i have a simple clojure syntax problem (bc i am new to the language). for both examples i have a list lst of (1 2 3 4):
in Lisp i can write:
=>`(first of list is ,(first lst))
(first of list is 1)
in Clojure, if i write the same thing (with the language translation of , to ~ as i THOUGHT i read somewhere) i get:
=>'(first of list is ~(first lst))
(first of list is (clojure.core/unquote (first lst)))
i was hoping i can do what i want to in Clojure as well, and that i just have the syntax wrong. all the examples i find though have functions first and use a ` (backtick). i dont want to call a function like:
`(my-function ~(first lst))
i just want to return '(some list with ,(first lst) replaced in it)
can i do such a thing in Clojure?
EDIT: i gave a poor example seeing as my ACTUAL problem dealt with strings. let me try another example...
=>(def color-lst '(red green blue))
what i wanted to return was:
=>`(the color i want is ~(first color-lst))
this yeilded all the strange returns i saw. the other way to do this is
=>(format "the color i want is %s" (first color-lst))
this is how i solved my problem.
Even if your problem is solved, there are some fundamental differences between CL and Clojure worth mentioning:
The main difference concerning symbols in backquotes between CL and Clojure is, that Clojure resolves quasiquoted symbols; yielding a namespace qualified symbol (take a look at the reader section of the Clojure docs):
user> `foo
user/foo
So, in CL:
CL-USER> (let ((list '(foo bar baz)))
`(first is ,(first list)))
(FIRST IS FOO)
But in Clojure:
user> (let [lst '(foo bar baz)]
`(first is ~(first lst)))
(clojure.core/first user/is foo)
In order to get a non-qualified symbol in Clojure (within backquotes), you'd have to use something like this:
user> `~'foo
foo
So, to get the same result as the CL version (ignoring readtable-case), you'd have to use:
user> (let [lst '(foo bar baz)]
`(~'first ~'is ~(first lst)))
(first is foo)
It's slightly unclear to me what you mean by "first of list is" in your code:
'(first of list is ,(first lst))
But if you meant by that just a placeholder for the beginning of the list, then the following should work fine (make sure to use backtick!):
`(1 2 3 ~(first [1 2]))
=> (1 2 3 1)
Alternatively, using quote will result in the following (which seems to be your problem):
'(1 2 3 ~(first [1 2]))
=> (1 2 3 (clojure.core/unquote (first [1 2])))
found another way. perhaps the whole time i was just not doing things the 'Clojure' way?
(formate "first of list is %d" (first lst)) yeilds what i was looking for.
i think it was partly my fault in the problem description, i gave a bad example. it seems the problem arose bc i was using strings (possibly improperly?) in the list. that and 'first' is also an acceptable function call. all of that confused the issue i believe.
Related
Why learning Clojure, I sometimes need to see what a function does at each step. For example:
(defn kadane [coll]
(let [pos+ (fn [sum x] (if (neg? sum) x (+ sum x)))
ending-heres (reductions pos+ 0 coll)]
(reduce max ending-heres)))
Should I insert println here and there (where, how); or is there a suggested workflow/tool?
This may not be what you're after at the level of a single function (see Charles Duffy's comment below), but if you wanted to do get an overview of what's going on at the level of a namespace (or several), you could use tools.trace (disclosure: I'm a contributor):
(ns foo.core)
(defn foo [x] x)
(defn bar [x] (foo x))
(in-ns 'user) ; standard REPL namespace
(require '[clojure.tools.trace :as trace])
(trace/trace-ns 'foo.core)
(foo.core/bar 123)
TRACE t20387: (foo.core/bar 123)
TRACE t20388: | (foo.core/foo 123)
TRACE t20388: | => 123
TRACE t20387: => 123
It won't catch inner functions and such (as pointed out by Charles), and might be overwhelming with large code graphs, but when exploring small-ish code graphs it can be quite convenient.
(It's also possible to trace individually selected Vars if the groups of interest aren't perfectly aligned with namespaces.)
If you use Emacs with CIDER as most Clojurians do, you already have a built-in debugger:
https://docs.cider.mx/cider/debugging/debugger.html
Chances are your favorite IDE/Editor has something built-in or a plugin already.
