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dplyr/rlang: parse_expr with multiple expressions
(3 answers)
Closed 2 years ago.
I often create a "vector" of the variables I use most often while I'm coding. Usually if I just input the vector object in select it works perfectly. Is there any way I can use in the helper functions in a string?
For example I could do
library(dplyr)
x = c('matches("cyl")')
mtcars %>%
select_(x)
but this is not preferable because 1) select_ is deprecated and 2) it's not scalable (i.e., x = c('hp', 'matches("cyl")') will not grab both the relevant columns.
Is there anyway I could use more tidyselect helper functions in as part of a vector?
Note: if I do something like:
x = c(matches("cyl"))
#> Error: `matches()` must be used within a *selecting* function.
#> ℹ See <https://tidyselect.r-lib.org/reference/faq-selection-context.html>.
I get an error, so I'll definitely need to enquo it somehow.
You are trying to turn a string into code which might not be the best approach. However, you can use parse_exprs with !!!.
library(dplyr)
library(rlang)
x = c('matches("cyl")')
mtcars %>% select(!!!parse_exprs(x))
# Cyl
#Mazda RX4 6
#Mazda RX4 Wag 6
#Datsun 710 4
#Hornet 4 Drive 6
#Hornet Sportabout 8
#...
x = c('matches("cyl")', 'hp')
mtcars %>% select(!!!parse_exprs(x))
# cyl hp
#Mazda RX4 6 110
#Mazda RX4 Wag 6 110
#Datsun 710 4 93
#Hornet 4 Drive 6 110
#Hornet Sportabout 8 175
#....
Related
I have some dataframes named as:
1_patient
2_patient
3_patient
Now I am not able to access its variables. For example:
I am not able to obtain:
2_patient$age
If I press tab when writing the name, it automatically gets quoted, but I am still unable to use it.
Do you know how can I solve this?
It is not recommended to name an object with numbers as prefix, but we can use backquote to extract the value from the object
`1_patient`$age
If there are more than object, we can use mget to return the objects in a list and then extract the 'age' column by looping over the list with lapply
mget(ls(pattern = "^\\d+_mtcars$"))
#$`1_mtcars`
# mpg cyl disp hp drat wt qsec vs am gear carb
#Mazda RX4 21 6 160 110 3.9 2.620 16.46 0 1 4 4
#Mazda RX4 Wag 21 6 160 110 3.9 2.875 17.02 0 1 4 4
lapply(mget(ls(pattern = "^\\d+_patient$")), `[[`, 'age')
Using a small reproducible example
data(mtcars)
`1_mtcars` <- head(mtcars, 2)
1_mtcars$mpg
Error: unexpected input in "1_"
`1_mtcars`$mpg
#[1] 21 21
I have a table full of data, where one of the columns is scraped from a webpage and thus is full of HTML tags I don't need. I was looking to remove the HTML tags. I found this thread: Removing html tags from a string in R
I eventually got the regex version working (so my actual problem is solved), but originally tried implementing David Robinson's answer which utilised the rvest package. However, when I tried that, I had an issue where instead of running the function across each table row's string, it just performed it on the first row and copied the result down. I'm curious as to what I was doing wrong, so I know how to fix my calls next time I hit this kind of issue. Here's an example:
library(dplyr)
library(tibble)
library(rvest)
mtcars %>%
rownames_to_column("Car") %>%
select(Car) %>%
mutate(html_string = paste0("<a>",Car,"</a>")) %>%
mutate(cleaned_string = html_text(read_html(html_string)))
#thelatemail is correct I believe, read_html works only for single url, for it to work for multiple url's you need to either use rowwise or use some kind of looping.
library(dplyr)
library(rvest)
mtcars %>%
rownames_to_column("Car") %>%
select(Car) %>%
mutate(html_string = paste0("<a>",Car,"</a>")) %>%
rowwise() %>%
mutate(cleaned_string = html_text(read_html(html_string)))
# Car html_string cleaned_string
# <chr> <chr> <chr>
# 1 Mazda RX4 <a>Mazda RX4</a> Mazda RX4
# 2 Mazda RX4 Wag <a>Mazda RX4 Wag</a> Mazda RX4 Wag
# 3 Datsun 710 <a>Datsun 710</a> Datsun 710
# 4 Hornet 4 Drive <a>Hornet 4 Drive</a> Hornet 4 Drive
# 5 Hornet Sportabout <a>Hornet Sportabout</a> Hornet Sportabout
# 6 Valiant <a>Valiant</a> Valiant
# 7 Duster 360 <a>Duster 360</a> Duster 360
# 8 Merc 240D <a>Merc 240D</a> Merc 240D
# 9 Merc 230 <a>Merc 230</a> Merc 230
#10 Merc 280 <a>Merc 280</a> Merc 280
# … with 22 more rows
Or using purrr::map_chr
mtcars %>%
rownames_to_column("Car") %>%
select(Car) %>%
mutate(html_string = paste0("<a>",Car,"</a>")) %>%
mutate(cleaned_string = purrr::map_chr(html_string, ~html_text(read_html(.))))
