Is there an R function which can pass all the elements of a list as the arguments of a function?
library(tidyr)
a <- c(1,2,3)
b <- c(4,5,6)
c <- c(7,8,9)
d <- list(a,b,c)
crossing(d[[1]],d[[2]],d[[3]])
Instead of specifying d[[1]],d[[2]],d[[3]], i'd like to just include d
Expected result:
> crossing(d[[1]],d[[2]],d[[3]])
# A tibble: 27 x 3
`d[[1]]` `d[[2]]` `d[[3]]`
<dbl> <dbl> <dbl>
1 1 4 7
2 1 4 8
3 1 4 9
4 1 5 7
5 1 5 8
6 1 5 9
7 1 6 7
8 1 6 8
9 1 6 9
10 2 4 7
# ... with 17 more rows
You can use do.call to executes a function call and a list of arguments to be passed to it.
c(d[[1]],d[[2]],d[[3]])
#[1] 1 2 3 4 5 6 7 8 9
do.call("c", d)
#[1] 1 2 3 4 5 6 7 8 9
And for crossing, which needs not duplicated Column names:
library(tidyr)
names(d) <- seq_along(d)
do.call(crossing, d)
## A tibble: 27 x 3
# `1` `2` `3`
# <dbl> <dbl> <dbl>
# 1 1 4 7
# 2 1 4 8
# 3 1 4 9
# 4 1 5 7
# 5 1 5 8
# 6 1 5 9
# 7 1 6 7
# 8 1 6 8
# 9 1 6 9
#10 2 4 7
## … with 17 more rows
Related
Hello I have a data frame of 245 columns but to add some sets and generate new columns try to do it recursively as follows
cl1<-sample(1:4,10,replace=TRUE)
cl2<-sample(1:4,10,replace=TRUE)
cl3<-sample(1:4,10,replace=TRUE)
cl4<-sample(1:4,10,replace=TRUE)
cl5<-sample(1:4,10,replace=TRUE)
cl6<-sample(1:4,10,replace=TRUE)
dat<-data.frame(cl1,cl2,cl3,cl4,cl5,cl6)
my intention is to add column 1 with column 3 and 5, likewise column 2 with 4 and 6 and in the end obtain a dataframe with two columns
and you should pay me something like that
I have programmed the following code
revisar<- function(a){
todos = list()
i=1
j=3
l=5
k=1
while(i<=2 ){
cl<-a[,i]
cl2<-a[,j]
cl3<-a[,l]
cl[is.na(cl)] <- 0
cl2[is.na(cl2)] <- 0
cl3[is.na(cl3)] <- 0
colu<-cl+cl2+cl3
col<-cbind(colu,colu)
i<-i+1
j<-j+1
l<-l+1
k<-k+1
}
return(col)
}
it turns out that it only returns column 2 repeated twice and I must replicate the same thing to join those 245 columns.7
I would like to know what is failing the example
base R
Literal programming:
with(dat, data.frame(s1 = cl1+cl3+cl5, s2 = cl2+cl4+cl6))
# s1 s2
# 1 7 11
# 2 7 7
# 3 4 11
# 4 4 10
# 5 9 8
# 6 12 5
# 7 7 6
# 8 7 10
# 9 4 9
# 10 6 5
Programmatically,
L <- list(s1 = c(1,3,5), s2 = c(2,4,6))
out <- data.frame(lapply(L, function(z) do.call(rowSums, list(as.matrix(dat[,z])))))
out
# s1 s2
# 1 7 11
# 2 7 7
# 3 4 11
# 4 4 10
# 5 9 8
# 6 12 5
# 7 7 6
# 8 7 10
# 9 4 9
# 10 6 5
dplyr
library(dplyr)
dat %>%
transmute(
s1 = rowSums(cbind(cl1, cl3, cl5)),
s2 = rowSums(cbind(cl2, cl4, cl6))
)
or programmatically using purrr:
purrr::map_dfc(L, ~ rowSums(dat[, .]))
