dplyr/rlang: parse_expr with multiple expressions
For example if i want to parse some string to mutate i can
e1 = "vs + am"
mtcars %>% mutate(!!parse_expr(e1))
But when i want to parse any text with special characters like "," it will give me an error,
e2 = "vs + am , am +vs"
mtcars %>% mutate(!!parse_expr(e2))
Error in parse(text = x) : <text>:1:9: unexpected ','
1: vs + am ,
^
Are there any ways to work around this?
Thanks
We can use the triple-bang operator with the plural form parse_exprs and a modified e2 expression to parse multiple expressions (see ?parse_quosures):
Explanation:
Multiple expressions in e2 need to be separated either by ; or by new lines.
From ?quasiquotation: The !!! operator unquotes and splices its argument. The argument should represents a list or a vector.
e2 = "vs + am ; am +vs";
mtcars %>% mutate(!!!parse_exprs(e2))
# mpg cyl disp hp drat wt qsec vs am gear carb vs + am am + vs
#1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 1 1
#2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 1 1
#3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 2 2
#4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 1 1
#5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 0 0
#6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 1 1
#7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 0 0
#8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 1 1
#9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 1 1
#10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 1 1
#11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 1 1
#12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 0 0
#13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 0 0
#14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 0 0
#15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 0 0
#16 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 0 0
#17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 0 0
#18 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 2 2
#19 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 2 2
#20 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 2 2
#21 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 1 1
#22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 0 0
#23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 0 0
#24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 0 0
#25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 0 0
#26 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 2 2
#27 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 1 1
#28 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 2 2
#29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 1 1
#30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 1 1
#31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 1 1
#32 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 2 2
You could always split them outside the expressions for example:
e2 = "vs + am"
e3 = "am +vs"
mtcars %>% mutate(!!parse_expr(e2),!!parse_expr(e3))
You can do this with parse_exprs and a semicolon instead of a comma thanks #Maurits Evers.
!!! takes a list of elements and splices them into to the current call.
e2 = "vs + am ; am +vs"
mtcars %>% mutate(!!!parse_exprs(e2))
Here a little trick I use to name variables (as Genom asked)
Exemple with 2 named expressions :
across_funs <- function(x, .fns, .cols) {
stopifnot(length(.fns) == length(.cols))
stopifnot(all(sapply(.fns, class) == "call"))
for (i in 1:length(.fns)) {
x <- x %>% mutate(!!.cols[i] := !!.fns[[i]])
}
return(x)
}
funs = parse_exprs(c("vs+am", "am+vs"))
cols = c("var1", "var2")
mtcars %>% across_funs(.fns = funs, .cols = cols)
Related
I would like to add new columns to an existing data frame. The column names are generated in a FOR loop so that they are numerically sequential. Here is the code:
NewColumn <- paste("return_date", as.character(i), sep = "_")
When I display NewColumn, this is what I want:
[1] "return_date_2"
When I execute:
mutate(Cima, NewColumn = "01-01-01")
The name of the column is: NewColumn
I can rename it, but is there a way to avoid this step?
Why does R not recognize that NewColumn holds a string?
Do you have to use mutate in your code?
If not, replace mutate(Cima, NewColumn = "01-01-01") with Cima[NewColumn] <- "01-01-01"
Because mutate consider the left part of the equal sign to be already the column name. U can get over it with the code below:
library(dplyr)
library(rlang)
i <- 1
NewColumn <- paste("return_date", as.character(i), sep = "_")
> mutate(mtcars, !!NewColumn := 5)
mpg cyl disp hp drat wt qsec vs am gear carb return_date_1
1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 5
2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 5
3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 5
4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 5
5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 5
6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 5
7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 5
8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 5
9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 5
10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 5
11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 5
12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 5
13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 5
14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 5
15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 5
16 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 5
17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 5
18 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 5
19 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 5
20 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 5
21 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 5
22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 5
23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 5
24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 5
25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 5
26 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 5
27 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 5
28 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 5
29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 5
30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 5
31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 5
32 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 5
Take a look into this one to understand it better:
Use dynamic variable names in `dplyr`
You can also check advanced R from Hadley Wickham and take a look at the bang bang operator and see what it does.
https://adv-r.hadley.nz/
This question already has answers here:
Use dynamic name for new column/variable in `dplyr`
(10 answers)
Closed 2 years ago.