I'm wondering if there is a way to select a group of columns by the name of the first column in the group and then all the next columns either a) to the end of the data frame, or b) to another column, also using its name.
a) As an example for the first question, in the mtcars dataset, is there a way to select the columns from drat to the end of the data frame? (Something like mtcars[,'drat':ncol(mtcars)])
b) For the second question, is there a way to select the columns starting at cyl and ending at wt? (Something like mtcars[,'cyl':'wt'])
Many elegant solutions already provided but one can even use base-R to get the desired result using which as:
Ans a:
mtcars[,which(names(mtcars) == "drat"):ncol(mtcars)]
Ans b:
mtcars[,which(names(mtcars) == "cyl"):which(names(mtcars) == "wt")]
# cyl disp hp drat wt
#Mazda RX4 6 160.0 110 3.90 2.620
#Mazda RX4 Wag 6 160.0 110 3.90 2.875
#Datsun 710 4 108.0 93 3.85 2.320
#Hornet 4 Drive 6 258.0 110 3.08 3.215
#Hornet Sportabout 8 360.0 175 3.15 3.440
#......so on
We can do with this with select from dplyr
Answer a)
mtcars %>% select(drat:get(last(names(.))))
Answer b)
mtcars %>% select(cyl:wt)
In dplyr, the select function does exactly this (no quotes needed):
mtcards %>%
select(cyl:wt)
If we need to use a quoted string, convert it to sym (symbol) and then do the evaluation (!!
mtcars %>%
select(!! (rlang::sym("cyl")): !!(rlang::sym("wt")))
It would be when these are stored in an object
a <- "cyl"
b <- "wt"
mtcars %>%
select(!! (rlang::sym(a)): !!(rlang::sym(b)))
Or another option is
mtcars %>%
select(!! rlang::parse_expr(glue::glue("{a}:{b}")))
I'm trying to write my function and need to pass argument inside.
Use mtcars dataset as an example:
get.param <- function(data, var){
data %>% select(eval(var)) %>%
head()
}
get.param(mtcars, 'hp')
In the above function, replacing eval() with get() gave me error.
I'm little bit confused which one should I use. I use get() i some other functions and work. What is the difference between these two?
You can get it to work via
get.param <- function(data, var){
var <- enquo(var)
data %>% select(!!var) %>%
head()
}
get.param(mtcars, hp)
hp
Mazda RX4 110
Mazda RX4 Wag 110
Datsun 710 93
Hornet 4 Drive 110
Hornet Sportabout 175
Valiant 105
Normally one does not use get or eval with dplyr. See the vignette in the rlang package for how it is done with that package; however, in this particular case one can just pass var directly to select adding parentheses around it so that it does not confuse it with a column called "var" should it exist. If you are not worried about that edge case you could omit the parentheses.
get.param <- function(data, var) {
data %>% select((var)) %>% head
}
get.param(mtcars, 'hp')
giving:
hp
Mazda RX4 110
Mazda RX4 Wag 110
Datsun 710 93
Hornet 4 Drive 110
Hornet Sportabout 175
Valiant 105
Another possibility is to use ... like this and giving the same answer. In this variation we don't need to add the parentheses to eliminate an edge case. It also allows multiple columns to be specified.
get.param <- function(data, ...) {
data %>% select(...) %>% head
}
get.param(mtcars, 'hp')
EXAMPLE DATASET:
mtcars
mpg cyl disp hp drat wt ...
Mazda RX4 21.0 6 160 110 3.90 2.62 ...
Mazda RX4 Wag 21.0 6 160 110 3.90 2.88 ...
Datsun 710 22.8 4 108 93 3.85 2.32 ...
............
Recommended ggplot way:
ggplot(mtcars,aes(x=mpg)) + geom_histogram
They way I want to do it:
ggplot(mtcars,aes(x=[,1]) +geom_histogram
or
ggplot(mtcars,aes(x=[[1]]))+geom_histogram
Why can't ggplot let me call out my variable by its column? I need to call it out by column number not name. Why is ggplot so strict here? Any work around for this?
The problem you're facing is that the ggplot aes argument evaluates within the data.frame that you pass it. A column name is a string, and can't be properly evaluated the same way.
Fortunately, there is a solution: use the aes_string option, as follows:
library(ggplot2)
my_data <- mtcars
names(my_data)
ggplot(my_data, aes_string(x=names(my_data)[1]))+
geom_histogram()
This works because names(my_data)[1] returns a string, and is perfectly acceptable for the aes_string option.