I'm using the http-kit library to make some webcalls and it returns a promise for each.
When I try to deref any of the promises in the vector I get the following error
ArityException Wrong number of args (1) passed to: core/eval5473/fn--5474 clojure.lang.AFn.throwArity (AFn.ja
va:429)
Simplest way to reproduce in a repl without http-kit is as follows
Create collection
(def x [ [1 (promise)] [2 (promise)] [3 (promise)]])
Simple Test
(map first x)
;user=> (1 2 3)
My Test
(map #(vector % #%2) x)
;user=> ArityException Wrong number of args (1) passed to: user/eval109/fn--110 clojure.lang.AFn.throwArity (AFn.java
:429)
Update
I should probably delete this question. The problem had nothing to do with promises as Valentin noted below.
I was typing %2 and thinking second argument. When what i needed was #(second %). i.e second entry in first and only argument.
The function that is the second argument of map must accept only 1 argument in this case (which is meant to be an element of the seq that is being walked through).
You seem to be mistaking passing 2 arguments to a function and passing 1 argument that is a vector of 2 elements.
What you want to write is
(map (fn [[a b]] (vector a #b)) x)
...whereas what you're currently writing is equivalent to:
(map (fn [a b] (vector a #b)) x)
So this is not a problem about promises in fact.
Related
I'm trying to do a really basic problem in clojure and having some trouble wrapping my head around how vectors/lists work.
First off when I am defining the arguments of a function that has a vector as an argument, how do you represent that as an argument.
Would you just have it as a single variable say
(defn example [avector] (This is where the function goes) )
Or do you have to list each element of a vector or list beforehand?
(defn example [vectorpart1 vectorpart2 vectorpart3 vectorpart4 ] (This is where the function goes) )
Also, in terms of vectors and lists, does anyone know of commands that allow you to figure out the length of a vector or get the first/last/or nth element?
To remove the element at index n from vector v:
(defn remove-indexed [v n]
(into (subvec v 0 n) (subvec v (inc n))))
For example,
(remove-indexed (vec (range 10)) 5)
;[0 1 2 3 4 6 7 8 9]
Lots can go wrong:
v might not be a vector.
n might not be a whole number.
n might be out of range for v (we require (contains? v n).
Clojure detects all these errors at run time. A statically typed language would detect 1 and 2 but not 3 at compile time.
Your first example defines a function that takes a single argument, regardless of type. If you pass a vector then that argument will be set to a vector.
(example [1 2 3 4]) ;; (= avector [1 2 3 4])
Your second example defines a function which takes four arguments. You need to pass four separate values for calls to this function to be valid.
(example [1] [2] [3] [4])
;; (= vectorpart1 [1])
;; (= vectorpart2 [2])
;; (= vectorpart3 [3])
;; (= vectorpart4 [4])
It sounds like you might be thinking about the destructuring syntax, which allows you to destructure values directly from an argument vector.
(defn example [[a b c d]]
())
The literal vector syntax in the argument definition describes a mapping between the items in the first argument and symbols available in the function scope.
(example [1 2 3 4])
;; (= a 1)
;; (= b 2)
;; (= c 3)
;; (= d 4)
The other function that also sits in this space is apply. Apply takes a list or vector of arguments and calls a function with them in-place.
(defn example [a b c]
(assert (= a 1))
(assert (= b 2))
(assert (= c 3)))
If we call this function with one vector, you'll get an arity exception.
(example [1 2 3])
;; ArityException Wrong number of args (1) passed ...
Instead we can use apply to pass the vector as arguments.
(apply example [1 2 3])
;; no errors!
You'll find all the methods you need to work with vectors in the Clojure docs.
If you want to remove a specific element, simply take the elements before it and the elements after it, then join them together.
(def v [1 2 3])
(concat (subvec v 0 1) (subvec v 2))
The short answer is that your first example is correct. You don't want to have to name every piece of your vector because you will commonly work with vectors of indeterminate length. If you want to do something with that vector where you need its parts to be assigned, you can do so by destructuring.