There is also (in no particular order):
spyscope
timbre/spy
tupelo/spyx
sayid
tools.trace
good old println
I would look at the above first. However there were/are other possibilities:
https://gist.github.com/ato/252421
https://github.com/philoskim/debux
https://github.com/pallet/ritz/tree/develop/nrepl-core
https://github.com/hozumi/eyewrap
probably many more
Also, if the function is simple enough you can add defs at development-time to peek inside the bindings at a given time inside your function.
Sayid is a tool presented at Clojure Conj 2016 that's directly appropriate to the purpose and comes with an excellent Emacs plugin. See the talk at which it was presented.
To see inside invocations of transient functions, see ws-add-inner-trace-fn (previously, ws-add-deep-trace-fn).
I frequently use the spyx and related functions like spy-let from the Tupelo library for this purpose:
(ns tst.clj.core
(:require [tupelo.core :as t] ))
(t/refer-tupelo)
(defn kadane [coll]
(spy-let [ pos+ (fn [sum x] (if (neg? sum) x (+ sum x)))
ending-heres (reductions pos+ 0 coll) ]
(spyx (reduce max ending-heres))))
(spyx (kadane (range 5)))
will produce output:
pos+ => #object[tst.clj.core$kadane$pos_PLUS___21786 0x3e7de165 ...]
ending-heres => (0 0 1 3 6 10)
(reduce max ending-heres) => 10
(kadane (range 5)) => 10
IMHO it is hard to beat a simple println or similar for debugging. Log files are also invaluable as you get closer to production.
I was a bit surprised by this racket code printing nay when I expected yeah:
(define five 5)
(case 5
[(five) "yeah"]
[else "nay"])
Looking at the racket documentation for case makes it clearer:
The selected clause is the first one with a datum whose quoted form is equal? to the result of val-expr.
So it's about quotation. I'm pretty sure that I did not yet fully grasp what quotation in lisps can buy me. I understand it in the viewpoint of macros and AST transformation. However I'm confused why is it helpful in the case of case for instance..?
I'm also curious, with this specification of case, can I use it to achieve what I wanted to (compare the actual values, not the quoted value), or should I use another construct for that? (cond, while strictly more powerful, is more verbose for simple cases, since you must repeat the predicate at each condition).
The problem is that case introduces implicit quote forms, which cause your example to work for 'five (whose value is 'five), instead of five (whose value is 5).
I almost never use case because of exactly this problem. Instead I use racket's match form with the == pattern:
(define five 5)
(define (f x)
(match x
[(== five) "yeah"]
[_ "nay"]))
(f 5) ; "yeah"
(f 6) ; "nay"
This produces "yeah" on only the value 5, just like you expected. If you wanted it to return "yeah" when it's equal to either five or six, you can use an or pattern:
(define five 5)
(define six 6)
(define (f x)
(match x
[(or (== five) (== six)) "yeah"]
[_ "nay"]))
(f 5) ; "yeah"
(f 6) ; "yeah"
(f 7) ; "nay"
And if you really want to match against quoted datums, you can do that by writing an explicit quote form.
(define (f x)
(match x
[(or 'five 'six) "yeah"]
[_ "nay"]))
(f 5) ; "nay"
(f 6) ; "nay"
(f 7) ; "nay"
(f 'five) ; "yeah"
(f 'six) ; "yeah"
These quote forms are implicit and invisible when you use case, lurking there waiting to cause confusion.
The Racket documentation gives this grammar:
(case val-expr case-clause ...)
where
case-clause = [(datum ...) then-body ...+]
| [else then-body ...+]
Let's compare to your example:
(define five 5)
(case 5 ; (case val-expr
[(five) "yeah"] ; [(datum) then-body1]
[else "nay"]) ; [else then-body2])
We see that (five) is interpreted as (datum). This means that five is
a piece of data (here a symbol), not an expression (later to be evaluated).
Your example of case is evaluated like this:
First the expression 5 is evaluated. The result is the value 5.
Now we look at a clause at a time. The first clause is [(five) "yeah"].
