empty string as column name in tibble - r

tibble::tibble(`` = 1:10)
Error: attempt to use zero-length variable name
tibble::tibble("" = 1:10)
Error: attempt to use zero-length variable name
How can I get around this? I need to have a column with precisely "" as the name.

My first thought is that this sounds like a report-representation thing, since one generally doesn't need nameless columns while developing or working with data. In that regard, I suggest you look at changing names in whatever reporting system you might be using (knitr, kableExtra, etc).
Having said that, R is not going to let you define a zero-length column name, but it'll let you update it later:
setNames(data.frame(" "=1),"")
#
# 1 1
setNames(tibble(" "=1),"")
# # A tibble: 1 x 1
# ``
# <dbl>
# 1 1

This can be achieved by directly modifying the names attribute of a tibble, though it's not a recommended practice. Do something like this:
attr(df, "names") <- c("", "cyl", "disp", "hp", "drat", "wt", "qsec", "vs", "am", "gear", "carb")
Tested with this dataset
df <- tibble::as_tibble(mtcars)
# A tibble: 32 x 11
mpg cyl disp hp drat wt qsec vs am gear carb
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1
4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2
9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2
10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
# ... with 22 more rows
Output
# A tibble: 32 x 11
`` cyl disp hp drat wt qsec vs am gear carb
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1
4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2
9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2
10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
# ... with 22 more rows

Related

Mutate a dynamic column name with conditions using other dynamic column names

I'm trying to use dplyr::mutate to change a dynamic column with conditions using other columns dynamically.
I've got this bit of code:
d <- mtcars %>% tibble
fld_name <- "mpg"
other_fld_name <- "cyl"
d <- d %>% mutate(!!fld_name := ifelse(!!other_fld_name < 5,NA,!!fld_name))
which sets mpg to
mpg
<chr>
1 mpg
2 mpg
3 mpg
4 mpg
5 mpg
6 mpg
7 mpg
8 mpg
9 mpg
10 mpg
it seems to select the field on the LHS of assignment operator, but just pastes the field name on the RHS.
Removing the unquotes on the RHS yields the same result.
Any help is much appreciated.
use get to retreive column value instead
library(tidyverse)
d <- mtcars %>% tibble
fld_name <- "mpg"
other_fld_name <- "cyl"
d %>% mutate(!!fld_name := ifelse(get(other_fld_name) < 5 ,NA, get(fld_name)))
#> # A tibble: 32 x 11
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
#> 2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
#> 3 NA 4 108 93 3.85 2.32 18.6 1 1 4 1
#> 4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
#> 5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
#> 6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
#> 7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
#> 8 NA 4 147. 62 3.69 3.19 20 1 0 4 2
#> 9 NA 4 141. 95 3.92 3.15 22.9 1 0 4 2
#> 10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
#> # ... with 22 more rows
Created on 2021-06-22 by the reprex package (v2.0.0)
We can also use ensym function to quote variable name stored as string and unquote it with !! like the following:
library(rlang)
d <- mtcars %>% tibble
fld_name <- "mpg"
other_fld_name <- "cyl"
d %>%
mutate(!!ensym(fld_name) := ifelse(!!ensym(other_fld_name) < 5, NA, !!ensym(fld_name)))
# A tibble: 32 x 11
mpg cyl disp hp drat wt qsec vs am gear carb
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
3 NA 4 108 93 3.85 2.32 18.6 1 1 4 1
4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
8 NA 4 147. 62 3.69 3.19 20 1 0 4 2
9 NA 4 141. 95 3.92 3.15 22.9 1 0 4 2
10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
# ... with 22 more rows
We could also use .data
library(dplyr)
d %>%
mutate(!! fld_name := case_when(.data[[other_fld_name]] >=5 ~
.data[[fld_name]]))
-output
# A tibble: 32 x 11
mpg cyl disp hp drat wt qsec vs am gear carb
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
3 NA 4 108 93 3.85 2.32 18.6 1 1 4 1
4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
8 NA 4 147. 62 3.69 3.19 20 1 0 4 2
9 NA 4 141. 95 3.92 3.15 22.9 1 0 4 2
10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
# … with 22 more rows
data
d <- mtcars %>%
as_tibble
fld_name <- "mpg"
other_fld_name <- "cyl"

Why is `select_if(., is_double)` returning dates? [duplicate]

