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)
Related
Is there a method in R, to substitute the values of a vector using a dictionary (2 column dataframe with old and new value)
The only method I know is to extract the old value into a dataframe and merge it with, what I call,the dictionary (which is a two column dataframe with old and new values). Afterwards reassign the new value to the original old value. However, it seems when using merge (at least since R v4.1, the order of the x value is not maintained, so I am using join now which keeps the original order of dataframe x intact. I am thinking that there must be an easier way, I just have not found it. Hope this is understandable, I appreciate any help.
cheers Hermann
You could use a named character vector as a dict for replacement by unquoting with !!! inside of dplyr::recode. If you have your "dict" stored as a two-column dataframe, then tidyr::deframe might be handy.
library(tidyverse)
x <- c("a", "b", "c")
dict <- tribble(
~old, ~new,
"a", "d",
"b", "e",
"c", "f"
)
recode(x, !!!deframe(dict))
#> [1] "d" "e" "f"
Created on 2021-06-14 by the reprex package (v1.0.0)
You can use match to substitute the values of a vector using a dictionary:
D$new[match(x, D$old)]
#[1] "d" "e" "f"
You can also use the names to get the new values:
L <- setNames(D$new, D$old)
L[x]
#"d" "e" "f"
Data:
x <- c("a", "b", "c")
D <- data.frame(old = c("a", "b", "c"), new = c("d", "e", "f"))
suppose I have 3 vectors:
a = c("A", "B", "C")
b = c("D", "E", "F")
c = c("G", "H", "I")
then I have an element:
element = "E"
I want to find which list does my element belongs to. In this case, list b.
It will be appreciated if the solution to this problem is more general because my real data set have more than a hundred lists.
element = "E"
names(our_lists)[sapply(our_lists, `%in%`, x = element)]
# [1] "b"
Data
our_lists <- list(
a = c("A", "B", "C"),
b = c("D", "E", "F"),
c = c("G", "H", "I")
)
Using grep.
element <- "E"
l <- mget(c("a", "b", "c"))
names(l)[grep(element, l)]
# [1] "b"
If you keep the data in individual objects, you need to check for the element in each one individually. Get them in a list.
list_data <- mget(c('a', 'b', 'c'))
names(Filter(any, lapply(list_data, `==`, element)))
#[1] "b"
If all your vectors have the same length then a vectorised idea can be,
c('a', 'b', 'c')[ceiling(which(c(a, b, c) == 'E') / length(a))]
#[1] "b"
You can use dplyr::lst that creates named list from variable names. Then purrr::keep to keep only the vectors that contain your element.
require(tidyverse)
lst(a, b, c) %>%
keep(~ element %in% .x) %>%
names()
output:
[1] "b"
Is there a way to use the recode function of dpylr together with a lookup table (data.frame or list)?
What I would like to have would look something like this:
# Recode values with list of named arguments
data <- sample(c("a", "b", "c", "d"), 10, replace = T)
lookup <- list(a = "Apple", b = "Pear")
dplyr::recode(data, lookup)
I found the mapvalues and revalue functions from the plyr package. Combining them is possible as explained here.
However, I am wondering whether something similar is possible with dplyr only.
do.call(dplyr::recode, c(list(data), lookup))
[1] "Pear" "c" "d" "c" "Pear" "Pear" "d" "c" "d" "c"
We can use base R
v1 <- unlist(lookup)[data]
ifelse(is.na(v1), data, v1)
It works like this:
dplyr::recode(data, !!!lookup)
Also useful with mutate in a dataframe tibble:
df <- tibble(code = data)
df %>%
mutate(fruit = recode(code, !!!lookup))
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.
I would like to make a subset of a data frame in R that is based on one OR another value in a column of factors but it seems I cannot use | with factor values.
Example:
# fake data
x <- sample(1:100, 9)
nm <- c("a", "a", "a", "b", "b", "b", "c", "c", "c")
fake <- cbind(as.data.frame(nm), as.data.frame(x))
# subset fake to only rows with name equal to a or b
fake.trunk <- fake[fake$nm == "a" | "b", ]
produces the error:
Error in fake$nm == "a" | "b" :
operations are possible only for numeric, logical or complex types
How can I accomplish this?
Obviously my actual data frame has more than 3 values in the factor column so just using != "c" won't work.
You need fake.trunk <- fake[fake$nm == "a" | fake$nm == "b", ]. A more concise way of writing that (especially with more than two conditions) is:
fake[ fake$nm %in% c("a","b"), ]
Another approach would be to use subset() and write
fake.trunk = subset(fake, nm %in% c('a', 'b'))