The slightly longer answer is that the list of parameters sent into any clojure defn already is a vector. Notice that the parameter list uses [] to wrap its list of args. This is because in Clojure code and data are the same thing. From this article...
Lisps are homoiconic, meaning code written in the language is encoded as data structures that the language has tools to manipulate.
This might be more than you're looking for but it's an important related concept.
Here'a a quick example to get you going... Pass a vector (of strings in this case) to a functions and it returns the vector. If you map over it however, it passes the contents of the vector to the function in succession.
user=> (def params ["bar" "baz"])
#'user/params
user=> (defn foo [params] (println params))
#'user/foo
user=> (foo params)
[bar baz]
nil
user=> (map foo params)
bar
baz
(nil nil)
Additionally, look at the Clojure cheatsheet to find more about things you can do with vectors (and everything else in Clojure).
I understand how apply works in a simple expression like this:
(apply + '(1 2 3))
I have come across a more complex example in a book I am reading.
(def make
(fn [class & args]
(let [seeded {:__class_symbol__ (:__own_symbol__ class)}
constructor (:add-instance-values (:__instance_methods__ class))]
(apply constructor seeded args))))
In the above example, seeded is a map and args is an ArraySeq.
Can anyone explain how apply works in this context?
In this case (apply constructor seeded args) is equivalent to calling (constructor seeded arg0 arg1 arg2 ...). It unwraps the last argument (which must be seqable) and appends them one by one to the list before evaluation.
For example, this: (apply + 1 [2 3]) unrolls to (+ 1 2 3).
It seems to be analogous to:
((make) MyClass) is equivalent to new MyClass()
((make) MyClass "foo" "bar" 3) is equivalent to new MyClass("foo", "bar", 3)
In clojure, I would like to apply a function to all the elements of a sequence and return a map with the results where the keys are the elements of the sequence and the values are the elements of the mapped sequence.
I have written the following function function. But I am wondering why such a function is not part of clojure. Maybe it's not idiomatic?
(defn map-to-object[f lst]
(zipmap lst (map f lst)))
(map-to-object #(+ 2 %) [1 2 3]) => {1 3, 2 4, 3 5}
Your function is perfectly idiomatic.
For a fn to be part of core, I think it has to be useful to most people. What is part of the core language and what is not is quite debatable. Just think about the amount of StringUtils classes that you can find in Java.
My comments were going to get too long winded, so...
Nothing wrong with your code whatsoever.
You might also see (into {} (map (juxt identity f) coll))
One common reason for doing this is to cache the results of a function over some inputs.
There are other use-cases for what you have done, e.g. when a hash-map is specifically needed.
If and only if #3 happens to be your use case, then memoize does this for you.
If the function is f, and the resultant map is m then (f x) and (m x) have the same value in the domain. However, the values of (m x) have been precalculated, in other words, memoized.
Indeed memoize does exactly the same thing behind the scene, it just doesn't give direct access to the map. Here's a tiny modification to the source of memoize to see this.
(defn my-memoize
"Exactly the same as memoize but the cache memory atom must
be supplied as an argument."
[f mem]
(fn [& args]
(if-let [e (find #mem args)]
(val e)
(let [ret (apply f args)]
(swap! mem assoc args ret)
ret))))
Now, to demonstrate
(defn my-map-to-coll [f coll]
(let [m (atom {})
g (my-memoize f m)]
(doseq [x coll] (g x))
#m))
And, as in your example
(my-map-to-coll #(+ 2 %) [1 2 3])
;=> {(3) 5, (2) 4, (1) 3}
But note that the argument(s) are enclosed in a sequence as memoize handles multiple arity functions as well.
I am doing the closure tutorial at http://clojurescriptkoans.com and I am stuck here: http://clojurescriptkoans.com/#functions/9
It looks like this
Higher-order functions take function arguments
(= 25 ( _ (fn [n] (* n n))))
I am supposed to fill in something at the underscore to make the expression true. I have no clue what to do.