This question already has answers here:
Why is Date is being returned as type 'double'?
(2 answers)
Closed 2 years ago.
When using is_double with select_if, the return value includes columns of lubridate's date data type. Why is this?
Here is a simple example using the today() function.
library(tidyverse)
library(lubridate)
mtcars %>%
as_tibble() %>% # Convert to tibble
mutate(today = today()) %>% # Create a date column
select_if(is_double) # Select double columns
Output:
# A tibble: 32 x 12
mpg cyl disp hp drat wt qsec vs am gear carb today
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <date>
1 21 6 160 110 3.9 2.62 16.5 0 1 4 4 2020-06-25
2 21 6 160 110 3.9 2.88 17.0 0 1 4 4 2020-06-25
3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1 2020-06-25
4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1 2020-06-25
5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2 2020-06-25
6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1 2020-06-25
7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4 2020-06-25
8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2 2020-06-25
9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2 2020-06-25
10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4 2020-06-25
# ... with 22 more rows
Hopefully I'm missing something simple, are dates recognized as type double?
Because, date is internally stored as double
typeof(today())
#[1] "double"
though its class is 'Date'
class(today())
#[1] "Date"
An option is to add another condition in select_if
library(dplyr)
mtcars %>%
as_tibble %>%
mutate(today = today()) %>%
select_if(~ is_double(.) && !inherits(., "Date"))
# A tibble: 32 x 10
# mpg disp hp drat wt qsec vs am gear carb
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 21 160 110 3.9 2.62 16.5 0 1 4 4
# 2 21 160 110 3.9 2.88 17.0 0 1 4 4
# 3 22.8 108 93 3.85 2.32 18.6 1 1 4 1
# 4 21.4 258 110 3.08 3.22 19.4 1 0 3 1
# 5 18.7 360 175 3.15 3.44 17.0 0 0 3 2
# 6 18.1 225 105 2.76 3.46 20.2 1 0 3 1
# 7 14.3 360 245 3.21 3.57 15.8 0 0 3 4
# 8 24.4 147. 62 3.69 3.19 20 1 0 4 2
# 9 22.8 141. 95 3.92 3.15 22.9 1 0 4 2
#10 19.2 168. 123 3.92 3.44 18.3 1 0 4 4
# … with 22 more rows
In the dplyr 1.0.0, we can also use where with select
mtcars %>%
as_tibble %>%
mutate(today = today()) %>%
select(where(~is_double(.) && !inherits(., "Date")))

group_by variable and sum in dplyr [duplicate]

This question already has answers here:
Why does summarize or mutate not work with group_by when I load `plyr` after `dplyr`?
(2 answers)
Closed 2 years ago.
I know this question has answers in multiple places, but I am unable to figure out where I am going wrong. Suppose I want to find the sum of hp for each group in cyl:
mtcars%>%
group_by(cyl) %>%
mutate(
sum_hp = sum(hp)
)
sum_hp is giving me 4694 for every value. I want the sum for each value of cyl.
It could be a case of plyr::mutate masking dplyr::mutate when both the packages are loaded. We can specify dplyr::<functionname> to correct this
library(dplyr)
mtcars%>%
group_by(cyl) %>%
dplyr::mutate(sum_hp = sum(hp))
# A tibble: 32 x 12
# Groups: cyl [3]
# mpg cyl disp hp drat wt qsec vs am gear carb sum_hp
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 21 6 160 110 3.9 2.62 16.5 0 1 4 4 856
# 2 21 6 160 110 3.9 2.88 17.0 0 1 4 4 856
# 3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1 909
# 4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1 856
# 5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2 2929
# 6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1 856
# 7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4 2929
# 8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2 909
# 9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2 909
#10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4 856
# … with 22 more rows
If we use plyr::mutate, the OP's output can be reproduced
mtcars%>%
group_by(cyl) %>%
plyr::mutate(
sum_hp = sum(hp)
)
# A tibble: 32 x 12
# Groups: cyl [3]
# mpg cyl disp hp drat wt qsec vs am gear carb sum_hp
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 21 6 160 110 3.9 2.62 16.5 0 1 4 4 4694
# 2 21 6 160 110 3.9 2.88 17.0 0 1 4 4 4694
# 3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1 4694
# 4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1 4694
# 5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2 4694
# 6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1 4694
# 7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4 4694
# 8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2 4694
# 9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2 4694
#10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4 4694
# … with 22 more rows

