I have several similar data frames with many columns in common. I would like to select and rename a subset of those columns from any table.
library(tidyverse)
mtcars %>%
select(my_mpg = mpg,
cylinders = cyl,
gear)
Is it possible to do something like
my_select_rename <- c("my_mpg"="mpg","cylinders"="cyl","gear")
mtcars %>%
select_(.dots = my_select_rename)
but using the tidyeval framework instead?
I think you want:
my_select <- c("mpg","cyl","gear")
my_select_rename <- c("my_mpg","cylinders","gear")
mtcars %>%
select_at(vars(my_select)) %>%
setNames(., my_select_rename)
my_mpg cylinders gear
Mazda RX4 21.0 6 4
Mazda RX4 Wag 21.0 6 4
Datsun 710 22.8 4 4
Hornet 4 Drive 21.4 6 3
Hornet Sportabout 18.7 8 3
lionel's answer to this question group_by by a vector of characters using tidy evaluation semantics provides the answer
mtcars %>%
select(!!! rlang::syms(my_select_rename))
Related
I need to rename multiple variables using a replacement dataframe. This replacement dataframe also includes regex. I would like to use a similar solution proposed here, .e.g
df %>% rename_with(~ newnames, all_of(oldnames))
MWE:
df <- mtcars[, 1:5]
# works without regex
replace_df_1 <- tibble::tibble(
old = df %>% colnames(),
new = df %>% colnames() %>% toupper()
)
df %>% rename_with(~ replace_df_1$new, all_of(replace_df_1$old))
# with regex
replace_df_2 <- tibble::tibble(
old = c("^m", "cyl101|cyl", "disp", "hp", "drat"),
new = df %>% colnames() %>% toupper()
)
old new
<chr> <chr>
1 ^m MPG
2 cyl101|cyl CYL
3 disp DISP
4 hp HP
5 drat DRAT
# does not work
df %>% rename_with(~ replace_df_2$new, all_of(replace_df_2$old))
df %>% rename_with(~ matches(replace_df_2$new), all_of(replace_df_2$old))
EDIT 1:
The solution of #Mael works in general, but there seems to be index issue, e.g. consider the following example
replace_df_2 <- tibble::tibble(
old = c("xxxx", "cyl101|cyl", "yyy", "xxx", "yyy"),
new = mtcars[,1:5] %>% colnames() %>% toupper()
)
mtcars[, 1:5] %>%
rename_with(~ replace_df_2$new, matches(replace_df_2$old))
Results in
mpg MPG disp hp drat
<dbl> <dbl> <dbl> <dbl> <dbl>
1 21 6 160 110 3.9
meaning that the rename_with function correctly finds the column, but replaces it with the first item in the replacement column. How can we tell the function to take the respective row where a replacement has been found?
So in this example (edit 1), I only want to substitute the second column with "CYL", the rest should be left untouched. The problem is that the function takes the first replacement (MPG) instead of the second (CYL).
Thank you for any hints!
matches should be on the regex-y column:
df %>%
rename_with(~ replace_df_2$new, matches(replace_df_2$old))
MPG CYL DISP HP DRAT
Mazda RX4 21.0 6 160.0 110 3.90
Mazda RX4 Wag 21.0 6 160.0 110 3.90
Datsun 710 22.8 4 108.0 93 3.85
Hornet 4 Drive 21.4 6 258.0 110 3.08
Hornet Sportabout 18.7 8 360.0 175 3.15
Valiant 18.1 6 225.0 105 2.76
#...
If the task is simply to set all col names to upper-case, then this works:
sub("^(.+)$", "\\U\\1", colnames(df), perl = TRUE)
[1] "MPG" "CYL" "DISP" "HP" "DRAT"
In dplyr:
df %>%
rename_with( ~sub("^(.+)$", "\\U\\1", colnames(df), perl = TRUE))
I found a solution using the idea of non standard evaluation from this question and #Maël's answer.
Using map_lgl we create a logical vector that returns TRUE if the column in replace_df_2$old can be found inside the dataframe df. Then we pass this logical vector to replace_df_2$new to get the correct replacement.
df <- mtcars[, 1:5]
df %>%
rename_with(.fn = ~replace_df_2$new[map_lgl(replace_df_2$old,~ any(str_detect(., names(df))))],
.cols = matches(replace_df_2$old))
Result:
mpg CYL disp hp drat
Mazda RX4 21.0 6 160.0 110 3.90
I want to rename a variable in my dataframe using dplyr to have spaces but this variable name is a concatenation of a dynamic variable and a static string. In the following example, I'd need "Test1" to be a dynamic variable
df <- mtcars %>% select(`Test1 mpg` = "mpg")
So when I try this, I end up with an error:
var <- "Test1"
df <- mtcars %>% select(paste0(var, " mpg") = "mpg")
How could I go about making those new variable names dynamic?
Using the special assignment operator := you could do:
library(dplyr)
df <- mtcars %>% select(`Test1 mpg` = "mpg")
var <- "Test1"
mtcars %>%
select("{var} mpg" := "mpg")
#> Test1 mpg
#> Mazda RX4 21.0
#> Mazda RX4 Wag 21.0
#> Datsun 710 22.8
#> Hornet 4 Drive 21.4
or using !!sym():
mtcars %>%
select(!!sym(paste(var, " mpg")) := "mpg")
#> Test1 mpg
#> Mazda RX4 21.0
#> Mazda RX4 Wag 21.0
#> Datsun 710 22.8
#> Hornet 4 Drive 21.4
I want to have the year column selected too in the output but I dont understand how to add another column in the select statement. The column I want to select is a simple column named "Year".
