I would like to use mutate to add new columns to a data.frame based on specific colums divided by another column and keep the originalname plus a fixed pattern.
mtcars$mpg_HorsePower = mtcars$mpg / mtcars$hp
mtcars$cyl_HorsePower = mtcars$cyl / mtcars$hp
mtcars$disp_HorsePower = mtcars$disp / mtcars$hp
head(mtcars)
# mpg cyl disp hp drat wt qsec vs am gear carb mpg_HorsePower cyl_HorsePower disp_HorsePower
# Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 0.1909091 0.05454545 1.454545
# Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 0.1909091 0.05454545 1.454545
# Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 0.2451613 0.04301075 1.161290
# Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 0.1945455 0.05454545 2.345455
# Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 0.1068571 0.04571429 2.057143
# Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 0.1723810 0.05714286 2.142857
I was hoping that something like this
mtcars %>%
mutate_at(vars(mpg:disp), funs(. / hp))
would work but does nothing.
Using dplyr::across you could achieve your desired result like so:
library(dplyr, w = FALSE)
mtcars2 <- mtcars %>%
mutate(across(mpg:disp, list(Horsepower = ~. / hp)))
head(mtcars2)
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#> Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#> Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#> Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#> Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
#> Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
#> mpg_Horsepower cyl_Horsepower disp_Horsepower
#> Mazda RX4 0.1909091 0.05454545 1.454545
#> Mazda RX4 Wag 0.1909091 0.05454545 1.454545
#> Datsun 710 0.2451613 0.04301075 1.161290
#> Hornet 4 Drive 0.1945455 0.05454545 2.345455
#> Hornet Sportabout 0.1068571 0.04571429 2.057143
#> Valiant 0.1723810 0.05714286 2.142857
Related
I recently dropped data from my meansleep data set using [-c(3,12), ]
I now am realizing I want that those columns to still be in my data set. How do I re add this data?
The variables were Id:
User, TotalMinutesAsleep, sleep_hours
3, 412, 8.44
12, 429, 7.15
If you delete your data, and overwrite the name of your old data like this...
head(mtcars)
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#> Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#> Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#> Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#> Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
#> Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
mtcars <- mtcars[-c(1:10)]
head(mtcars)
#> carb
#> Mazda RX4 4
#> Mazda RX4 Wag 4
#> Datsun 710 1
#> Hornet 4 Drive 1
#> Hornet Sportabout 2
#> Valiant 1
Created on 2022-04-02 by the reprex package (v2.0.1)
... then it's gone. You need to import your original data again and start your analysis over.
This is why it's a good idea to:
Use R scripts for your analysis code, so it's easy to rerun if you need to start over
Save your data with different names if you're not aboslutely sure you want to delete it:
head(mtcars)
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#> Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#> Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#> Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#> Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
#> Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
new_data <- mtcars[-c(1:10)]
# New dataframe:
head(new_data)
#> carb
#> Mazda RX4 4
#> Mazda RX4 Wag 4
#> Datsun 710 1
#> Hornet 4 Drive 1
#> Hornet Sportabout 2
#> Valiant 1
# Original data still here
head(mtcars)
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#> Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#> Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#> Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#> Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
#> Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
Created on 2022-04-02 by the reprex package (v2.0.1)
I am trying to write a function using rlang so that I can subset data based on supplied expression. Although the actual function is complicated, here is a minimal version of it that illustrates the problem.
minimal version of needed function
library(rlang)
# define a function
foo <- function(data, expr = NULL) {
if (!quo_is_null(enquo(expr))) {
dplyr::filter(data, !!enexpr(expr))
} else {
data
}
}
# does the function work? yes
head(foo(mtcars, NULL)) # with NULL
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#> Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#> Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#> Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#> Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
#> Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
head(foo(mtcars, mpg > 20)) # with expression
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
#> Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
#> Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
#> Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
#> Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
#> Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
problems with purrr::pmap
When used with purrr::pmap(), it works as expected when expr is NULL, but not otherwise. Instead of list, I also tried using alist to supply the input.
library(purrr)
# works when expression is `NULL`
pmap(
.l = list(data = list(head(mtcars)), expr = list(NULL)),
.f = foo
)
#> [[1]]
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#> Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#> Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#> Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#> Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
#> Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
# but not otherwise
pmap(
.l = list(data = list(head(mtcars)), expr = list("mpg > 20")),
.f = foo
)
#> Error: Problem with `filter()` input `..1`.
#> ℹ Input `..1` is `"mpg > 20"`.
#> x Input `..1` must be a logical vector, not a character.
