Pie chart 3 options

Question

I am completely lost right now with R. Been trying to teach it to myself for the past week now but I've been stuck with a pie chart problem for two days that I can't seem to figure out.

I have a table that has several columns. With two of which I want to create a pie chart. The first is called "Rabatt" and the second one is called "Gewinnspiel" now they can both either bei 0 for False or 1 for True but neither of the rows can have True in both columns. So the combinations can either be 00 10 or 01. Now I want a pie chart that shows the percentage for how many are 00, 10 or 01. But how do I tell R to only show percentages from both rows for when they contain the value 1 (since when they contain the value 0 they are False and irrelevant until they are both False), add these two up and then also show the percentage of those cases where both are 0?

I hope my explanation was somehow understandable.

Here is a part of the data (in total it's more then 100 rows).

Rabatt  Gewinnspiel
0       0
0       0
0       0
1       0
1       0
1       0
0       0
1       0
1       0
1       0
1       0
1       0
1       0
1       0
0       1
0       1
1       0
1       0
0       0

Answer 1

Here is an example how it can be done with package dplyr and table function

your data

  df = read.table(header = T, stringsAsFactors = F,  text  = "Rabatt  Gewinnspiel
0       0
0       0
0       0
1       0
1       0
1       0
0       0
1       0
1       0
1       0
1       0
1       0
1       0
1       0
0       1
0       1
1       0
1       0
0       0")

you create new variable called ny_var by joining Rabatt and Gewinnspiel

df <- df %>% mutate(ny_var = paste(Rabatt, Gewinnspiel, sep = ""))

you plot your new variable as a pie chart

pie(table(df$ny_var))

you can take a look on your summarised data

table(df$ny_var)

Answer 2

If you don't want to use an additional package for this task and have your %-numbers in the label as well, try this (where df is your data frame):

df <- read.table(text = "Rabatt  Gewinnspiel
0       0
                 0       0
                 0       0
                 1       0
                 1       0
                 1       0
                 0       0
                 1       0
                 1       0
                 1       0
                 1       0
                 1       0
                 1       0
                 1       0
                 0       1
                 0       1
                 1       0
                 1       0
                 0       0", header = TRUE)

pie_tbl <- table(with(df, paste0(Rabatt, Gewinnspiel)))
pie(x = pie_tbl, labels = paste0(names(pie_tbl), ": ", round(pie_tbl/sum(pie_tbl)*100), "%"))

Answer 3

Here is a way:

#data
df <- read.table(text = "Rabatt  Gewinnspiel
    0       0
                     0       0
                     0       0
                     1       0
                     1       0
                     1       0
                     0       0
                     1       0
                     1       0
                     1       0
                     1       0
                     1       0
                     1       0
                     1       0
                     0       1
                     0       1
                     1       0
                     1       0
                     0       0", header = T, stringsAsFactors = F)

#Create a factor based on combination
df$Data <- as.factor(paste(df$Rabatt, df$Gewinnspiel, sep = ""))

library(dplyr)
#Calculate number of occurrence for each combination using summarise
df_mod <- df %>%
  group_by(Data) %>%
  summarise(Count = n())

#Draw pie
pie(df_mod$Count, df_mod$Data)