The syntax simply consists of binding the function, and then calling it.
Since this is an exercise, I will show a similar situation rather than showing the exercise's solution:
user> ((fn [f] (f "abc")) (fn [s] (str s s s)))
"abcabcabc"
here I bind the argument of the first function to f, and call f with the argument "abc".
or you can use the short-hand notation:
#(%1 5)
Higher order functions takes functions as arguments.
Defining two functions
user=> (defn multiply [n] (* n n))
#'user/multiply
user=> (defn add [n] (+ n n))
#'user/add
Defining higher order function
user=> (defn highorderfn [fn number] (fn number))
#'user/highorderfn
Calling the higher order function
user=> (highorderfn multiply 5)
25
user=> (highorderfn add 5)
10
The following expression in clojure works great:
(doseq [x '(1 2 3 4)] (println x))
This one gives me a nullpointer:
(doseq [x '(1 2 3 4)] ((println x)(println "x")))
It produces the following output:
user=> (doseq [x '(1 2 3 4)] ((println x)(println "x")))
1
x
java.lang.NullPointerException (NO_SOURCE_FILE:0)
user=> (.printStackTrace *e)
java.lang.NullPointerException (NO_SOURCE_FILE:0)
at clojure.lang.Compiler.eval(Compiler.java:4639)
at clojure.core$eval__5182.invoke(core.clj:1966)
at clojure.main$repl__7283$read_eval_print__7295.invoke(main.clj:180)
at clojure.main$repl__7283.doInvoke(main.clj:197)
at clojure.lang.RestFn.invoke(RestFn.java:426)
at clojure.main$repl_opt__7329.invoke(main.clj:251)
at clojure.main$legacy_repl__7354.invoke(main.clj:292)
at clojure.lang.Var.invoke(Var.java:359)
at clojure.main.legacy_repl(main.java:27)
at clojure.lang.Repl.main(Repl.java:20)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
at java.lang.reflect.Method.invoke(Method.java:597)
at jline.ConsoleRunner.main(ConsoleRunner.java:69)
Caused by: java.lang.NullPointerException
at user$eval__266.invoke(NO_SOURCE_FILE:26)
at clojure.lang.Compiler.eval(Compiler.java:4623)
... 14 more
nil
Just adding an extra set of parentheses around the body of a doseq gives me that nullpointer.
What am I doing wrong?
Well, you already figured out the solution, so just a few hints to explain the behavior:
In Clojure (just like in Lisp, Scheme, etc) everything is an expression and an expression is either an atom or a list. With regard to lists, the Clojure manual says
Non-empty Lists are considered calls
to either special forms, macros, or
functions. A call has the form
(operator operands*).
In your example, the body ((println x) (println x)) is a list and the operator is itself an expression which Clojure has to evaluate to obtain the actual operator. That is, you're saying "evaluate the first expression and take its return value as a function to invoke upon the second expression". However, println returns, as you noticed, only nil. This leads to the NullPointerException if nil is interpreted as an operator.
Your code works with (do (println x) (println x)) because do is a special form which evaluates each expression in turn and returns the value of the last expression. Here do is the operator and the expressions with println ar the operands.
To understand the usefulness of this behavior, note that functions are first-class objects in Clojure, e.g., you could return a function as a result from another function. For instance, take the following code:
(doseq [x '(1 2 3 4)] ((if (x > 2)
(fn [x] (println (+ x 2)))
(fn [x] (println (* x 3)))) x))
Here, I am dynamically figuring out the operator to invoke upon the element in the sequence. First, the if-expression is evaluated. If x is larger than two, the if evalutes to the function that prints x + 2, else it evaluates to the function that prints x * 3. This function is than applied to the x of the sequence.
I see you've already realised the problem, however please note you don't need a do:
(doseq [x '(1 2 3 4)] (println x) (println "x"))
doseq is (as the name suggests) a do already :)