问当试图构建饼图时，r不显示百分比

EN

#your example data
Survey<-structure(list(
    Horodateur = structure(c(1619171956.596, 1619172695.039, 1619173104.83, 1619174548.534, 1619174557.538, 1619174735.457), class = c("POSIXct", "POSIXt"), tzone = "UTC"), 
    `To_which_gender_you_identify_the_most?` = c("Male", "Female", "Male", "Female", "Female", "Female"),
    What_is_your_age_group = c("[18-24[", "[10,18[", "[18-24[", "[18-24[", "[18-24[", "[25,34["), 
    How_much_time_do_you_spend_on_social_media = c("1-5 hours", "1-5 hours", ">10 hours", "5-10 hours", "5-10 hours", "1-5 hours")), 
row.names = c(NA, 6L), class = c("tbl_df", "tbl", "data.frame"))
#make a dataframe of a gender table
gender<-data.frame(table(Survey$`To_which_gender_you_identify_the_most?`))
#add a percentages to the labels. Don't use 'count' to label unless you are sure it mataches the order of your data. More on that below. 
pie(gender$Freq,labels=c(paste0(gender$Var1," ",round(100*gender$Freq/length(Survey$`To_which_gender_you_identify_the_most?`), digits = 0),"%")))

#make a dataframe out of the table of your gender object  
your_table<-data.frame(table(gender))
#add a percentages to the labels. Don't use 'count' to label unless you are sure it mataches the order of your data. More on that below. 
pie(your_table$Freq,labels=c(paste0(your_table$gender," ",round(100*your_table$Freq/length(gender), digits = 0),"%")))

#This is an example of how you might mislabel your data.
Survey<-data.frame(`To_which_gender_you_identify_the_most?`=c("Female","Male","Female","Male","Female", "Prefer not to say"))
gender=Survey$`To_which_gender_you_identify_the_most`
gender<-as.factor(gender)
count=c("Prefer not to say","Male","Female")
your_table<-data.frame(table(gender))
pie(your_table$Freq,labels=c(paste0(count," ",round(100*your_table$Freq/length(gender), digits = 0),"%")))