I want to rename multiple columns that starts with the same string.
However, all the codes I tried did not change the columns.
For example this:
df %>% rename_at(vars(matches('^oldname,\\d+$')), ~ str_replace(., 'oldname', 'newname'))
And also this:
df %>% rename_at(vars(starts_with(oldname)), funs(sub(oldname, newname, .))
Are you familiar with a suitable code for rename?
Thank you!
Take iris for example, you can use rename_with() to replace those column names started with "Petal" with a new string.
head(iris) %>%
rename_with(~ sub("^Petal", "New", .x), starts_with("Petal"))
Sepal.Length Sepal.Width New.Length New.Width Species
1 5.1 3.5 1.4 0.2 setosa
2 4.9 3.0 1.4 0.2 setosa
3 4.7 3.2 1.3 0.2 setosa
4 4.6 3.1 1.5 0.2 setosa
5 5.0 3.6 1.4 0.2 setosa
6 5.4 3.9 1.7 0.4 setosa
You can also use rename_at() in this case, although rename_if(), rename_at(), and rename_all() have been superseded by rename_with().
head(iris) %>%
rename_at(vars(starts_with("Petal")), ~ sub("^Petal", "New", .x))
Related
Not sure why the first one has an error but the second line works? My understanding was using names(.) in the formulas tells R to use the data before pipe operator. It seems to work for .cols argument but not for formula.
iris%>%rename_with(~gsub("Petal","_",names(.)),all_of(names(.)))
iris%>%rename_with(~~gsub("Petal","_",names(iris)),all_of(names(.)))
rename_with applies a function to the names of the passed data frame. The function should be one that, given the vector of names, returns the altered names, so the syntax is much simpler than you are trying to make it:
iris %>%
rename_with(~ gsub("Petal", "_", .x))
#> Sepal.Length Sepal.Width _.Length _.Width Species
#> 1 5.1 3.5 1.4 0.2 setosa
#> 2 4.9 3.0 1.4 0.2 setosa
#> 3 4.7 3.2 1.3 0.2 setosa
#> 4 4.6 3.1 1.5 0.2 setosa
#> 5 5.0 3.6 1.4 0.2 setosa
#> 6 5.4 3.9 1.7 0.4 setosa
#... etc
I would like to use dplyr to divide a subset of variables by the IQR. I am open to ideas that use a different approach than what I've tried before, which is a combination of mutate_if and %in%. I want to reference the list bin instead of indexing the data frame by position. Thanks for any thoughts!
contin <- c("age", "ct")
data %>%
mutate_if(%in% contin, function(x) x/IQR(x))
You should use:
data %>%
mutate(across(all_of(contin), ~.x/IQR(.x)))
Working example:
data <- head(iris)
contin <- c("Sepal.Length", "Sepal.Width")
data %>%
mutate(across(all_of(contin), ~.x/IQR(.x)))
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 15.69231 7.777778 1.4 0.2 setosa
2 15.07692 6.666667 1.4 0.2 setosa
3 14.46154 7.111111 1.3 0.2 setosa
4 14.15385 6.888889 1.5 0.2 setosa
5 15.38462 8.000000 1.4 0.2 setosa
6 16.61538 8.666667 1.7 0.4 setosa
I'm looking to use a non-across function from mutate to create multiple columns. My problem is that the variable in the function will change along with the crossed variables. Here's an example:
needs=c('Sepal.Length','Petal.Length')
iris %>% mutate_at(needs, ~./'{col}.Width')
This obviously doesn't work, but I'm looking to divide Sepal.Length by Sepal.Width and Petal.Length by Petal.Width.
I think your needs should be something which is common in both the columns.
You can select the columns based on the pattern in needs and divide the data based on position. !! and := is used to assign name of the new columns.
library(dplyr)
library(rlang)
needs = c('Sepal','Petal')
purrr::map_dfc(needs, ~iris %>%
select(matches(.x)) %>%
transmute(!!paste0(.x, '_divide') := .[[1]]/.[[2]]))
# Sepal_divide Petal_divide
#1 1.457142857 7.000000000
#2 1.633333333 7.000000000
#3 1.468750000 6.500000000
#4 1.483870968 7.500000000
#...
#...
If you want to add these as new columns you can do bind_cols the above with iris.
Here is a base R approach based that the columns you want to divide have a similar name pattern,
res <- sapply(split.default(iris[-ncol(iris)], sub('\\..*', '', names(iris[-ncol(iris)]))), function(i) i[1] / i[2])
iris[names(res)] <- res
head(iris)
# Sepal.Length Sepal.Width Petal.Length Petal.Width Species Petal.Petal.Length Sepal.Sepal.Length
#1 5.1 3.5 1.4 0.2 setosa 7.00 1.457143
#2 4.9 3.0 1.4 0.2 setosa 7.00 1.633333
#3 4.7 3.2 1.3 0.2 setosa 6.50 1.468750
#4 4.6 3.1 1.5 0.2 setosa 7.50 1.483871
#5 5.0 3.6 1.4 0.2 setosa 7.00 1.388889
#6 5.4 3.9 1.7 0.4 setosa 4.25 1.384615
I'm trying to create a binary set of variables that uses data across multiple columns.