using `rlang` NSE to group by multiple variables

I am trying to write a custom function that uses rlang's non-standard evaluation to group a dataframe by more than one variable.
This is what I've-
library(rlang)
# function definition
tryfn <- function(data, groups, ...) {
# preparing data
df <- dplyr::group_by(data, !!!rlang::enquos(groups))
print(head(df))
# applying some function `.f` on df that absorbs `...`
# .f(df, ...)
}
This works with a single grouping variable-
# works
tryfn(mtcars, am)
#> # A tibble: 6 x 11
#> # Groups: am [2]
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
#> 2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
#> 3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1
#> 4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
#> 5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
#> 6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
But if try to use more than one grouping variable, this doesn't work-
# doesn't work
tryfn(mtcars, c(am, cyl))
#> Error: Column `c(am, cyl)` must be length 32 (the number of rows) or one, not 64
# doesn't work
tryfn(mtcars, list(am, cyl))
#> Error: Column `list(am, cyl)` must be length 32 (the number of rows) or one, not 2
We could parse as an expression with enexpr and use !!!
tryfn <- function(data, groups, ...) {
groups <- as.list(rlang::enexpr(groups))
groups <- if(length(groups) > 1) groups[-1] else groups
group_by(data, !!!groups)
}
-testing
tryfn(mtcars, am)
# A tibble: 32 x 11
# Groups: am [2]
# mpg cyl disp hp drat wt qsec vs am gear carb
# * <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
# 2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
# 3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1
# 4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
# 5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
# 6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
# 7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
# 8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2
# 9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2
#10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
# … with 22 more rows
tryfn(mtcars, c(am, cyl))
# A tibble: 32 x 11
# Groups: am, cyl [6]
# mpg cyl disp hp drat wt qsec vs am gear carb
# * <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
# 2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
# 3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1
# 4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
# 5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
# 6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
# 7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
# 8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2
# 9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2
#10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
# … with 22 more rows

How do I selectively change variable data type automatically in the tidyverse?

I would like to change some of the variables from numerical to factor types, leaving other types as they are. I know how to do this one variable at a time, but I would like to automate the process for larger datasets.
I've changed variables in the mtcars dataset one by one, copying and pasting the code. I've used mapply to successfully automate this, but I've only managed to do it on a subset of mtcars. I'm not sure how I would keep the entire dataset intact with the new variable types, though. Reprex below.
#before
as_tibble(mtcars)
#> # A tibble: 32 x 11
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
#> 2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
#> 3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1
#> 4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
#> 5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
#> 6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
#> 7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
#> 8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2
#> 9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2
#> 10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
#> # ... with 22 more rows
#copy + paste job
mtcars$cyl <- factor(as.character(mtcars$cyl))
mtcars$hp <- factor(as.character(mtcars$hp))
mtcars$vs <- factor(as.character(mtcars$vs))
#after
as_tibble(mtcars)
#> # A tibble: 32 x 11
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> <dbl> <fct> <dbl> <fct> <dbl> <dbl> <dbl> <fct> <dbl> <dbl> <dbl>
#> 1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
#> 2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
#> 3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1
#> 4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
#> 5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
#> 6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
#> 7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
#> 8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2
#> 9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2
#> 10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
#> # ... with 22 more rows
Created on 2019-05-17 by the reprex package (v0.2.1)
I managed to change the variable types successfully. I would hate to do this something like 30-50 times though. What are some ways to automate this? Thank you.
library(dplyr)
as_tibble(mtcars) %>%
mutate_at(.vars = vars(cyl, hp, vs),
.funs = ~ factor(as.character(.)))
Hope this helps.
Using base R:
vars_to_make_f <- c("cyl", "hp", "vs")
mtcars[vars_to_make_f] <-
lapply(mtcars[vars_to_make_f], function(x) as.factor(as.character(x)))
mtcars
# A tibble: 32 x 11
mpg cyl disp hp drat wt qsec vs am gear carb
<dbl> <fct> <dbl> <fct> <dbl> <dbl> <dbl> <fct> <dbl> <dbl> <dbl>
1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1
4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2
9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2
10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
# ... with 22 more rows
You can use mutate_at:
mtcars %>%
mutate_at(c("cyl","hp","vs"),function(x) factor(as.character(x)))
Or use purrr modify_at:
mtcars %>%
modify_at(c("cyl","hp","vs"),function(x) factor(as.character(x)))
An option is mutate_at. The as.factor(as.character is not needed, we can directly convert to factor. But, the reverse route would be `factor -> character -> numeric)
library(dplyr)
mtcars %>%
as_tibble %>%
mutate_at(vars(cyl, hp, vs), factor)
# A tibble: 32 x 11
# mpg cyl disp hp drat wt qsec vs am gear carb
# <dbl> <fct> <dbl> <fct> <dbl> <dbl> <dbl> <fct> <dbl> <dbl> <dbl>
# 1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
# 2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
# 3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1
# 4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
# 5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
# 6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
# 7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
# 8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2
# 9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2
#10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
# … with 22 more rows

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