Prevent Clojure from simplifying ratios - clojure

Is there any way to prevent clojure from making for example a 2/5 from 6/15? I need for a function to have the original denominators of ratios, hence the question.

There is no way to prevent clojure from making 2/5 from 6/15. This is most readily apparent from the equality of clojure.lang.Ratio defined here. Preserving the original unreduced version would break equality.
This sounds like a datatype problem. You are putting information into a type that doesn't preserve the amount of data that you need. Fundamentally you are putting two numbers into a ratio datatype which is a single scalar value. You'll (most probably) need to thread more information through or delay the conversion into a ratio.

The calculation of GCD is not conditional:
https://github.com/clojure/clojure/blob/master/src/jvm/clojure/lang/Numbers.java#L355
You can create a clojure.lang.Ratio type directly:
user=> (def x (clojure.lang.Ratio.
(java.math.BigInteger. "6") (java.math.BigInteger. "15")))
user=> (type x)
clojure.lang.Ratio
user=> x
6/15
But compareTo assumes the reduction has occurred and checks the numerator and denominator values individually:
user=> (def y (/ 6 15))
#'user/y
user=> (type y)
clojure.lang.Ratio
user=> y
2/5
user=> (= x y)
false
And other operations will wind up reducing:
user=> (* 3 x)
6/5
user=> (* 3 y)
6/5

Strange requirement. A simple solution is NOT to calculate, i.e. store them as is
{:n 6 :d 15}
The only time you calculate is at the end, when you want a result, or if you want to check for equal.

Related

Implementation of Newton's Method in Clojure

I'm trying to implement Newton's Method in clojure to solve the equation f(x)=0. Function takes these arguments: f(function) f'(function's derivative) n(# of iterations)=10 and x0(initial guess)=10.
(defn newtons-method [f f' n x0]
(if (<= (f x0))
n
(newtons-method f f' (- x0 (/ (f x0) (f' x0))) (+ n 1)))
)
I'm getting an output of 10, but instead want the final solution of x and result of f(x) and I know 10 is wrong, because my function f and its derivative give the correct answer, so I assume I'm messing up somewhere in the iterations
Well, you seem to have at least two issues.
Firstly, your conditional will always return true (assuming (f x0) returns a numeric value), so that is likely not what you want to do.
Also, in order to properly implement recursive functions in clojure, you should have a look at recur, otherwise you could run into stack overflows (unlikely in this particular case, but still)
Another minor thing, instead of doing (+ n 1), it's idiomatic to use (inc n)

Removing the nth element of a vector in clojure

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).

How to apply values to multiple functions in Clojure?

Basically, I need to do something like map, but instead of applying a function to all elements in a collection, I need to apply the same (set of) value(s) to a collection of functions (does this operation have a name?). This might seem like a simple question, but I haven't found an idiomatic way to do it in Clojure. For the special case where I need to apply only one value to each function, for example, I have used
(for [f funs] (f value))
where value is, of course, the value I need each function to take as an argument, and funs is the collection of functions which need to be called with value as the argument.
My question is, then, is there a function in Clojure that does this, but is also generalised for arbitrary numbers of arguments? Or is the above indeed idiomatic Clojure?
You're looking for juxt
juxt
Takes a set of functions and returns a fn that is the juxtaposition
of those fns. The returned fn takes a variable number of args, and
returns a vector containing the result of applying each fn to the
args (left-to-right).
((juxt a b c) x) => [(a x) (b x) (c x)]
From a section of CLOJURE for the BRAVE and TRUE
Another fun thing you can do with map is pass it a collection of
functions. You could use this if you wanted to perform a set of
calculations on different collections of numbers, like so:
(def sum #(reduce + %))
(def avg #(/ (sum %) (count %)))
(defn stats
[numbers]
(map #(% numbers) [sum count avg]))
(stats [3 4 10])
; => (17 3 17/3)
(stats [80 1 44 13 6])
; => (144 5 144/5)

Clojure :: arity-overloaded functions calling each other

Examples of Clojure arity-overloading on functions like the following (taken from the cookbook):
(defn argcount
([] 0) ; Zero arguments
([x] 1) ; One argument
([ x & args] (inc (count args)))) ; List of arguments
... use a form that doesn't seem to allow the functions of lower arity to simply call the functions of higher arity with some default values (that's a common idiom in Java).
Is some other special form used for that ?
There's usually a good way to express the higher arity arguments in a way that doesn't need to refer to other arities using higher order functions and map / reduce. In this case it's pretty simple:
(defn argcount
([] 0)
([x] 1)
([x & args]
(reduce + 1 (map (constantly 1) args))))
Notice the general form of the expression is:
(reduce reducing-function arity-1-value (map mapping-function rest-of-args))
You can't do everything this way, but this works for a surprisingly large proportion of multi-argument functions. It also gains the advnatages of laziness using map, so you can do crazy things like pass ten million arguments to a function with little fear:
(apply argcount (take 10000000 (range)))
=> 10000000
Try that in most other languages and your stack will be toast :-)
mikera's answer is awesome; I'd just add an additional method.
When the a default value is needed for an overloaded function, a local can be used.
In the example division below, the local requires numbers and precision. The defined function overloads the precision with a default value.
(def overloaded-division
(let [divide-with-precision
(fn [divisor dividend precision]
(with-precision precision (/ (bigdec divisor) (bigdec dividend))))]
(fn
;lower-arity calls higher with a default precision.
([divisor dividend] (divide-with-precision divisor dividend 10))
;if precision is supplied it is used.
([divisor dividend precision] (divide-with-precision divisor dividend precision)))
)
)
When called at lower-arity, the default it applied:
user=> (overloaded-division 3 7)
0.4285714286M
user=> (overloaded-division 3 7 40)
0.4285714285714285714285714285714285714286M

Check the class of something in clojure?

I'm learning clojure and have a very basic question: given that clojure has type inference, how can you tell what class was inferred?
For instance, these would each result in different data types:
(2)
(/ 2 3)
(/ 2.0 3)
Is there some kind of class function that will return the data type? Also, is there a normal way of casting something to be a specific type? So in the second example above, what would I do if I wanted the result to be float?
There is a type function in the clojure.core library.
user> (type 2)
java.lang.Integer
user> (type (/ 2 3))
clojure.lang.Ratio
user> (type (/ 2.0 3))
java.lang.Double
If you want to convert a given number into a float then use float.
user> (float 10)
10.0
Similarly you may not need to cast because the following works:
user> (Double/toString (/ 2 3))
"0.6666666666666667"
However, this does too:
user> (str (/ 2 3))
"0.6666666666666667"