Is the value 5 equal (in the sense of equal?) to one of the datums in (five)? No, so we look at the next clause: [else "nay"]. It is an else-clause so the expression "nay" is evaluated and the result is the value "nay".
The result of the case-expression is thus the value "nay".
Note 1: The left-hand sides of case-clauses are datums (think: they are implicitly quoted).
Note 2: The result of val-expr is compared to the clause datums using equal?. (This is in contrast to Scheme, which uses eqv?.
UPDATE
Why include case? Let's see how one can write the example using cond:
(define five 5)
(let ([val five])
(cond
[(member val '(five)) "yeah"]
[(member val '(six seven)) "yeah"] ; added
[else "nay"])
This shows that one could do without case and just use cond.
However - which version is easier to read?
For a case expression it is easy to see which datums the value is compared to.
Here one must look closely to find the datums. Also in the example we know beforehand that we are trying to find the value among a few list of datums. In general we need to examine a cond-expression more closely to see that's what's happening.
In short: having a case-expression increases readability of your code.
For the historically interested: https://groups.csail.mit.edu/mac/ftpdir/scheme-mail/HTML/rrrs-1986/msg00080.html disussed whether to use eqv? or equal? for case.
UPDATE 2
I'll attempt to given an answer to:
I'm still not clear on the quotation vs working simply on the values though.
I'm wondering specifically why doing the quotation, why working on datum instead
of working on values. Didn't get that bit yet.
Both approaches make sense.
Let's for the sake of argument look at the case where case uses expressions rather than datums in the left hand side of a clause. Also following the Scheme tradition, let's assume eqv? is used for the comparison. Let's call such a
case-expression for ecase (short for expression-case).
The grammar becomes:
(ecase val-expr ecase-clause ...)
where
ecase-clause = [(expr ...) then-body ...+]
| [else then-body ...+]
Your example now becomes:
(define five 5)
(ecase five
[('five) "yeah"]
[else "nay")
This doesn't look too bad and the result is what we are used to.
However consider this example:
(ecase '(3 4)
[('five (list 3 4) "yeah"]
[else "nay")
The result of this would be "nay". The two lists resulting from evaluating the expressions '(3 4) and (list 3 4) are not equal in the sense of eqv?.
This shows that if one chooses to use eqv? for comparisions, having expressions available on the left hand side won't be helpful. The only values that work with eqv? atomic values - and therefore one could just as well use implicit quotations and restrict the left hand side to datums.
Now if equal? was used it would make much more sense to use expressions on the left hand side. The original Racket version of case was the same as the one in Scheme (i.e. it used eq?) later on it was changed to used equal?. If case was designed from scratch, I think, expressions would be allowed rather than datums.
The only remaining issue: Why did the authors of Scheme choose eqv? over equal? for comparisons? My intuition is that the reason were performance (which back in the day was more important than now). The linked to post from the rrrs-authors mailing list gives two options. If you dig a little further you might be able to find responses.
I can't find a reference right now, but case statements use literal, unevaluated data in their different clauses because it is both a frequent use-case and more easily subject to efficient compilation.
You could probably write your own version of Clojure's condp macro or a custom conditional operator to handle your use case.
There is some in clojure, which can be expressed in chicken scheme as following:
(define (some pred coll)
(lazy-head (lazy-filter pred (list->lazy-seq coll))))
Is there something built-in?
Take a look at find from SRFI 1—it works just like your implementation:
(find even? '(3 1 4 1 5 9)) ; => 4
Note, however, that is is slightly different from Clojure's some: Clojure returns the value of pred applied to the element, whereas find returns the element itself.
See also any, also from SRFI 1, which is preferable if you don't need the value, just a check for existence.
so i am trying to understand this piece of code, and after staring at it for far too long i decided to ask here if anyone could help me understand how and why it works
(define knock-knock
(letrec ([dig (lambda (i)
(cons (* i (list-ref knock-knock (- i 1)))
(dig (+ i 1))))])
(cons 1 (dig 1))))
the function is then called by name with the value:
(list-ref knock-knock 5)
So my main problem is that i can not see where the letrec would end. the other thing is that i am not given a list, so what is the 4th element in the list that i am supposed to reference in line 3?
First, a note: this is not normal Scheme, as it requires lazy evaluation.
