Standard Evaluation in Dplyr: Summarise a Variable Given as a Character String

standard evaluation in dplyr: summarise a variable given as a character string

dplyr 1.0 has changed pretty much everything about this question as well as all of the answers. See the dplyr programming vignette here:

https://cran.r-project.org/web/packages/dplyr/vignettes/programming.html

The new way to refer to columns when their identifier is stored as a character vector is to use the .data pronoun from rlang, and then subset as you would in base R.

library(dplyr)

key <- "v3"
val <- "v2"
drp <- "v1"

df <- tibble(v1 = 1:5, v2 = 6:10, v3 = c(rep("A", 3), rep("B", 2)))

df %>% 
    select(-matches(drp)) %>% 
    group_by(.data[[key]]) %>% 
    summarise(total = sum(.data[[val]], na.rm = TRUE))

#> `summarise()` ungrouping output (override with `.groups` argument)
#> # A tibble: 2 x 2
#>   v3    total
#>   <chr> <int>
#> 1 A        21
#> 2 B        19

If your code is in a package function, you can @importFrom rlang .data to avoid R check notes about undefined globals.

How do I do Standard evaluation with dplyr's arrange?

I don't know why !! does not work with arrange but you can still use get

a %>% arrange(get(meh))


#  date ok
#1    1  1
#2    2  2
#3    3  3

dplyr standard evaluation: summarise_ with variable name for summed variable

After poring over the NSE vignette for awhile and poking at things, I found you can use setNames within summarise_ if you use the .dots argument and put the interp work in a list.

a %>%
    filter_(~x == tag) %>%
    group_by_(tag) %>%
    summarise_(.dots = setNames(list(interp(~sum(var, na.rm = TRUE),
                                            var = as.name(paste0(metric,"_",run1)))), 
                                                            paste0(metric,"_",run1)))

Source: local data frame [1 x 2]

  2011 y_zm
1 2011   30

You could also add a rename_ step to do the same thing. I could see this being less ideal, as it relies on knowing the name you used in summarise_. But if you always use the same name, like variable_name, this does seem like a viable alternative for some situations.

a %>%
    filter_(~x == tag) %>%
    group_by_(tag) %>%
    summarise_(variable_name = interp(~sum(var, na.rm = T), 
                                         var = as.name(paste0(metric,"_",run1)))) %>%
    rename_(.dots = setNames("variable_name", paste0(metric,"_",run1)))

Source: local data frame [1 x 2]

  2011 y_zm
1 2011   30

Use I string to refer to a variable inside dplyr?

As it is a string, convert it to symbol (sym from rlang) and evaluate (!!)

test_df %>%
     summarise(y = mean(!! rlang::sym(string_outcome)))

Or use summarise_at which can take strings in vars parameter

test_df %>%
    summarise_at(vars(string_outcome), list(y = ~  mean(.)))

---

Or if we need a single value without any attributes, even pull with mean can be used

test_df %>% 
       pull(string_outcome) %>%
       mean

standard eval with `dplyr::count()`

To create a list of symbols from strings, you want rlang::syms (not rlang::sym). For unquoting a list or a vector, you want to use !!! (not !!). The following will work:

library(magrittr)

variables <- c("cyl", "vs")

vars_sym <- rlang::syms(variables)
vars_sym
#> [[1]]
#> cyl
#> 
#> [[2]]
#> vs

mtcars %>%
  dplyr::count(!!! vars_sym)
#> # A tibble: 5 x 3
#>     cyl    vs     n
#>   <dbl> <dbl> <int>
#> 1     4     0     1
#> 2     4     1    10
#> 3     6     0     3
#> 4     6     1     4
#> 5     8     0    14

using variable column names in dplyr summarise

You can use base::get:

df %>% summarise(mean(get(x[1])) - mean(get(x[2])))

# # A tibble: 2 x 2
#        c `mean(a) - mean(b)`
#    <dbl>               <dbl>
# 1     1                  -1
# 2     2                  -1

get will search in current environment by default.

As the error message says, mean expects a logical or numeric object, as.name returns a name:

class(as.name("a")) # [1] "name"

You could evaluate your name, that would work as well :

df %>% summarise(mean(eval(as.name(x[1]))) - mean(eval(as.name(x[2]))))
# # A tibble: 2 x 2
#       c `mean(eval(as.name(x[1]))) - mean(eval(as.name(x[2])))`
#   <dbl>                                                   <dbl>
# 1     1                                                      -1
# 2     2                                                      -1

Using strings as arguments in custom dplyr function using non-standard evaluation

You can either use sym to turn "y" into a symbol or parse_expr to parse it into an expression, then unquote it using !!:

library(rlang)

testFun(data.frame(x = c("a", "b", "c"), y = 1:3), !!sym(myVar))

testFun(data.frame(x = c("a", "b", "c"), y = 1:3), !!parse_expr(myVar))

Result:

  x   y
1 a   0
2 b 100
3 c 200

Check my answer in this question for explanation of difference between sym and parse_expr.

Pass column names as strings to group_by and summarize

For this you can now use _at versions of the verbs

df %>%  
  group_by_at(cols2group) %>% 
  summarize_at(.vars = col2summarize, .funs = min)

Edit (2021-06-09):

Please see Ronak Shah's answer, using

mutate(across(all_of(cols2summarize), min))

Now the preferred option

Conditional Evaluation in Dplyr

As there are multiple statements, wrap it inside a {}

r <- c()
iris %>%
    {if(length(r) > 0) {
        mutate(., Test = 1)
      } else .}
    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
...

-testing with r length > 0

r <- 5
iris %>%
     {if(length(r) > 0) {
         mutate(., Test = 1)
       } else .}
    Sepal.Length Sepal.Width Petal.Length Petal.Width    Species Test
1            5.1         3.5          1.4         0.2     setosa    1
2            4.9         3.0          1.4         0.2     setosa    1
3            4.7         3.2          1.3         0.2     setosa    1
...

---

However, this can be easily modified without a loop i.e. convert the logical vector to numeric index by adding 1 (as indexing in R starts from 1). Use that to select a list with values 1 and NULL. If the length is 0, then NULL is selected and thus no column is created

iris %>%
    mutate(Test =  list(NULL, 1)[[1 + (length(r) > 0)]])

---

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