This is probably a simple thing I'm missing, but I'm trying to get the cdr of a pair and every call to say (cdr (cons 'a '5)) comes back as (5). I sort of get why that is, but how can I get the it to return without the parens?
I don't want to use flatten because what I'm trying to get (i.e. the cdr) might itself be another procedure expression already wrapped in parens, so I don't want to flatten the list.
(If it matters, I'm working on transforming a let expression into a lambda expression, and this is one of the steps I'm taking, trying to break apart the lambda bindings so I can move them around).
When applied to a proper list, cdr will always return another list (including '(), the empty list).
With proper list I mean a list which ends with the empty list. For instance, when you do this (define lst '(4 5)) under the hood this is what gets assigned to lst: (cons 4 (cons 5 '())), so when you evaluate (cdr lst) you get the second element of the first cons, which happens to be (cons 5 '()), which in turn gets printed as (5).
For extracting only the second element in the list (not the second element of the first cons, which is what cdr does) you could:
As has been pointed in the comments, use (car (cdr lst)) or just (cadr lst) for short
Even simpler: use (second lst)
Another possibility - if the list only has two elements and it's ok to replace it with an improper list, use (define cell (cons 4 5)) or (define cell '(4 . 5)) to build a cons cell and then you can use (car cell) to extract the first element and (cdr cell) to extract the second element.
Related
In The Little Schemer (4th Ed.) it is claimed that a list for which null? is false contains at least one atom, or so I understand from my reading of the text.
This doesn't make sense to me, since (atom '()) is false, and we can stick those into a list to make it non-null:
> (null? '(()))
#f
So my question is, is this a mistake in my reading, or a matter of definitions? Since it's not in the errata I assume such a well-studied book wouldn't have a mistake like this.
If we considered (()) to be the same as (() . ()) or even (cons '() '()) and then considered cons an atom then I could see how you can get there, but I don't think that's what's going on.
(this was tested in Racket 7.0, with the definition of atom? given in the book, i.e.
(define atom?
(lambda (x)
(and (not (pair? x)) (not (null? x)))))
I know this doesn't cover funny Racket features, but should be sufficient here.)
lat is assumed to be a list of atoms at that point in the book.
If it's not empty, by definition it contains some atoms in it.
It's not about Lisp, it's about the book's presentation.
I think lat indicates list of atoms. Thus if lat is not null?, then it needs to contain at least one atom.
There is a procedure called lat? defined as such:
(define lat?
(lambda (l)
(cond
((null? l) #t)
((atom? (car l))
(lat? (cdr l)))
(else #f))))
(lat? '(()) ; ==> #f so by definition '(()) is not a lat and thus the statement does not apply to that list.
A list can contain any type of elements, including empty and other lists, both which are not atoms. lat is a restricted to a flat list with only atomic elements.
As a concept an “atom” is something that can not be broken into smaller parts. A number 42 is an atom, a list (42 43) is not an atom since it contains two smaller parts (namely the numbers 42 and 43). Since an empty list does not contain any smaller parts, it is by this logic an atom.
Now let’s attempt to implement an atom? predicate, that determines whether it’s input is an atom.
(define (atom? x)
(cond
[(number? x) #t]
[(symbol? x) #t]
[(char? x) #t]
...
[else #f]))
Here the ... needs to be replaced with a test for every atomic data type supported by the implementation. This can potentially be a long list. In order to avoid this, we can try to be clever:
(define (atom? x)
(not (list? x)))
This will correctly return false for non-empty lists, and true for numbers, characters etc. However it will return false for the empty list.
Since it is up to the authors of the book to define the term “atom” (the word does not appear in the language standard) they might have opted for the above simple definition.
Note that the definition as non-list is misleading when the language contains other compound data structures such as vectors and structures. If I recall correctly the only compound data structure discussed in the book is lists.
I know cons is for building pairs, like (cons 2 (cons 3 empty)), but I don't understand the code here that use cons after empty. Isn't cons built for numbers? How can cons take 2 arguments? If they can take 2 arguments, how does it evaluate them? and can someone please translate the code after [(empty? lst) empty] please.
(define (removed2 lst)
(cond
[(empty? lst) empty]
[(not (member? (first lst) (rest lst)))
(cons (first lst) (removed2 (rest lst)))]
[else (removed2 (rest lst))])
I suggest you read a good book or a tutorial on Scheme, you're asking for explanations of some of the most basic concepts, that should be well-understood before starting to code in Scheme. For instance, this chapter is a great starting point. I'll address your questions:
I know cons is for building pairs, like (cons 2 (cons 3 empty))
Indeed, you can build pairs of anything you want… pairs of numbers, pairs of pairs, you name it.