I've used curly-curly with group_by and summarise as described in the rlang announcement. But I can't get it to work when mutating a variable in place. What's the best way to do this currently with dplyr?
Say I want to supply an unquoted column name and have it mutated, here's a toy example function that doesn't work:
my_fun <- function(dat, var_name){
dat %>%
mutate({{var_name}} = 1)
}
my_fun(mtcars, cyl)
What should that mutate line be to change any column in mtcars to be a constant?
You need to use the assignment operator (:=) if you want to use the curly-curly to specify a name on the left hand side of an assignment in mutate:
my_fun <- function(dat, var_name){
dat %>%
mutate({{var_name}} := 1)
}
Which allows:
my_fun(mtcars, cyl)
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> 1 21.0 1 160.0 110 3.90 2.620 16.46 0 1 4 4
#> 2 21.0 1 160.0 110 3.90 2.875 17.02 0 1 4 4
#> 3 22.8 1 108.0 93 3.85 2.320 18.61 1 1 4 1
#> 4 21.4 1 258.0 110 3.08 3.215 19.44 1 0 3 1
#> 5 18.7 1 360.0 175 3.15 3.440 17.02 0 0 3 2
#> 6 18.1 1 225.0 105 2.76 3.460 20.22 1 0 3 1
#> 7 14.3 1 360.0 245 3.21 3.570 15.84 0 0 3 4
#> 8 24.4 1 146.7 62 3.69 3.190 20.00 1 0 4 2
#> 9 22.8 1 140.8 95 3.92 3.150 22.90 1 0 4 2
#> 10 19.2 1 167.6 123 3.92 3.440 18.30 1 0 4 4
#> 11 17.8 1 167.6 123 3.92 3.440 18.90 1 0 4 4
#> 12 16.4 1 275.8 180 3.07 4.070 17.40 0 0 3 3
#> 13 17.3 1 275.8 180 3.07 3.730 17.60 0 0 3 3
#> 14 15.2 1 275.8 180 3.07 3.780 18.00 0 0 3 3
#> 15 10.4 1 472.0 205 2.93 5.250 17.98 0 0 3 4
#> 16 10.4 1 460.0 215 3.00 5.424 17.82 0 0 3 4
#> 17 14.7 1 440.0 230 3.23 5.345 17.42 0 0 3 4
#> 18 32.4 1 78.7 66 4.08 2.200 19.47 1 1 4 1
#> 19 30.4 1 75.7 52 4.93 1.615 18.52 1 1 4 2
#> 20 33.9 1 71.1 65 4.22 1.835 19.90 1 1 4 1
#> 21 21.5 1 120.1 97 3.70 2.465 20.01 1 0 3 1
#> 22 15.5 1 318.0 150 2.76 3.520 16.87 0 0 3 2
#> 23 15.2 1 304.0 150 3.15 3.435 17.30 0 0 3 2
#> 24 13.3 1 350.0 245 3.73 3.840 15.41 0 0 3 4
#> 25 19.2 1 400.0 175 3.08 3.845 17.05 0 0 3 2
#> 26 27.3 1 79.0 66 4.08 1.935 18.90 1 1 4 1
#> 27 26.0 1 120.3 91 4.43 2.140 16.70 0 1 5 2
#> 28 30.4 1 95.1 113 3.77 1.513 16.90 1 1 5 2
#> 29 15.8 1 351.0 264 4.22 3.170 14.50 0 1 5 4
#> 30 19.7 1 145.0 175 3.62 2.770 15.50 0 1 5 6
#> 31 15.0 1 301.0 335 3.54 3.570 14.60 0 1 5 8
#> 32 21.4 1 121.0 109 4.11 2.780 18.60 1 1 4 2
I can not get my head around this. I have a dataset which contains a data.frame in per day for 3 years, so i have a list with 1000 dataframes.
I want to filter all dataframes like in the example below. I know I could easily filter (or use rbindlist), first and then do the split, but I desire a way to apply a filter function to multiple dataframes. Can you help me? The code below does not work, but hope it helps to make clear what I want to archieve.
dflist <- mtcars %>%
split(.$cyl)
lapply(dflist, function(x) dplyr::filter(x[["mpg"]] > 10))
The filter works on a data.frame/tbl_df. Instead, we are extracting a vector (x[["mpg"]])
library(tidyverse)
filter(mtcars$mpg > 10)
Error in UseMethod("filter_") : no applicable method for 'filter_'
applied to an object of class "logical"
and apply filter on it.
