:
x <- c("a", "b", "c", "d", "e")
x <- x[-length(x)]
Work for lists and vectors
Removing Null elements from a list in single line :
x=x[-(which(sapply(x,is.null),arr.ind=TRUE))]
Cheers
If you have a named list and want to remove a specific element you can try:
lst <- list(a = 1:4, b = 4:8, c = 8:10)
if("b" %in% names(lst)) lst <- lst[ - which(names(lst) == "b")]
This will make a list lst with elements a, b, c. The second line removes element b after it checks that it exists (to avoid the problem #hjv mentioned).
or better:
lst$b <- NULL
This way it is not a problem to try to delete a non-existent element (e.g. lst$g <- NULL)
Use - (Negative sign) along with position of element, example if 3rd element is to be removed use it as your_list[-3]
Input
my_list <- list(a = 3, b = 3, c = 4, d = "Hello", e = NA)
my_list
# $`a`
# [1] 3
# $b
# [1] 3
# $c
# [1] 4
# $d
# [1] "Hello"
# $e
# [1] NA
Remove single element from list
my_list[-3]
# $`a`
# [1] 3
# $b
# [1] 3
# $d
# [1] "Hello"
# $e
[1] NA
Remove multiple elements from list
my_list[c(-1,-3,-2)]
# $`d`
# [1] "Hello"
# $e
# [1] NA
my_list[c(-3:-5)]
# $`a`
# [1] 3
# $b
# [1] 3
my_list[-seq(1:2)]
# $`c`
# [1] 4
# $d
# [1] "Hello"
# $e
# [1] NA
There's the rlist package (http://cran.r-project.org/web/packages/rlist/index.html) to deal with various kinds of list operations.
Example (http://cran.r-project.org/web/packages/rlist/vignettes/Filtering.html):
library(rlist)
devs <-
list(
p1=list(name="Ken",age=24,
interest=c("reading","music","movies"),
lang=list(r=2,csharp=4,python=3)),
p2=list(name="James",age=25,
interest=c("sports","music"),
lang=list(r=3,java=2,cpp=5)),
p3=list(name="Penny",age=24,
interest=c("movies","reading"),
lang=list(r=1,cpp=4,python=2)))
list.remove(devs, c("p1","p2"))
Results in:
# $p3
# $p3$name
# [1] "Penny"
#
# $p3$age
# [1] 24
#
# $p3$interest
# [1] "movies" "reading"
#
# $p3$lang
# $p3$lang$r
# [1] 1
#
# $p3$lang$cpp
# [1] 4
#
# $p3$lang$python
# [1] 2
Don't know if you still need an answer to this but I found from my limited (3 weeks worth of self-teaching R) experience with R that, using the NULL assignment is actually wrong or sub-optimal especially if you're dynamically updating a list in something like a for-loop.
To be more precise, using
myList[[5]] <- NULL
will throw the error
myList[[5]] <- NULL : replacement has length zero
or
more elements supplied than there are to replace
What I found to work more consistently is
myList <- myList[[-5]]
Just wanted to quickly add (because I didn't see it in any of the answers) that, for a named list, you can also do l["name"] <- NULL. For example:
l <- list(a = 1, b = 2, cc = 3)
l['b'] <- NULL
In the case of named lists I find those helper functions useful
member <- function(list,names){
## return the elements of the list with the input names
member..names <- names(list)
index <- which(member..names %in% names)
list[index]
}
exclude <- function(list,names){
## return the elements of the list not belonging to names
member..names <- names(list)
index <- which(!(member..names %in% names))
list[index]
}
aa <- structure(list(a = 1:10, b = 4:5, fruits = c("apple", "orange"
)), .Names = c("a", "b", "fruits"))
> aa
## $a
## [1] 1 2 3 4 5 6 7 8 9 10
## $b
## [1] 4 5
## $fruits
## [1] "apple" "orange"
> member(aa,"fruits")
## $fruits
## [1] "apple" "orange"
> exclude(aa,"fruits")
## $a
## [1] 1 2 3 4 5 6 7 8 9 10
## $b
## [1] 4 5
Using lapply and grep:
lst <- list(a = 1:4, b = 4:8, c = 8:10)
# say you want to remove a and c
toremove<-c("a","c")
lstnew<-lst[-unlist(lapply(toremove, function(x) grep(x, names(lst)) ) ) ]
#or
pattern<-"a|c"
lstnew<-lst[-grep(pattern, names(lst))]
You can also negatively index from a list using the extract function of the magrittr package to remove a list item.
a <- seq(1,5)
b <- seq(2,6)
c <- seq(3,7)
l <- list(a,b,c)
library(magrittr)
extract(l,-1) #simple one-function method
[[1]]
[1] 2 3 4 5 6
[[2]]
[1] 3 4 5 6 7
There are a few options in the purrr package that haven't been mentioned:
pluck and assign_in work well with nested values and you can access it using a combination of names and/or indices:
library(purrr)
l <- list("a" = 1:2, "b" = 3:4, "d" = list("e" = 5:6, "f" = 7:8))
# select values (by name and/or index)
all.equal(pluck(l, "d", "e"), pluck(l, 3, "e"), pluck(l, 3, 1))
[1] TRUE
# or if element location stored in a vector use !!!
pluck(l, !!! as.list(c("d", "e")))
[1] 5 6
# remove values (modifies in place)
pluck(l, "d", "e") <- NULL
# assign_in to remove values with name and/or index (does not modify in place)
assign_in(l, list("d", 1), NULL)
$a
[1] 1 2
$b
[1] 3 4
$d
$d$f
[1] 7 8
Or you can remove values using modify_list by assigning zap() or NULL:
all.equal(list_modify(l, a = zap()), list_modify(l, a = NULL))
[1] TRUE
You can remove or keep elements using a predicate function with discard and keep:
# remove numeric elements
discard(l, is.numeric)
$d
$d$e
[1] 5 6
$d$f
[1] 7 8
# keep numeric elements
keep(l, is.numeric)
$a
[1] 1 2
$b
[1] 3 4
Here is a simple solution that can be done using base R. It removes the number 5 from the original list of numbers. You can use the same method to remove whatever element you want from a list.
#the original list
original_list = c(1:10)
#the list element to remove
remove = 5
#the new list (which will not contain whatever the `remove` variable equals)
new_list = c()
#go through all the elements in the list and add them to the new list if they don't equal the `remove` variable
counter = 1
for (n in original_list){
if (n != ){
new_list[[counter]] = n
counter = counter + 1
}
}
The new_list variable no longer contains 5.
new_list
# [1] 1 2 3 4 6 7 8 9 10
How about this? Again, using indices
> m <- c(1:5)
> m
[1] 1 2 3 4 5
> m[1:length(m)-1]
[1] 1 2 3 4
or
> m[-(length(m))]
[1] 1 2 3 4
You can use which.
x<-c(1:5)
x
#[1] 1 2 3 4 5
x<-x[-which(x==4)]
x
#[1] 1 2 3 5
if you'd like to avoid numeric indices, you can use
a <- setdiff(names(a),c("name1", ..., "namen"))
to delete names namea...namen from a. this works for lists
> l <- list(a=1,b=2)
> l[setdiff(names(l),"a")]
$b
[1] 2
as well as for vectors
> v <- c(a=1,b=2)
> v[setdiff(names(v),"a")]
b
2