This is an example from the book I am reading:
1 (define (length items)
2 (define (length-iter a count)
3 (if (null? a)
4 count
5 (length-iter (cdr a)(+ 1 count))))
6 (length-iter items 0))
What I am not understanding is how can length-iter know about count? The first time this
procedure is called with a list, it will in turn define another procedure with two argumenets, I get that much. But how does it know that a is the list items? It hasnt reached line 6 yet, where items is passed to length-iter as the argument a. Somehow though it already knows this and is able to make the computation. Any help in clarifying this a bit is appreciated!
There are two parts in the length function:
Definition of the inner function length-iter;
Invocation of the inner function length-iter.
In the invocation, i.e., line 6, you pass the original items list to the inner function as an argument. This is where the inner function gets called. Previously, you are just defining the function, not calling it:
(length-iter items 0)
Thus, items will be bound to a, and 0 to count. You can think of the inner function as a separate function:
(define (length-iter a count)
(if (null? a)
count
(length-iter (cdr a)(+ 1 count))))
And then, think of your length function as if it just delegated all the work to the length-iter function:
(define (length items)
(length-iter items 0))
That's what's being done in your function. The difference, is that the length-iter function is only known to length.
Related
I need to remove the last two elements from a list in common list, but I can remove only one. What's the way?
(defun my-butlast (list)
(loop for l on list
while (rest l)
collect (first l)))
Simple: reverse, pop, pop, reverse ;-) 1
More efficiently, the following works too:
(let ((list '(a b c d)))
(loop
for x in list
for y in (cddr list)
collect x))
This can also be written, for some arbitrary L and N:
(mapcar #'values L (nthcdr N L))
It works because iteration over multiple lists is bounded by the shortest one. What matters here is the length of the second list (we don't care about its values), which is the length of the original list minus N, which must be a non-negative integer. Notice that NTHCDR conveniently works with sizes greater than the length of the list given in argument.
With the second example, I use the VALUES function as a generalized identity function; MAPCAR only uses the primary value of the computed values, so this works as desired.
The behavior is consistent with the actual BUTLAST2 function, which returns nil for N larger than the number of elements in the list. The actual BUTLAST function can also deal with improper (dotted) lists, but the above version cannot.
1. (alexandria:compose #'nreverse #'cddr #'reverse)
2. BUTLAST is specified as being equivalent to (ldiff list (last list n)). I completely forgot about the existence of LDIFF !
There's a function in the standard for this: butlast, or if you're willing to modify the input list, nbutlast.
butlast returns a copy of list from which the last n conses have been omitted. If n is not supplied, its value is 1. If there are fewer than n conses in list, nil is returned and, in the case of nbutlast, list is not modified.
nbutlast is like butlast, but nbutlast may modify list. It changes the cdr of the cons n+1 from the end of the list to nil.
Examples:
CL-USER> (butlast '(1 2 3 4 5) 2)
(1 2 3)
CL-USER> (nbutlast (list 6 7 8 9 10) 2)
(6 7 8)
The fact that you called your function my-butlast suggests that you might know about this function, but you didn't mention wanting to not use this function, so I assume it's still fair game. Wrapping it up is easy:
CL-USER> (defun my-butlast (list)
(butlast list 2))
MY-BUTLAST
CL-USER> (my-butlast (list 1 2 3 4))
(1 2)
I've been thinking about the following problem. Suppose I'm dealing with a function returning multiple values, such as truncate. Is there a clever way to reverse the order of values that get returned? I'm talking about something more clever than e.g.
(multiple-value-bind (div rem) (truncate x y)
(values rem div))
I don't know how clever this is, but here's what you want:
(reverse (multiple-value-list (the-function-that-returns-multiple-values)))
multiple-value-list being the key, here.
To return these again as separate values, use values-list:
(values-list (reverse (multiple-value-list (the-function-that-returns-multiple-values))))
This whole page may be enlightening.
This problem can be solved more cleverly by writing a higher order function whose input is a function that returns some (values a b), and which returns a function which calls that function, but returns (values b a). In other words a value reversing combinator:
(defun val-rev (function)
(lambda (&rest args)
(multiple-value-bind (a b) (apply function args)
(values b a))))
Though inside the definition of this function we are doing the cumbersome thing you don't want (capturing the values with m-v-bind and reversing with values) this is encapsulated in the combinator and just an implementation detail. It's probably more efficient than consing up a value list and reversing it. Also, it specifically targets the first two values. If a function returns four values, A B C D, then reversing the multiple-value-list means that the first two return values will be C D. However, if we just bind the first two and reverse them, then we bet B A. Reversing the first two (or only two) values is clearly not the same as reversing all values.