We need to apply filter on the data.frame itself
map(dflist, ~ .x %>%
filter(mpg > 10))
#$`4`
# mpg cyl disp hp drat wt qsec vs am gear carb
#1 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
#2 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
#3 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
#4 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
#5 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2
#6 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1
#7 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
#8 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1
#9 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
#10 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
#11 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2
#$`6`
# mpg cyl disp hp drat wt qsec vs am gear carb
#1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
#2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
#3 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
#4 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
#5 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4
#6 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4
#7 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
#$`8`
# mpg cyl disp hp drat wt qsec vs am gear carb
#1 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
#2 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
#3 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
#4 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3
#5 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
#6 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
#7 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4
#8 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4
#9 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
#10 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2
#11 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4
#12 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
#13 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
#14 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
Or using lapply
lapply(dflist, function(x) x %>%
filter(mpg > 10))
I feel there's smarter/more efficient way than this code:
df <- mtcars
df$somename <- as.array(rep(c(0), 32))
for (i in 1:32){
df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
}
maybe with %>%? but how?
An option using purrr::map2
library(tidyverse)
mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))
# mpg cyl disp hp drat wt qsec vs am gear carb somename
#1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 9.786358
#2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 10.00203
#3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 11.51877
#4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 11.47281
#5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 9.60251
#6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 11.85111
#7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 8.6762
#8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 11.88646
#9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 13.96536
#10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 10.50761
#11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 10.93187
#12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 9.425733
#13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 9.807571
#14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 10.05506
#15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 9.001469
#16 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 8.765296
#17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 8.538314
#18 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 12.21173
#19 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 11.95364
#20 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 12.77388
#21 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 12.40619
#22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 9.439876
#23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 9.804036
#24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 8.181225
#25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 9.337345
#26 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 11.99607
#27 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 10.29547
#28 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 10.88025
#29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 8.01152
#30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 9.001469
#31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 7.799388
#32 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 11.18643
Update
I re-ran #42-'s microbenchmark analysis using a larger dataset
library(microbenchmark)
df <- do.call(rbind, lapply(1:100, function(x) mtcars))
res <- microbenchmark(
orig = {
df$somename <- as.array(rep(c(0), nrow(df)))
for (i in 1:nrow(df)) {
df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))}},
tidy = {
df <- df %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))},
mapply = {
df$somename <- mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)},
rowMeans = {
df$rm <- rowMeans(df[,c("wt","qsec")])
df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 )})
res
#Unit: microseconds
# expr min lq mean median uq max
# orig 331092.86 349754.808 360716.6501 357229.3920 366635.2820 446581.924
# tidy 168701.28 181079.910 189710.1927 187026.6290 194392.5190 273725.354
# mapply 161711.77 172457.395 179326.5484 177263.3045 183688.5365 266102.901
# rowMeans 228.08 315.854 343.9151 334.8975 358.5915 807.847
library(ggplot2)
autoplot(res)
More of a comment than an answer:
> library(microbenchmark)
> microbenchmark( orig = {df <- mtcars
+
+ df$somename <- as.array(rep(c(0), 32))
+
+ for (i in 1:32){
+ df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
+ }}, tidy = {
+ mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))}, mapply = { mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)})
#------------------------------------
Unit: microseconds
expr min lq mean median uq max neval cld
orig 5069.391 5161.9270 5555.5886 5236.769 5490.7365 12400.502 100 b
tidy 910.071 943.9685 986.4419 970.541 998.8075 1241.711 100 a
mapply 744.639 761.1875 805.6328 773.426 807.2545 2206.393 100 a
Code:
df$somename <- apply(matrix(c(df$wt, df$qsec), ncol=2), MARGIN = 1, FUN=sd)
Output:
> head(df$somename)
somename
1 9.786358
2 10.002025
3 11.518769
4 11.472808
5 9.602510
6 11.851110
7 8.676200
8 11.886465
9 13.965359
10 10.507607
The code is from r4ds's exercise
trans <- list(
disp = function(x) x * 0.0163871,
am = function(x) {
factor(x, labels = c("auto", "manual"))
}
)
for (var in names(trans)) {
mtcars[[var]] <- trans[[var]](mtcars[[var]])
}
I studied the next section here, and have a question that
How can I remake this code using purrr and dplyr?