Created on 2021-07-20 by the reprex package (v2.0.0)
One way to make this work is by wrapping with quote
purrr::pmap(
.l = list(data = list(head(mtcars)), expr = list(quote(mpg > 20))),
.f = foo
)
-output
[[1]]
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
which also works with the NULL
pmap(
.l = list(data = list(head(mtcars)), expr = list(quote(NULL))),
.f = foo
)
[[1]]
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
Same output with subset
subset(head(mtcars), mpg > 20)
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Or another option is to modify the function by changing the enexpr to parse_expr
foo1 <- function(data, expr = NULL) {
if (!quo_is_null(enquo(expr))) {
dplyr::filter(data, !!parse_expr(expr))
} else {
data
}
}
-testing
> pmap(
+ .l = list(data = list(head(mtcars)), expr = list(NULL)),
+ .f = foo1
+ )
[[1]]
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
>
> pmap(
+ .l = list(data = list(head(mtcars)), expr = list("mpg > 20")),
+ .f = foo1
+ )
[[1]]
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
I want to subset dataframe such that no of rows needed to get mpg value is at least 100.
library(datasets)
data(mtcars)
head(mtcars)
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
The output should be top 5 values
here mpg sum is >100 after Hornet Sportabout
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
I want to the checksum at each row for the mpg column and then the output as no of rows it took to get that sum of at least 100
I would use cumsum in association with lag
library(dplyr)
mtcars %>%
filter(cumsum(lag(mpg, default = 0)) < 100)
This should solve it
library(tidyverse)
df_answer <- mtcars %>%
rownames_to_column() %>%
tibble() %>%
mutate(cum_sum = cumsum(mpg)) %>%
filter(cum_sum < 100)
df_answer %>%
nrow() + 1
A base R option using subset + cumsum
subset(mtcars, c(TRUE, cumsum(mpg) <= 100)[-nrow(mtcars)])
gives
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
You can also use the purrr library:
library(purrr)
which.max(purrr::accumulate(mtcars$mpg, `+`) > 100)
# 5
If you want the whole dataset you can use dplyr::slice:
library(tidyverse)
dplyr::slice(mtcars, 1 : which.max(purrr::accumulate(mtcars$mpg, `+`) > 100))
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
You can use a filter condition with dplyr:
library(tidyverse)
mtcars %>%
filter(row_number() %in% 1:(max(which(cumsum(mpg) < 100)) + 1))
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
The code can be shortened with slice:
mtcars %>%
slice(1:(max(which(cumsum(mpg) < 100)) + 1))
And packaged as a function:
fnc = function(data, var, cutoff) {
data %>%
slice(1:(max(which(cumsum({{var}}) < cutoff)) + 1))
}
fnc(mtcars, mpg, 100)
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#> Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#> Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#> Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#> Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
fnc(iris, Sepal.Width, 10)
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#> 1 5.1 3.5 1.4 0.2 setosa
#> 2 4.9 3.0 1.4 0.2 setosa
#> 3 4.7 3.2 1.3 0.2 setosa
#> 4 4.6 3.1 1.5 0.2 setosa
Can you sort a df based on object class? Say
data("mtcars")
mtcars$cyl <- as.factor(mtcars$cyl)
mtcars$vs <- as.factor(mtcars$vs)
mtcars$am <- as.factor(mtcars$am)
sapply(mtcars,class)
and I want all numeric variables first and then all factors at the end? I want to be able to do this on a much larger dataset so I prefer solutions that do not rely on subsetting by column number. Cheers.
Maybe this one?
head(mtcars)
# mpg cyl disp hp drat wt qsec vs am gear carb
# Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
# Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
# Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
# Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
# Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
# Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
x <- mtcars[,names(sort(unlist(lapply(mtcars, class)), decreasing = T))]
head(x)
# mpg disp hp drat wt qsec gear carb cyl vs am
# Mazda RX4 21.0 160 110 3.90 2.620 16.46 4 4 6 0 1
# Mazda RX4 Wag 21.0 160 110 3.90 2.875 17.02 4 4 6 0 1
# Datsun 710 22.8 108 93 3.85 2.320 18.61 4 1 4 1 1
# Hornet 4 Drive 21.4 258 110 3.08 3.215 19.44 3 1 6 1 0
# Hornet Sportabout 18.7 360 175 3.15 3.440 17.02 3 2 8 0 0
# Valiant 18.1 225 105 2.76 3.460 20.22 3 1 6 1 0
In x, as you see, the columns cyl, vs and am that are of class factor are place at the end and those of class numeric first.
I would like to define a string
string<- "modelName"
That could be used to name an object later. Something like
paste0(string) <- mtcars
cat(string) <- mtcars
print(string) <- mtcars
get(string) <- mtcars
The needed result is the dataset called "modelName". None of the examples above work, obviously.
Question:
How can create one create an object which name is defined by the sourced string?
As #Spacedman notes this is not generally the way things are done but you can use assign
string<- "modelName"
assign(string, mtcars)
> head(modelName)
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
In general it may be perferable to use sometthing like a list:
x <- list()
x[[string]] <- mtcars
> head(x$modelName)
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1