Following is the code I am trying to use but it gives me an error. The select_if condition is correct because if I do not add the "select("Year")" line the code works fine.
Please help
Subset <-df %>%
select("Year") & select_if(grepl("Apple", names(.)))
library(dplyr)
mtcars %>%
select(cyl, contains("gea")) %>%
head()
cyl gear
Mazda RX4 6 4
Mazda RX4 Wag 6 4
Datsun 710 4 4
Hornet 4 Drive 6 3
Hornet Sportabout 8 3
Valiant 6 3
For each group, I'm trying to get the top_n car_names to appear in a new column comma separated.
For example, when you run the code below you'll see the top 2 mpg cars per group (cyl). Next, I want to extract the top two cars (or more, if there is a tie) and store them together into a new column called car_summary.
mtcars2 %>%
select(mpg, cyl, car_name) %>%
group_by(cyl) %>%
mutate(Score = rank(mpg, ties.method = "max")) %>%
arrange(desc(Score)) %>% top_n(2,Score)
The expected output looks like below
cyl <- c(8,4,6)
car_summary <- c("Pontiac Firebird, Hornet Sportabout", "Toyota Corolla,
Fiat 128", "Hornet 4 Drive, Mazda RX4, Mazda RX4 Wag")
data.frame(cyl, car_summary)
cyl car_summary
1 8 Pontiac Firebird, Hornet Sportabout
2 4 Toyota Corolla, Fiat 128
3 6 Hornet 4 Drive, Mazda RX4, Mazda RX4 Wag
You need toString from base R -
mtcars2 %>%
select(mpg, cyl, car_name) %>%
group_by(cyl) %>%
mutate(Score = rank(mpg, ties.method = "max")) %>%
arrange(desc(Score)) %>%
top_n(2,Score) %>%
summarize(car_summary = toString(car_name))
This question already has answers here:
Grouping functions (tapply, by, aggregate) and the *apply family
(10 answers)
Closed 6 years ago.
I have trouble applying a split to a data.frame and then assembling some aggregated results back into a different data.frame. I tried using the 'unsplit' function but I can't figure out how to use it properly to get the desired result. Let me demonstrate on the common 'mtcars' data: Let's say that my ultimate result is to get a data frame with two variables: cyl (cylinders) and mean_mpg (mean over mpg for group of cars sharing the same count of cylinders).
So the initial split goes like this:
spl <- split(mtcars, mtcars$cyl)
The result of which looks something like this:
$`4`
mpg cyl disp hp drat wt qsec vs am gear carb
Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
...
$`6`
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
...
$`8`
mpg cyl disp hp drat wt qsec vs am gear carb
Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
...
Now I want to do something along the lines of:
df <- as.data.frame(lapply(spl, function(x) mean(x$mpg)), col.names=c("cyl", "mean_mpg"))
However, doing the above results in:
X4 X6 X8
1 26.66364 19.74286 15.1
While I'd want the df to be like this:
cyl mean_mpg
1 4 26.66364
2 6 19.74286
3 8 15.10000
Thanks, J.
If you are only interested in reassembling a split then look at (2), (4) and (4a) but if the actual underlying question is really about the way to perform aggregations over groups then they all may be of interest:
1) aggregate Normally one uses aggregate as already mentioned in the comments. Simplifying #alistaire's code slightly:
aggregate(mpg ~ cyl, mtcars, mean)
2) split/lapply/do.call Also #rawr has given a split/lapply/do.call solution in the comments which we can also simplify slightly:
spl <- split(mtcars, mtcars$cyl)
do.call("rbind", lapply(spl, with, data.frame(cyl = cyl[1], mpg = mean(mpg))))
3) do.call/by The last one could alternately be rewritten in terms of by:
do.call("rbind", by(mtcars, mtcars$cyl, with, data.frame(cyl = cyl[1], mpg = mean(mpg))))
4) split/lapply/unsplit Another possibility is to use split and unsplit:
spl <- split(mtcars, mtcars$cyl)
L <- lapply(spl, with, data.frame(cyl = cyl[1], mpg = mean(mpg), row.names = cyl[1]))
unsplit(L, sapply(L, "[[", "cyl"))
4a) or if row names are sufficient:
spl <- split(mtcars, mtcars$cyl)
L <- lapply(spl, with, data.frame(mpg = mean(mpg), row.names = cyl[1]))
unsplit(L, sapply(L, rownames))
The above do not use any packages but there are also many packages that can do aggregations including dplyr, data.table and sqldf:
5) dplyr
library(dplyr)
mtcars %>%
group_by(cyl) %>%
summarize(mpg = mean(mpg)) %>%
ungroup()
6) data.table
library(data.table)
as.data.table(mtcars)[, list(mpg = mean(mpg)), by = "cyl"]
7) sqldf
library(sqldf)
sqldf("select cyl, avg(mpg) mpg from mtcars group by cyl")