Demo:
[1]> (truncate 17 3)
5 ;
2
[2]> (funcall (val-rev #'truncate) 17 3)
2 ;
5
Note that in a Lisp-1 dialect, the invocation loses the added noise of #' and funcall, reducing simply to: ((val-rev truncate) 17 3).
val-rev is kind of a dual of the flip higher order function which you see in some functional languages, which takes a binary function and returns a binary function which is that function, but with the arguments reversed.
To have it as clean/consistent as multiple-value-bind, you could define a macro such as this:
(defmacro reverse-multiple-value-bind (args f &rest body)
`(multiple-value-bind ,(reverse args)
,f
,#body))
Then you have
>> (multiple-value-bind (x y) (floor 3.7) (print x) (print y))
3
0.70000005
and
> (reverse-multiple-value-bind (x y) (floor 3.7) (print x) (print y))
0.70000005
3
I'm trying to write a recursive function that prints to the screen a list (each number in a new line) of elements of the Fibonacci series for the given parameter n. I need to use the display function to print those numbers and use a helper method as well.
Example:
(fibo 5)
1
1
2
3
5
Can anyone help me please? Thanks!!
Simply call your procedure inside a looping function, taking care of printing the elements in the right order and breaking with new lines. Notice that the point where we call the recursion is very important to obtain the desired behavior! (to see this, move the recursive call after the line break - the printing order will change).
(define print-fibo
(lambda (n)
(cond ((> n 0) ; keep iterating if we haven't reached zero
(print-fibo (- n 1)) ; advance the recursion
(display (fibo n)) ; display current value of fibo
(newline))))) ; print a new line
For example:
(print-fibo 5)
1
1
2
3
5
(setf list (loop for i from 1 to 12 collect i))
(defun removef (item seq)
(setf seq (remove item seq)))
CL-USER> (removef 2 list)
(1 3 4 5 6 7 8 9 10 11 12)
CL-USER> (removef 3 list)
(1 2 4 5 6 7 8 9 10 11 12)
Why doesn't removef really modify the variable?
In Common Lisp, parameters are passed "by identity" (this term goes back to D. Rettig, one of the developers of the Allegro Common Lisp implementation). Think of pointers (to heap objects) being passed by values, which is true for most Lisp objects (like strings, vectors, and, of course, lists; things are slightly more complicated, since implementations might also have immediate values, but that's beside the point here).
The setf of seq modifies the (private, lexical) variable binding of the function. This change is not visible outside of removef.
In order for removef to be able to affect the surrounding environment at the point of the call, you need to make it a macro:
(defmacro removef (element place)
`(setf ,place (remove ,element ,place)))
You might want to take at look at setf and the concept of generalized references. Note, that the macro version of removef I provided above is not how it should actually be done! For details, read about get-setf-expansion and its ugly details.
If all you want to do is to destructively modify the list, consider using delete instead of remove, but be aware, that this might have unintended consequences:
(delete 2 '(1 2 3 4))
is not allowed by the ANSI standard (you are destructively modifying a literal object, i.e., part of your code). In this example, the mistake is easy to spot, but if you are 7 frames deep in some callstack, processing values whose origin is not entirely clear to you, this becomes a real problem. And anyway, even
(setf list (list 1 2 3 4))
(delete 1 list)
list
might be surprising at first, even though
(setf list (list 1 2 3 4))
(delete 2 list)
list
seems to "work". Essentially, the first example does not work as intended, as the function delete has the same problem as your original version of removef, namely, it cannot change the caller's notion of the list variable, so even for the destructive version, the right way to do it is:
(setf list (delete 1 (list 1 2 3 4)))
Here is an example of an implementation of removef that is "able to affect the surrounding environment at the point of the call", as stated by #Dirk.
(defmacro removef (item place &rest args &key from-end test test-not start end count key &environment env)
(declare (ignore from-end test test-not start end count key))
(multiple-value-bind (vars vals store-vars writer-form reader-form)
(get-setf-expansion place env)
(assert (length= store-vars 1) ()
"removef only supports single-value places")
(let ((v.args (make-gensym-list (length args)))
(store-var (first store-vars)))
(once-only (item)
`(let* (,#(mapcar #'(lambda (var val)
`(,var ,val))
vars vals)
,#(mapcar #'(lambda (v.arg arg)
`(,v.arg ,arg))
v.args args)
(,store-var (remove ,item ,reader-form ,#v.args)))
,writer-form)))))
The utilities length= , make-gensym-list and once-only are available at Project Alexandria.