In lazy evaluation, values are only computed when they are needed. So, for defining knock-knock, we can just do
(cons 1 <thunk: (dig 1)>)
i.e., we generate a pair, but we don't need the second element, so we defer its evaluation until later.
When we actually want to evaluate the second element, we will already have knock-knock defined, so we can reference it.
The next element is computed by taking the previous (i-1-st) element, and multiplies it by i. So this will generate the series {n!}: 1,1,2,6,24,...
A straightforward translation of this code to the (normally lazy) Haskell language goes like this:
knock :: [Int]
knock = 1 : dig 1
where dig i = (i * knock !! (i-1)) : dig (i+1)
try to figure out how to use "append" in Scheme
the concept of append that I can find like this:
----- part 1: understanding the concept of append in Scheme-----
1) append takes two or more lists and constructs a new list with all of their elements.
2) append requires that its arguments are lists, and makes a list whose elements are the elements of those lists. it concatenates the lists it is given. (It effectively conses the elements of the other lists onto the last list to create the result list.)
3) It only concatenates the top-level structure ==> [Q1] what does it mean "only concatenates the top-level"?
4) however--it doesn't "flatten" nested structures.
==> [Q2] what is "flatten" ? (I saw many places this "flatten" but I didn't figure out yet)
==> [Q3] why append does not "flatten" nested structures.
---------- Part 2: how to using append in Scheme --------------------------------
then I looked around to try to use "append" and I saw other discussion
based on the other discussion, I try this implementation
[code 1]
(define (tst-foldr-append lst)
(foldr
(lambda (element acc) (append acc (list element)))
lst
'())
)
it works, but I am struggling to understand that this part ...(append acc (list element)...
what exactly "append" is doing in code 1, to me, it just flipping.
then why it can't be used other logics e.g.
i) simply just flip or
iii).... cons (acc element).....
[Q4] why it have to be "append" in code 1??? Is that because of something to do with foldr ??
again, sorry for the long question, but I think it is all related.
Q1/2/3: What is this "flattening" thing?
Scheme/Lisp/Racket make it very very easy to use lists. Lists are easy to construct and easy to operate on. As a result, they are often nested. So, for instance
`(a b 34)
denotes a list of three elements: two symbols and a number. However,
`(a (b c) 34)
denotes a list of three elements: a symbol, a list, and a number.
The word "flatten" is used to refer to the operation that turns
`(3 ((b) c) (d (e f)))
into
`(3 b c d e f)
That is, the lists-within-lists are "flattened".
The 'append' function does not flatten lists; it just combines them. So, for instance,
(append `(3 (b c) d) `(a (9)))
would produce
`(3 (b c) d a (9))
Another way of saying it is this: if you apply 'append' to a list of length 3 and a list of length 2, the result will be of length 5.
Q4/5: Foldl really has nothing to do with append. I think I would ask a separate question about foldl if I were you.
Final advice: go check out htdp.org .
Q1: It means that sublists are not recursively appended, only the top-most elements are concatenated, for example:
(append '((1) (2)) '((3) (4)))
=> '((1) (2) (3) (4))
Q2: Related to the previous question, flattening a list gets rid of the sublists:
(flatten '((1) (2) (3) (4)))
=> '(1 2 3 4)
Q3: By design, because append only concatenates two lists, for flattening nested structures use flatten.
Q4: Please read the documentation before asking this kind of questions. append is simply a different procedure, not necessarily related to foldr, but they can be used together; it concatenates a list with an element (if the "element" is a list the result will be a proper list). cons just sticks together two things, no matter their type whereas append always returns a list (proper or improper) as output. For example, for appending one element at the end you can do this:
(append '(1 2) '(3))
=> '(1 2 3)
But these expressions will give different results (tested in Racket):
(append '(1 2) 3)
=> '(1 2 . 3)
(cons '(1 2) '(3))
=> '((1 2) 3)
(cons '(1 2) 3)
=> '((1 2) . 3)
Q5: No, cons will work fine here. You wouldn't be asking any of this if you simply tested each procedure to see how they work. Please understand what you're using by reading the documentation and writing little examples, it's the only way you'll ever learn how to program.