Of course, I can do like this
mtcars %>%
mutate(
disp = disp * 0.0163871,
am = factor(am, labels = c("auto", "manual"))
)
But I want to make the best use of FP.
It is very hard to me because of combining variable assignment and purrr
Here is a purrr/dplyr option using imap_dfc
library(tidyverse)
imap_dfc(trans, ~mtcars %>% transmute_at(vars(.y), funs(.x))) %>%
bind_cols(mtcars %>% select(-one_of(names(trans)))) %>%
select(names(mtcars))
# mpg cyl disp hp drat wt qsec vs am gear carb
#1 21.0 6 2.621936 110 3.90 2.620 16.46 0 manual 4 4
#2 21.0 6 2.621936 110 3.90 2.875 17.02 0 manual 4 4
#3 22.8 4 1.769807 93 3.85 2.320 18.61 1 manual 4 1
#4 21.4 6 4.227872 110 3.08 3.215 19.44 1 auto 3 1
#5 18.7 8 5.899356 175 3.15 3.440 17.02 0 auto 3 2
#6 18.1 6 3.687098 105 2.76 3.460 20.22 1 auto 3 1
#7 14.3 8 5.899356 245 3.21 3.570 15.84 0 auto 3 4
#8 24.4 4 2.403988 62 3.69 3.190 20.00 1 auto 4 2
#9 22.8 4 2.307304 95 3.92 3.150 22.90 1 auto 4 2
#10 19.2 6 2.746478 123 3.92 3.440 18.30 1 auto 4 4
#11 17.8 6 2.746478 123 3.92 3.440 18.90 1 auto 4 4
#12 16.4 8 4.519562 180 3.07 4.070 17.40 0 auto 3 3
#13 17.3 8 4.519562 180 3.07 3.730 17.60 0 auto 3 3
#14 15.2 8 4.519562 180 3.07 3.780 18.00 0 auto 3 3
#15 10.4 8 7.734711 205 2.93 5.250 17.98 0 auto 3 4
#16 10.4 8 7.538066 215 3.00 5.424 17.82 0 auto 3 4
#17 14.7 8 7.210324 230 3.23 5.345 17.42 0 auto 3 4
#18 32.4 4 1.289665 66 4.08 2.200 19.47 1 manual 4 1
#19 30.4 4 1.240503 52 4.93 1.615 18.52 1 manual 4 2
#20 33.9 4 1.165123 65 4.22 1.835 19.90 1 manual 4 1
#21 21.5 4 1.968091 97 3.70 2.465 20.01 1 auto 3 1
#22 15.5 8 5.211098 150 2.76 3.520 16.87 0 auto 3 2
#23 15.2 8 4.981678 150 3.15 3.435 17.30 0 auto 3 2
#24 13.3 8 5.735485 245 3.73 3.840 15.41 0 auto 3 4
#25 19.2 8 6.554840 175 3.08 3.845 17.05 0 auto 3 2
#26 27.3 4 1.294581 66 4.08 1.935 18.90 1 manual 4 1
#27 26.0 4 1.971368 91 4.43 2.140 16.70 0 manual 5 2
#28 30.4 4 1.558413 113 3.77 1.513 16.90 1 manual 5 2
#29 15.8 8 5.751872 264 4.22 3.170 14.50 0 manual 5 4
#30 19.7 6 2.376130 175 3.62 2.770 15.50 0 manual 5 6
#31 15.0 8 4.932517 335 3.54 3.570 14.60 0 manual 5 8
#32 21.4 4 1.982839 109 4.11 2.780 18.60 1 manual 4 2
Explanation: imap_dfc(...) column-binds the two modified columns, which in turn are then column-bound to mtcars without the two columns that were modified; the last line re-arranges columns such that they correspond to the original mtcars column ordering.
A possible suggestion, but it is just a different color of the same paint!
result <- mtcars
walk(1:length(trans),
function(i) result <<- result %>% mutate_at(names(trans)[[i]],trans[[i]]))
result
A best one should be
result <- mtcars
pmap(list(names(trans),trans),
function(n,f) result <<- result %>% mutate_at(n,f))
result
And a shorter one :
result <- mtcars
iwalk(trans,
function(f,n) result <<- result %>% mutate_at(n,f))
result