Assuming:
aa = c('A','B','C')
bb = c('D','E','F')
list1 = list(aa,bb)
vect1 = c(aa,bb)
Is there a way to extract the variable names as strings ('aa', 'bb') from either list1 or vect1?
Is this information being stored in lists and vectors? If not, what would be the appropriate format?
Thanks in advance!
For the situation what you have done the answer is no. But if you are ready to do some changes in your code then you can easily get it,
list1 <- list( aa = aa, bb = bb)
Now you can easily access the string version of names of variables from which list is formed,
names(list1)
aa = c('A','B','C')
bb = c('D','E','F')
list1 = list(aa,bb)
vect1 = c(aa,bb)
The short answer is no. If you look at the results of dput(list1) and dput(vect1), you'll see that these objects don't contain that information any more:
list(c("A", "B", "C"), c("D", "E", "F"))
c("A", "B", "C", "D", "E", "F")
There is one way you can get this information, which is when the expression has been passed to a function:
f <- function(x) {
d <- substitute(x)
n <- sapply(d[-1],deparse)
return(n)
}
f(c(aa,bb))
## [1] "aa" "bb"
However, it would be good to have more context about what you want to do.
You can also get there by adapting vect1 using cbind:
vect1 = cbind(aa,bb)
colnames(vect1)
Related
I’d be grateful for suggestions as to how to remap letters in strings in a map-specified way.
Suppose, for instance, I want to change all As to Bs, all Bs to Ds, and all Ds to Fs. If I do it like this, it doesn’t do what I want since it applies the transformations successively:
"abc" %>% str_replace_all(c(a = "b", b = "d", d = "f"))
Here’s a way I can do what I want, but it feels a bit clunky.
f <- function (str) str_c( c(a = "b", b = "d", c = "c", d = "f") %>% .[ strsplit(str, "")[[1]] ], collapse = "" )
"abc" %>% map_chr(f)
Better ideas would be much appreciated.
James.
P.S. Forgot to specify. Sometimes I want to replace a letter with multiple letters, e.g., replace all As with the string ZZZ.
P.P.S. Ideally, this would be able to handle vectors of strings too, e.g., c("abc", "gersgaesg", etc.)
We could use chartr in base R
chartr("abc", "bdf", "abbbce")
#[1] "bdddfe"
Or a package solution would be mgsub which would also match and replace strings with number of characters greater than 1
library(mgsub)
mgsub("abbbce", c("a", "b", "c"), c("b", "d", "f"))
#[1] "bdddfe"
mgsub("abbbce", c("a", "b", "c"), c("ba", "ZZZ", "f"))
#[1] "baZZZZZZZZZfe"
Maybe this is more elegant? It will also return warnings when values aren't found.
library(plyr)
library(tidyverse)
mappings <- c(a = "b", b = "d", d = "f")
str_split("abc", pattern = "") %>%
unlist() %>%
mapvalues(from = names(mappings), to = mappings) %>%
str_c(collapse = "")
# The following `from` values were not present in `x`: d
# [1] "bdc"
I have a list and I would like to remove any list object with a sublist. In the example below, I would like to remove ob2 and ob5 and keep all other objects.
dat <- list(ob1 = c("a", "b", "c"),
ob2 = list(a = c("d")),
ob3 = c("e", "f", "g"),
ob4 = c("h", "i", "j"),
ob5 = list(1:3))
Can anyone offer a solution of how to do this?
We can create a condition with sapply (from base R)
dat[!sapply(dat, is.list)]
Or with Filter from base R
Filter(Negate(is.list), dat)
Or with discard
library(purrr)
discard(dat, is.list)
From the recode examples, what if I have two variables where I want to apply the same recode?
factor_vec1 <- factor(c("a", "b", "c"))
factor_vec2 <- factor(c("a", "d", "f"))
How can I recode the same answer without writing a recode for each factor_vec? These don't work, do I need to learn how to use purrr to do it, or is there another way?