But I don't understand the code here that use cons after empty.
In this case, empty means the empty list '(). So, we can build a pair where the first element is an atom (a number in this case) and the second is an empty list - and that's how we build proper lists in Scheme!
Isn't cons built for numbers?
No, it's for building pairs of anything, including other pairs.
How can cons take 2 arguments? If they can take 2 arguments, how does it evaluate them?
And why not? as you've already stated, cons is for building pairs. A pair is made up of two things, so cons takes 2 arguments. It evaluates each of them in turn and then sticks them together to form a pair. You should really, really read the documentation.
And can someone please translate the code after [(empty? lst) empty] please
That's just asking if the list we're recursively traversing is empty, if it is, then it returns an empty list - the base case of any recursive procedure that returns a list. The procedure is simply building a list as a result, for that is consing elements to pairs, and the last pair ends with an empty list, producing a proper list. Basically, this is how you build a list in Scheme:
(cons 1 (cons 2 (cons 3 empty)))
=> '(1 2 3)
I need to write a basic scheme procedure that can find the median of a list and another for the mean.
This is what I've come up with so far:
Mean:
(define (mean lst)
(if (null? lst) ()
(+ 1 (car lst) (mean (cdr lst))))
I know I need to divide my the length somewhere but not sure how to do so. My thought process for this is to add each element to the stack of the list and then divide my the length of the list?
Median:
I'm not sure where to start for median.I know I need to determine if the list has an odd number of elements or even, so to do that I've come up with
(define (median lst)
(if (integer? (/ (length lst) 2) ;which is the one for even
I don't know if I need another procedure to get me to the middle of the list?
The median procedure was already discussed here.
Calculating the mean is simple, just add all the elements and divide by the length of the list, the only special case to take care of is when the list is empty (because that will lead to a division by zero: the length is zero!), return an appropriate value indicating this.
By now you should definitely know how to add all the elements in a list, check with your instructor in case of doubts, but it's a basic operation, it shouldn't be a problem.
I am having trouble writing a function that will move the first element to the end of the list every time it is called. I have tried using a combination of reverse and cdr to cut off the elements at either end, but cannot figure out how to add the elements to the correct end. Any help would be appreciated. Thanks!
Correct outcomes:
(first_to_last '(1 2 3))
(2 3 1)
(first-to-last (first-to-last '(1 2 3)))
(3 1 2)
I think you're over-doing the reversing, personally.
What we want is a list consisting of cdr x with car x appended to the end. The one trick here is that car x isn't a list, so we want to convert it to a list before appending it:
(define (first-to-last x) (append (cdr x) (list (car x))))
If you wanted to stick to the fundamentals, cons is the really fundamental way to put things together into lists, but it would be a bit more work. You'd basically end up defining something essentially identical to append in terms of cons. That's pretty easy but kind of pointless, given that append already exists.
Edit: I guess if you want to use reverse for some reason or other, you could do something like this:
(define (first-to-last x) (reverse (cons (car x) (reverse (cdr x)))))
It's a bit longer and strikes me as kind of clumsy, but it ought to work anyway.
Is it possible to do so? Let's say I want to get the last element of a list, I would create a variable i = 0, and increment it until it equals to length. Any idea? An example would be greatly appreciated.
Thanks,
There are several ways to declare a variable; the cleanest one is let:
(let ((x some-expr))
; code block that uses x
But you don't need this to get the last element of a list. Just use recursion:
(define (last xs)
(if (null? (cdr xs))
(car xs)
(last (cdr xs))))
Note: if you want, you can use a variable to cache cdr's result:
(define (last xs)
(let ((tail (cdr xs)))
(if (null? tail)
(car xs)
(last tail))))
Yes, it's possible to define local variables in scheme, either using let or define inside a function. Using set!, it's also possible to reassign a variable like you imagine.
That being said, you should probably not solve your problem this way. In Scheme it's generally good practice to avoid set! when you don't need to (and in this case you definitely don't need to). Further iterating over a list using indices is usually a bad idea as scheme's lists are linked lists and as such random access O(n) (making the last function as you want to implement it O(n^2)).
So a simple recursive implementation without indices would be more idiomatic and faster than what you're planning to do and as such preferable.