Data
set.seed(42)
# your `dat` above
Here is an alternative general approach:
Here we sum all uneven columns -> s1 and
all even columns -> s2:
library(dplyr)
dat %>%
rowwise() %>%
mutate(s1 = sum(c_across(seq(1,ncol(dat),2)), na.rm = TRUE),
s2 = sum(c_across(seq(2,ncol(dat),2)), na.rm = TRUE))
cl1 cl2 cl3 cl4 cl5 cl6 s1 s2
<int> <int> <int> <int> <int> <int> <int> <int>
1 1 1 3 2 3 2 7 5
2 2 4 1 4 2 3 5 11
3 2 2 2 2 1 3 5 7
4 2 4 4 3 1 4 7 11
5 2 4 4 3 2 2 8 9
6 3 3 3 2 2 2 8 7
7 2 1 1 2 1 4 4 7
8 2 4 1 3 2 3 5 10
9 3 1 1 2 3 4 7 7
10 2 4 1 3 4 4 7 11
I'm using rename_at, but since it is superseded, I need to find a way to rename n last columns with some vector of strings using rename_with() or rename()
library(tidyverse)
df <- tibble(
a = 1:10,
b = 1:10,
c = 1:10,
d = 1:10,
e = 1:10
)
new_names <- c("1", "2", "4", "5", "10")
df %>%
rename_at(vars(names(.) %>% tail(5)), funs(paste0("", new_names))) # only `funs(new_names)` won't work
Base R approach :
n <- ncol(df)
names(df)[(n-4):n] <- new_names
df
# A tibble: 10 x 5
# `1` `2` `4` `5` `10`
# <int> <int> <int> <int> <int>
# 1 1 1 1 1 1
# 2 2 2 2 2 2
# 3 3 3 3 3 3
# 4 4 4 4 4 4
# 5 5 5 5 5 5
# 6 6 6 6 6 6
# 7 7 7 7 7 7
# 8 8 8 8 8 8
# 9 9 9 9 9 9
#10 10 10 10 10 10
Using rename_with
library(dplyr)
library(stringr)
df %>%
rename_with(~ str_c(., new_names), tail(names(.), 5))
# A tibble: 10 x 5
# a1 b2 c4 d5 e10
# <int> <int> <int> <int> <int>
# 1 1 1 1 1 1
# 2 2 2 2 2 2
# 3 3 3 3 3 3
# 4 4 4 4 4 4
# 5 5 5 5 5 5
# 6 6 6 6 6 6
# 7 7 7 7 7 7
# 8 8 8 8 8 8
# 9 9 9 9 9 9
#10 10 10 10 10 10
Or with rename
df %>%
rename(!!! setNames(tail(names(.), 5), new_names))
Or using rename_at directly on the tail of names
df %>%
rename_at(vars(tail(names(.), 5)), ~ str_c(., new_names))
-output
# A tibble: 10 x 5
# a1 b2 c4 d5 e10
# <int> <int> <int> <int> <int>
# 1 1 1 1 1 1
# 2 2 2 2 2 2
# 3 3 3 3 3 3
# 4 4 4 4 4 4
# 5 5 5 5 5 5
# 6 6 6 6 6 6
# 7 7 7 7 7 7
# 8 8 8 8 8 8
# 9 9 9 9 9 9
#10 10 10 10 10 10
if it is to just replace the names
df %>%
rename_at(vars(tail(names(.), 5)), ~ new_names)
# A tibble: 10 x 5
# `1` `2` `4` `5` `10`
# <int> <int> <int> <int> <int>
# 1 1 1 1 1 1
# 2 2 2 2 2 2
# 3 3 3 3 3 3
# 4 4 4 4 4 4
# 5 5 5 5 5 5
# 6 6 6 6 6 6
# 7 7 7 7 7 7
# 8 8 8 8 8 8
# 9 9 9 9 9 9
#10 10 10 10 10 10
In the example, there are only 5 columns. Suppose, if it is only the last 3 columns
df %>%
rename_at(vars(tail(names(.), 3)), ~ str_c(., tail(new_names, 3)))
funs take a function, so wrapping with paste0 or as.character does that instead of just a input vector
I am having some trouble getting mutate, across, and case_when to function properly, I've recreated a simple version of my problem here:
a <- c(1:10)
b <- c(2:11)
c <- c(3:12)
test <- tibble(a, b, c)
# A tibble: 10 x 3
a b c
<int> <int> <int>
1 1 2 3
2 2 3 4
3 3 4 5
4 4 5 6
5 5 6 7
6 6 7 8
7 7 8 9
8 8 9 10
9 9 10 11
10 10 11 12
My goal is to replace all of the 3's with 4's, and keep everything else the same. I have the following code:
test_1 <-
test %>%
mutate(across(a:c, ~ case_when(. == 3 ~ 4)))
# A tibble: 10 x 3
a b c
<dbl> <dbl> <dbl>
1 NA NA 4
2 NA 4 NA
3 4 NA NA
4 NA NA NA
5 NA NA NA
6 NA NA NA
7 NA NA NA
8 NA NA NA
9 NA NA NA
10 NA NA NA
It's close but I get NA values where I want to maintain the value in the original tibble. How do I maintain the original values using the mutate across structure?
Thank you in advance!
What about this?
> test %>%
+ mutate(across(a:c, ~ case_when(. == 3 ~ 4, TRUE ~ 1 * (.))))
