CODEWITHSUNDEEP
rshiny
Doug Fir
Doug Fir

Reputation: 21212

Using Shiny input variable within dplyr chain

My input radio buttons allow for several features that the user can filter on, including "Voice", "Format", "DeviceType".

Here is a data frame with hard coded values using the "Voice" option that allows my Shiny app to run when used:

combinations <- reactive({
  expand.grid(unique(combined_data$date),
                            unique(combined_data$Voice)) %>%
  rename_("date" = "Var1",
         "Voice" = "Var2") %>%
 mutate_("Voice" = as.character("Voice"))
})

I'm trying to transform combinations to use the input variable:

combinations <- reactive({
  the_breakdown <- input$breakdown
  expand.grid(unique(combined_data$date),
                            unique(combined_data[[input$breakdown]])) %>%
  rename_("date" = "Var1",
         the_breakdown = "Var2") %>%
 mutate_(the_breakdown = as.character(the_breakdown))
})

This gives me an error later in my code, and right now I'm having a hard time debugging it. However, it's clear that my attempt to make reactive data frame combinations() use the variable input$breakdown, somehting is not working.

I tried to incrementally go through the initial df by switching hard coded example "Voice" to input$breakdown but failed to isolate the problem.

How should I make df combinations use the input variable input$breakdown, given I know that the hard coded version with "Voice" does work?

Upvotes: 2

Views: 679

Answers (1)

akrun
akrun

Reputation: 887118

We can use rename and mutate with sym from rlang

combinations <- reactive({
 the_breakdown <- input$breakdown
  expand.grid(unique(combined_data$date),
                         unique(combined_data[[input$breakdown]])) %>%
  rename(date = Var1,
         !! the_breakdown := Var2) %>%
  mutate(!! the_breakdown :=  as.character(!! rlang::sym(the_breakdown)))
})

Using a small reproducible example 

the_breakdown <- 'digitnew'
df1 %>%
  rename(!!the_breakdown := digit, s9 = s3)  %>% #rename
  mutate(!! the_breakdown :=  as.character(!! rlang::sym(the_breakdown)))  %>% #transform
  str # get the structure
# 'data.frame':   4 obs. of  8 variables:
# $ digitnew: chr  "0" "1" "2" "3"  #####
# $ s1      : int  1 0 1 1
# $ s2      : int  1 0 0 0
# $ s9      : int  1 1 1 1
# $ s4      : int  1 0 0 1
# $ s5      : int  1 0 1 0
# $ s6      : int  1 1 0 1
# $ s7      : int  0 0 0 1

data

df1 <- structure(list(digit = 0:3, s1 = c(1L, 0L, 1L, 1L), s2 = c(1L, 
 0L, 0L, 0L), s3 = c(1L, 1L, 1L, 1L), s4 = c(1L, 0L, 0L, 1L), 
s5 = c(1L, 0L, 1L, 0L), s6 = c(1L, 1L, 0L, 1L), s7 = c(0L, 
0L, 0L, 1L)), .Names = c("digit", "s1", "s2", "s3", "s4",  
 "s5", "s6", "s7"), row.names = c("1", "2", "3", "4"), class = "data.frame")

Upvotes: 2

Related Questions