Converting Python Function into R Function

Question

I have the following Python function,

import random
import math

deck = [2,3,4,5,6,7,8,9,10,10,10,10,11]

def draw_card(deck):
        card = deck.pop(random.sample(range(len(deck)), 1)[0])
        return(card)

print(deck,'\n')
my_draw = draw_card(deck)
print(my_draw, '\n')
print(deck, '\n')

The output is below,

[2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10, 11] 

3 

[2, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10, 11]

I need to convert this to R but I understand there is no equivalent Python list.pop function in R. I am not worried about random pick as I can do a rep in R and am able to emulate the drawing of card in R by,

DrawCard <- function(card){
  card <- tail(deck,1)
  return(card)
}

Also I am able remove the same drawn card by head(deck, -1) but I do not know how to feed the 2 in 1 function like in Python that will return both the card value and also the shorter length deck.

Please help.

Thanks, Lobbie

R mostly doesn't do in-place mutation; i.e. a function doesn't alter an object, but instead returns a version of that object with which it can be overwritten. The R-esqe version of list.pop would be to subset twice. See ?sample for shuffling. — alistaire
– alistaire, Commented May 6, 2016 at 15:05

C_Z_ · Accepted Answer · 2016-05-06 16:50:01Z

1

I'm not suggesting that you do this (returning a list of the card and the new deck is a better idea), but you can have the function alter the global deck variable like so:

draw_card <- function(deck){
  card <- tail(deck,1)
  deck <<- head(deck, length(deck)-1)
  return(card)
}

print(deck)
# [1]  2  3  4  5  6  7  8  9 10 10 10 10 11

print(draw_card(deck))
# [1] 11

print(deck)
#  [1]  2  3  4  5  6  7  8  9 10 10 10 10

For random sampling (thanks to Ben Bolker) and a function that will work with any deck variable:

draw_card <- function(deck){
  s <- sample(length(deck),size=1)
  card <- deck[s]
  n <- deparse(substitute(deck))
  assign(n, deck[-s], envir=globalenv())
  return(card)
}

edited May 6, 2016 at 16:50

answered May 6, 2016 at 15:09

C_Z_

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7 Comments

Ben Bolker Over a year ago

this shouldn't be head/tail, to match OP's question it should be something like s <- sample(length(deck),size=1); card <- deck[s]; deck <- deck[-s]

Lobbie Over a year ago

Hi CactusWoman and Ben, one of my ideas was to create another ReDeck func for head(deck,-1) to be called after DrawCard() but both your suggestions are good and with Ben's, I can actually random draw, I think. will try it out and let you know how I go. thanks.

Lobbie Over a year ago

Hi again, thank you and both of your answers worked!

C_Z_ Over a year ago

@Lobbie Well, if you want to use this method (I still think you should return a list), check out the edit I made; the previous code I wrote would overwrite the global variable deck, not the variable you pass into the function.

Lobbie Over a year ago

thanks but am a bit confused now between the previous one and this latest one. both assign the value back to the global variable deck isn't it? also, why do u think I should return a list? any particular reasons? thanks.

|

Ananta · Accepted Answer · 2016-05-06 15:42:34Z

1

With your logic, return as list

DrawCard <- function(deck){
  card <- tail(deck,1)
  remain<-head(deck,9)
  return(list(card,remain))
}

cc<-1:10
DrawCard(cc)

A better logic may be,

DrawCard <- function(deck){
  ind<-1:length(deck)
  choose<-sample(ind,1)
  card<-deck[choose]
  remain<-deck[-choose]
  return(list(card,remain))
}

Appended: to answer loop

sum=0
ll<-length(cc)
deck_rem<-cc
for(x in 1:ll){

  a<-DrawCard(deck_rem)
  deck_rem=a[[2]]
  sum=sum+a[[1]]
  print(a[[1]])
  print(a[[2]])
  print(sum)

}

edited May 6, 2016 at 15:42

answered May 6, 2016 at 15:04

Ananta

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3 Comments

Lobbie Over a year ago

thanks but this function is to be in a loop and the card value will be use for summing and the remaining deck will be used for drawing over and over again so am not sure a list will be useful here. thank you no less for responding.

Ananta Over a year ago

I don't want to dig into whether algorithm is correct or not. but its doable. See my appended answer.

Lobbie Over a year ago

thank you, Ananta and yes, it is definitely doable. CatusWoman's and Ben's answers above are simpler and suit my needs. thanks again.

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