Output 1: recode(c(factor_vec1, factor_vec2), a = "Apple")
Output 2: recode(c(factor_vec2, factor_vec2), a = "Apple", b =
"Banana")
If there are not many items needed to be recoded, you can try a simple lookup table approach using base R.
v1 <- c("a", "b", "c")
v2 <- c("a", "d", "f")
# lookup table
lut <- c("a" ="Apple",
"b" = "Banana",
"c" = "c",
"d" = "d",
"f" = "f")
lut[v1]
lut[v2]
You can reuse the lookup table for any relevant variables. The results are:
> lut[v1]
a b c
"Apple" "Banana" "c"
> lut[v2]
a d f
"Apple" "d" "f"
Use lists to hold multiple vectors and then you can apply same function using lapply/map.
library(dplyr)
list_fac <- lst(factor_vec1, factor_vec2)
list_fac <- purrr::map(list_fac, recode, a = "Apple", b = "Banana")
You can keep the vectors in list itself (which is better) or get the changed vectors in global environment using list2env.
list2env(list_fac, .GlobalEnv)
This question already has answers here:
Create sequence of repeated values, in sequence?
(3 answers)
Closed 6 years ago.
I want to create the following vector using a, b, c repeating each letter thrice:
BB<-c("a","a","a","b","b","b","c","c","c")
This is my code:
Alphabet<-c("a","b","c")
AA<-list()
for(i in 1:3){
AA[[i]]<-rep(Alphabet[i],each=3)
}
BB<-do.call(rbind,AA)
But I am getting a dataframe:
dput(BB)
structure(c("a", "b", "c", "a", "b", "c", "a", "b", "c"), .Dim = c(3L,
3L))
What I am doing wrong?
As Akrun mentioned we can use the same rep function
create a vector which consists of letters a,b,c
A <- c("A","B","C")
Apply rep function for the same vector, use each as sub function
AA <- rep(A,each=3)
print(AA)
[1] "A" "A" "A" "B" "B" "B" "C" "C" "C"
You should use c function to concatenate, not the rbind. This will give you vector.
Alphabet<-c("a","b","c")
AA<-list()
for(i in 1:3){
AA[[i]]<-rep(Alphabet[i],each=3)
}
BB<-do.call(c,AA)
Akrun comment is also true, if thats what you want.
You can also concatenate the rep function like so:
BB <- c(rep("a", 3), rep("b", 3), rep("c", 3))
Here is a solution but note this form or appending is not efficient for large input arrays
Alphabet <- c("a","b","c")
bb <- c()
for (i in 1:length(Alphabet)) {
bb <- c(bb, rep(Alphabet[i], 3))
}
I have a list of 10,000 vectors, and each vector might have different elements and different lengths. I would like to know how many unique vectors I have and how often each unique vector appears in the list.
I guess the way to go is the function "unique", but I don't know how I could use it to also get the number of times each vector is repeated.
So what I would like to get is something like that:
"a" "b" "c" d" 301
"a" 277
"b" c" 49
being the letters, the contents of each unique vector, and the numbers, how often are repeated.
I would really appreciate any possible help on this.
thank you very much in advance.
Tina.
Maybe you should look at table:
Some sample data:
myList <- list(A = c("A", "B"),
B = c("A", "B"),
C = c("B", "A"),
D = c("A", "B", "B", "C"),
E = c("A", "B", "B", "C"),
F = c("A", "C", "B", "B"))
Paste your vectors together and tabulate them.
table(sapply(myList, paste, collapse = ","))
#
# A,B A,B,B,C A,C,B,B B,A
# 2 2 1 1
You don't specify whether order matters (that is, is A, B the same as B, A). If it does, you can try something like:
table(sapply(myList, function(x) paste(sort(x), collapse = ",")))
#
# A,B A,B,B,C
# 3 3
Wrap this in data.frame for a vertical output instead of horizontal, which might be easier to read.
Also, do be sure to read How to make a great R reproducible example? as already suggested to you.
As it is, I'm just guessing at what you're trying to do.