R-采样频率直方图：效率更高

Question

I'm a university student, beginning to explore R for an exam. Sorry for the vague title, as I have many questions related to this post.

I've run into the problem of sampling a population of people who are either Male (M) or Female (F). I wished to define a function that could take the number of Males and Females in this population, then create sample.number samples of size sample.size and return a data frame containing the sample proportions of females over the total size of the sample, with related frequencies.

I'm positive there is a simple and well-optimized way to do this, but I've written a small function that (barely) works:

senators <- function(Fem = 13, 
                 Mal = 87, 
                 sample.size = 10, 
                 sample.number = 100){

pop <- c(rep("F", Fem), rep("M", Mal)) # I create the population base

popsa <- list(NA)           # I make some empty variables used later
popsa.factor <- list(NA)    # Not sure if this passage is even needed...
popsa.proportion <- list(NA)

Here comes a for loop. I've read that for loops are really inefficient way to do this. Is there a better way?

for(i in 1:sample.number){
  popsa[[i]] <- sample(pop, sample.size, replace = TRUE)
  popsa.factor[[i]] <- table(factor(popsa[[i]], levels = c("M", "F")))
  popsa.proportion[[i]] <- popsa.factor[[i]][2]/sample.size
  }

I start by assigning each element of the list popsa with a sample, then I use popsa to create a table from each sample, and store it in popsa.factor . Then I calculate the proportions of females over the total and store it in popsa.proportion . This for loop seems super messy to me, and is really slow to process lots of samples. Is there a better, more efficient way to do what I've done here?

popsa.unlisted <- unlist(popsa.proportion)
popsa.frequency <- table(popsa.unlisted)

popsa.frame <- data.frame(Level = as.numeric(names(popsa.frequency)), 
                          Freq =  as.numeric(popsa.frequency))
return(popsa.frame)
} # This closes the function call

I then unlist popsa.proportion to get every proportion in a vector, and table those values to get the frequencies, storing them into popsa.frequency . Now I try to turn the factor popsa.frequency into a data frame, by cheating and converting the names of popsa.frequency as numeric and storing them as the first column of the data frame. The function then returns popsa.frame , as I wanted.

popsa.frame , though, still carries over the factor properties of popsa.frequency in its first column ( Level ). How can I change this? Should I?

Since these are frequencies of a sample distribution, I'd like to create an histogram from this dataframe, although hist() only accepts numeric vectors, so popsa.frame isn't a valid object. plot(popsa.frame) returns more or less what I want, though. How can I create such an histogram?

Edit: Following the marked answer below, I've also come up on how to simply convert the data frame the function creates into an object that hist() can actually use to create a frequency histogram (although using a barplot yields more or less the same graph, and possibly be a more statistically correct way to show such a result):

result <- senators(Fem=13,Mal=87,sample.size=50,sample.number=10000)

raw <- sapply(1:length(result$Level), function(x){
  rep(result$Level, result$Freq)
})

hist(raw)

Answer 1

Your function has some default values that leads to the creation of a data.frame by just doing senators() .

Following your data I would do:

df <- senators() # using default values
plot(df, type="h", lwd = 5, lend=1) # type changes your plot type while lwd changes line sizes, while lend would give squared aspect yo your bars.

Take a look at ?plot to see the types of plots you can do. Also, you can see how change parameters by doing ?par .

PS: look at this post for line width details.

Answer 2

The creation of the lists and the for loop has some performance bottlenecks. I was able to use sapply to remove the for loop and some of the temporary variables.

I am still returning the data fame and another option would return the vector answer just pass the result to the histogram plotting function for your final plot.

senators <- function(Fem = 13, 
                     Mal = 87, 
                     sample.size = 10, 
                     sample.number = 100){

  pop <- c(rep("F", Fem), rep("M", Mal)) # I create the population base

  answer<-sapply(1:sample.number, function(x){popsa <- sample(pop, sample.size, replace = TRUE);
                                            length(popsa[popsa=="F"])/sample.size})

popsa.frequency <- table(answer)

popsa.frame <- data.frame(Level = as.numeric(names(popsa.frequency)), 
                          Freq =  as.numeric(popsa.frequency))
return(popsa.frame)
} 

senators()