# A tibble: 10 x 3
a b c
<dbl> <dbl> <dbl>
1 1 2 4
2 2 4 4
3 4 4 5
4 4 5 6
5 5 6 7
6 6 7 8
7 7 8 9
8 8 9 10
9 9 10 11
10 10 11 12
or
> test %>%
+ replace(. == 3, 4)
# A tibble: 10 x 3
a b c
<int> <int> <int>
1 1 2 4
2 2 4 4
3 4 4 5
4 4 5 6
5 5 6 7
6 6 7 8
7 7 8 9
8 8 9 10
9 9 10 11
10 10 11 12
In base R, we can do
test[test ==3] <- 4
This also works:
a <- c(1:10)
b <- c(2:11)
c <- c(3:12)
tibble(a, b, c) %>%
modify(~ ifelse(. == 3, 4, .))
# A tibble: 10 x 3
a b c
<dbl> <dbl> <dbl>
1 1 2 4
2 2 4 4
3 4 4 5
4 4 5 6
5 5 6 7
6 6 7 8
7 7 8 9
8 8 9 10
9 9 10 11
10 10 11 12
The following data set is in the wide format and has repeated measures of "ql", "st" and "xy" prefixed by "a", "b" and "c";
df<-data.frame(id=c(1,2,3,4),
ex=c(1,0,0,1),
aql=c(5,4,NA,6),
bql=c(5,7,NA,9),
cql=c(5,7,NA,9),
bst=c(3,7,8,9),
cst=c(8,7,5,3),
axy=c(1,9,4,4),
cxy=c(5,3,1,4))
I'm looking for a way to insert dots after the prefixed letters "a", "b" and "c", while keeping other columns (i.e. id, ex) unchanged. I've been working around this using gsub function, e.g.
names(df) <- gsub("", "\\.", names(df))
but got undesired results. The expected output would look like
id ex a.ql b.ql c.ql b.st c.st a.xy c.xy
1 1 1 5 5 5 3 8 1 5
2 2 0 4 7 7 7 7 9 3
3 3 0 NA NA NA 8 5 4 1
4 4 1 6 9 9 9 3 4 4
Try
sub("(^[a-c])(.+)", "\\1.\\2", names(df))
# [1] "id" "ex" "a.ql" "b.ql" "c.ql" "b.st" "c.st" "a.xy" "c.xy"
or
sub("(?<=^[a-c])", ".", names(df), perl = TRUE)
# [1] "id" "ex" "a.ql" "b.ql" "c.ql" "b.st" "c.st" "a.xy" "c.xy"
You can do
setNames(df, sub("(ql$)|(st$)|(xy$)", "\\.\\1\\2\\3", names(df)))
#> id ex a.ql b.ql c.ql b.st c.st a.xy c.xy
#> 1 1 1 5 5 5 3 8 1 5
#> 2 2 0 4 7 7 7 7 9 3
#> 3 3 0 NA NA NA 8 5 4 1
#> 4 4 1 6 9 9 9 3 4 4
Another way you can try
library(dplyr)
df %>%
rename_at(vars(aql:cxy), ~ str_replace(., "(?<=\\w{1})", "\\."))
# id ex a.ql b.ql c.ql b.st c.st a.xy c.xy
# 1 1 1 5 5 5 3 8 1 5
# 2 2 0 4 7 7 7 7 9 3
# 3 3 0 NA NA NA 8 5 4 1
# 4 4 1 6 9 9 9 3 4 4
You can also try a tidyverse approach reshaping your data like this:
library(tidyverse)
#Data
df<-data.frame(id=c(1,2,3,4),
ex=c(1,0,0,1),
aql=c(5,4,NA,6),
bql=c(5,7,NA,9),
cql=c(5,7,NA,9),
bst=c(3,7,8,9),
cst=c(8,7,5,3),
axy=c(1,9,4,4),
cxy=c(5,3,1,4))
#Reshape
df %>% pivot_longer(-c(1,2)) %>%
mutate(name=paste0(substring(name,1,1),'.',substring(name,2,nchar(name)))) %>%
pivot_wider(names_from = name,values_from=value)
Output:
# A tibble: 4 x 9
id ex a.ql b.ql c.ql b.st c.st a.xy c.xy
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 5 5 5 3 8 1 5
2 2 0 4 7 7 7 7 9 3
3 3 0 NA NA NA 8 5 4 1
4 4 1 6 9 9 9 3 4 4
# A tibble: 12 x 3
x y z
<dbl> <int> <int>
1 1 1 2
2 3 2 3
3 2 3 4
4 3 4 5
5 2 5 6
6 4 6 7
7 5 7 8
8 2 8 9
9 1 9 10
10 1 10 11
11 3 11 12
12 4 12 13
The above is named df. It is one of ten tibbles that I want to store inside of:
t <- tibble(tbl=vector("list", 10))
If I do this:
t$tbl[1] <- df
or
t[1, 1] <- df
I get this warning message:
Warning message:
In t$tbl[1] <- df :
number of items to replace is not a multiple of replacement length
and the output of
t$tbl[1]
is
[[1]]
[1] 1 3 2 3 2 4 5 2 1 1 3 4