Reading racket code and interpreting in python3

Question

I want to understand what the code really translates into python3 as i cant follow along with it in racket. thank you!

(define set-equal?
  (lambda (s1 s2)
    (and (set-subset? s1 s2)
      (set-subset? s2 s1))))

(define set-subset?
  (lambda (s1 s2)
    (or (null? s1)
      (and (member (car s1) s2)
        (set-subset? (cdr s1) s2)))))

Answer 1

I'm not a Racket expert, but I can at least read that code and figure out what it does.

Let's start with set-subset? . This is a recursive function that takes two lists, s1 and s2 , and returns whether every item in s1 is within s2 . This function works by returning whether s1 is empty, or whether the first element of s1 is a member of s2 and the list of all elements in s1 other than the first is a subset of s2 .

(null? s1) returns whether the list s1 is empty, (car s1) returns the first element of s1 and (cdr s1) returns all elements of the list s1 other than the first.

You could perhaps translate this into Python as

def set_subset(s1, s2):
    return (not s1) or (s1[0] in s2 and set_subset(s1[1:], s2))

although this does assume that s1 and s2 are list s not set s. (If s1 is a set, you can't call s1[0] or s1[1:] .)

set-equal? builds on set-subset? : two sets are equal if both of them are a subset of each other. Again, this would translate into Python as

def set_equal(s1, s2):
    return set_subset(s1, s2) and set_subset(s2, s1)

These functions assume that the lists don't have any repeated values.

However, if you are using set objects in Python 3, you don't need either of these functions, as Python 3 already has these operations built in: (set-subset? s1 s2) in Racket is equivalent to s1.issubset(s2) in Python 3, and (set-equal? s1 s2) to s1 == s2 .