BTW exists at Alexandria a removef definition that uses define-modify-macro but requires an auxiliary definition. This version does not requires an auxiliary defintion.
I am fairly new to lisp and this is one of the practice problems.
First of all, this problem is from simply scheme. I am not sure how to answer this.
The purpose of this question is to write the function, count-odd that takes a sentence as its input and count how many odd digits are contained in it as shown below:
(count-odd'(234 556 4 10 97))
6
or
(count-odd '(24680 42 88))
0
If possible, how would you be able to do it, using higher order functions, or recursion or both - whatever gets the job done.
I'll give you a few pointers, not a full solution:
First of all, I see 2 distinct ways of doing this, recursion or higher order functions + recursion. For this case, I think straight recursion is easier to grok.
So we'll want a function which takes in a list and does stuff, so
(define count-odd
(lambda (ls) SOMETHING))
So this is recursive, so we'd want to split the list
(define count-odd
(lambda (ls)
(let ((head (car ls)) (rest (cdr ls)))
SOMETHING)))
Now this has a problem, it's an error for an empty list (eg (count-odd '())), but I'll let you figure out how to fix that. Hint, check out scheme's case expression, it makes it easy to check and deal with an empty list
Now something is our recursion so for something something like:
(+ (if (is-odd head) 1 0) (Figure out how many odds are in rest))
That should give you something to start on. If you have any specific questions later, feel free to post more questions.
Please take first into consideration the other answer guide so that you try to do it by yourself. The following is a different way of solving it. Here is a tested full solution:
(define (count-odd num_list)
(if (null? num_list)
0
(+ (num_odds (car num_list)) (count-odd (cdr num_list)))))
(define (num_odds number)
(if (zero? number)
0
(+ (if (odd? number) 1 0) (num_odds (quotient number 10)))))
Both procedures are recursive.
count-odd keeps getting the first element of a list and passing it to num_odds until there is no element left in the list (that is the base case, a null list).
num_odds gets the amount of odd digits of a number. To do so, always asks if the number is odd in which case it will add 1, otherwise 0. Then the number is divided by 10 to remove the least significant digit (which determines if the number is odd or even) and is passed as argument to a new call. The process repeats until the number is zero (base case).
Try to solve the problem by hand using only recursion before jumping to a higher-order solution; for that, I'd suggest to take a look at the other answers. After you have done that, aim for a practical solution using the tools at your disposal - I would divide the problem in two parts.
First, how to split a positive integer in a list of its digits; this is a recursive procedure over the input number. There are several ways to do this - by first converting the number to a string, or by using arithmetic operations to extract the digits, to name a few. I'll use the later, with a tail-recursive implementation:
(define (split-digits n)
(let loop ((n n)
(acc '()))
(if (< n 10)
(cons n acc)
(loop (quotient n 10)
(cons (remainder n 10) acc)))))
With this, we can solve the problem in terms of higher-order functions, the structure of the solution mirrors the mental process used to solve the problem by hand:
First, we iterate over all the numbers in the input list (using map)
Split each number in the digits that compose it (using split-digits)
Count how many of those digits are odd, this gives a partial solution for just one number (using count)
Add all the partial solutions in the list returned by map (using apply)
This is how it looks:
(define (count-odd lst)
(apply +
(map (lambda (x)
(count odd? (split-digits x)))
lst)))
Don't be confused if some of the other solutions look strange. Simply Scheme uses non-standard definitions for first and butfirst. Here is a solution, that I hope follows Simply Scheme friendly.
Here is one strategy to solve the problem:
turn the number into a list of digits
transform into a list of zero and ones (zero=even, one=odd)
add the numbers in the list
Example: 123 -> '(1 2 3) -> '(1 0 1) -> 2
(define (digit? x)
(<= 0 x 9))
(define (number->digits x)
(if (digit? x)
(list x)
(cons (remainder x 10)
(number->digits (quotient x 10)))))
(define (digit->zero/one d)
(if (even? d) 0 1))
(define (digits->zero/ones ds)
(map digit->zero/one ds))
(define (add-numbers xs)
(if (null? xs)
0
(+ (first xs)
(add-numbers (butfirst xs)))))
(define (count-odds x)
(add-numbers
(digits->zero/ones
(number->digits x))))
The above is untested, so you might need to fix a few typos.
I think this is a good way, too.
(define (count-odd sequence)
(length (filter odd? sequence)))
(define (odd? num)
(= (remainder num 2) 1))
(count-odd '(234 556 4 10 97))
Hope this will help~
The (length sequence) will return the sequence's length,
(filter proc sequence) will return a sequence that contains all the elements satisfy the proc.
And you can define a function called (odd? num)