I have a dataset where I'm trying to create a binary variable where any column with a specific name will be indexed for a certain value. I'll use iris as an example dataset.
Let's say I want to create a variable where any column with the string "Sepal" and any row in those columns with the values of 5.1, 3.0, and 4.7 will become "Class A" while values with 3.1, 5.0, and 5.4 will be "Class B". So let's look at the first few entries of iris
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 5.1 3.5 1.4 0.2 setosa
2 4.9 3.0 1.4 0.2 setosa
3 4.7 3.2 1.3 0.2 setosa
4 4.6 3.1 1.5 0.2 setosa
5 5.0 3.6 1.4 0.2 setosa
6 5.4 3.9 1.7 0.4 setosa
The first 3 rows should then be under "Class A" While rows 4-6 will be under "Class B". I tried writing this code to do that
mutate(iris, Class = if_else(
vars(contains("Sepal")), any_vars(. %in% c(5.1,3.0, 4.7))), "Class A",
ifelse(vars(contains("Sepal")), any_vars(. %in% c(3.1, 5.0, 5.4))), "Class B",NA)
and received the error
Error: `condition` must be a logical vector, not a `quosures/list` object
So I've realized I need lapply here, but I'm not even sure where to begin to write this because I'm not sure how to represent the entire part of selecting columns with "Sepal" in the name and also include the specific values in those rows as one list object to provide to lapply
This is clearly the wrong syntax
lapply(vars(contains("Sepal")), any_vars(. %in% c(5.1,3.0, 4.7)))
Examples using case_when will also be accepted as answers.
If you want to do this using dplyr, you can use rowwise with new c_across :
library(dplyr)
iris %>%
rowwise() %>%
mutate(Class = case_when(
any(c_across(contains("Sepal")) %in% c(5.1,3.0, 4.7)) ~ 'Class A',
any(c_across(contains("Sepal")) %in% c(3.1,5.0,5.4)) ~ 'Class B')) %>%
head
# Sepal.Length Sepal.Width Petal.Length Petal.Width Species Class
# <dbl> <dbl> <dbl> <dbl> <fct> <chr>
#1 5.1 3.5 1.4 0.2 setosa Class A
#2 4.9 3 1.4 0.2 setosa Class A
#3 4.7 3.2 1.3 0.2 setosa Class A
#4 4.6 3.1 1.5 0.2 setosa Class B
#5 5 3.6 1.4 0.2 setosa Class B
#6 5.4 3.9 1.7 0.4 setosa Class B
However, note that using %in% on numerical values is not accurate. If interested you may read Why are these numbers not equal?
I want to manipulate several columns to create new columns with names that are variants of the names of the columns being manipulating.
dplyr 1.0.0's across() function seems like the tool for the job, but the .names argument seems to have limited functionality. Here's what I want to do:
tmp <- iris %>%
mutate(across(starts_with('Sepal'),
~ .x - Petal.Length,
.names = gsub('Sepal', '', "{col}")))
but the gsub function doesn't work. I can work around this in the following way:
tmp <- iris %>%
mutate(across(starts_with('Sepal'),
~ .x - Petal.Length,
.names = "mod_{col}"))
names(tmp) <- gsub("mod_Sepal", "mod_", names(tmp))
but that requires more code and is harder to keep track of. Am I missing something here and is there a simpler way to set the new column names with across?
We can use rename_at after the mutate step
library(dplyr)
library(stringr)
iris %>%
mutate(across(starts_with('Sepal'),
~ .x - Petal.Length)) %>%
rename_at(vars(starts_with("Sepal")), ~ str_remove(., "Sepal"))
According to ?across
.names - The default (NULL) is equivalent to "{col}" for the single function case
And there is no option to remove the already existing column name, but, we can add a suffix or prefix
You can pass a function to .names as -
library(dplyr)
iris %>%
mutate(across(starts_with('Sepal'), ~ .x - Petal.Length,
.names = "{gsub('Sepal.', '', {col}, fixed = TRUE)}"))
# Sepal.Length Sepal.Width Petal.Length Petal.Width Species Length Width
#1 5.1 3.5 1.4 0.2 setosa 3.7 2.1
#2 4.9 3.0 1.4 0.2 setosa 3.5 1.6
#3 4.7 3.2 1.3 0.2 setosa 3.4 1.9
#4 4.6 3.1 1.5 0.2 setosa 3.1 1.6
#5 5.0 3.6 1.4 0.2 setosa 3.6 2.2
#6 5.4 3.9 1.7 0.4 setosa 3.7 2.2