I'm lost in clojure ratio. I can't understand it. Why do the equality and inequality tests behave this way?
(= 3/2 1.5)
;; false
(>= 3/2 1.5)
;; true
(> 3/2 1.5)
;;false
(not= 3/2 1.5)
;; true
Use == for numerical comparisons where you want to know if two numbers represent the same number regardless of types:
user> (= 3/2 1.5)
false
user> (== 3/2 1.5)
true
Though keep in mind that == is only for numbers and throws if given something not a number.
user> (== :1 :1)
ClassCastException clojure.lang.Keyword cannot be cast to java.lang.Number clojure.lang.Numbers.equiv (Numbers.java:206)
Clojure tries hard to keep you away from floats as they're susceptible to rounding errors:
user=> (+ 0.1 0.2)
0.30000000000000004
user=>
So the real reason the comparison fails is that there's no accurate way to represent 1.5 internally, and the moment the Clojure parser consumes "1.5" information is potentially lost.
By using fractions, there's no loss of information, so the fraction form 3/2 can be freely passed around to other functions without fear of runaway rounding errors as was the case with the infamous Patriot missile bug.
Simply put, in Clojure ratio is a type. It does not try to convert it to double or float or anything of that sort. That-way you can preserve what the number actually represents without losing any precision.
Check-out Rich Hickey's talk called "Clojure for Java Programmers" in YouTube, He explains it better.
Related
I wonder what the difference is between those operations in Scheme. I have seen similar questions in Stack Overflow but they are about Lisp, and there is not a comparison between three of those operators.
I am writing the different types of commands in Scheme, and I get the following outputs:
(eq? 5 5) -->#t
(eq? 2.5 2.5) -->#f
(equal? 2.5 2.5) --> #t
(= 2.5 2.5) --> #t
Why is this the case?
I'll answer this question incrementally. Let's start with the = equivalence predicate. The = predicate is used to check whether two numbers are equal. If you supply it anything else but a number then it will raise an error:
(= 2 3) => #f
(= 2.5 2.5) => #t
(= '() '()) => error
The eq? predicate is used to check whether its two parameters respresent the same object in memory. For example:
(define x '(2 3))
(define y '(2 3))
(eq? x y) => #f
(define y x)
(eq? x y) => #t
Note however that there's only one empty list '() in memory (actually the empty list doesn't exist in memory, but a pointer to the memory location 0 is considered as the empty list). Hence when comparing empty lists eq? will always return #t (because they represent the same object in memory):
(define x '())
(define y '())
(eq? x y) => #t
Now depending upon the implementation eq? may or may not return #t for primitive values such as numbers, strings, etc. For example:
(eq? 2 2) => depends upon the implementation
(eq? "a" "a") => depends upon the implementation
This is where the eqv? predicate comes into picture. The eqv? is exactly the same as the eq? predicate, except that it will always return #t for same primitive values. For example:
(eqv? 2 2) => #t
(eqv? "a" "a") => depends upon the implementation
Hence eqv? is a superset of eq? and for most cases you should use eqv? instead of eq?.
Finally we come to the equal? predicate. The equal? predicate is exactly the same as the eqv? predicate, except that it can also be used to test whether two lists, vectors, etc. have corresponding elements which satisfy the eqv? predicate. For example:
(define x '(2 3))
(define y '(2 3))
(equal? x y) => #t
(eqv? x y) => #f
In general:
Use the = predicate when you wish to test whether two numbers are equivalent.
Use the eqv? predicate when you wish to test whether two non-numeric values are equivalent.
Use the equal? predicate when you wish to test whether two lists, vectors, etc. are equivalent.
Don't use the eq? predicate unless you know exactly what you're doing.
There are a full two pages in the RnRS specification related to eq?, eqv?, equal? and =. Here is the Draft R7RS Specification. Check it out!
Explanation:
= compares numbers, 2.5 and 2.5 are numerically equal.
equal? for numbers reduces to =, 2.5 and 2.5 are numerically equal.
eq? compares 'pointers'. The number 5, in your Scheme implementation, is implemented as an 'immediate' (likely), thus 5 and 5 are identical. The number 2.5 may require an allocation of a 'floating point record' in your Scheme implementation, the two pointers are not identical.
eq? is #t when it is the same address/object. Normally one could expect #t for same symbol, boolean and object and #f for values that is of different type, with different values, or not the same structure Scheme/Lisp-implementations has a tradition to embed type in their pointers and to embed values in the same space if it's enough space. Thus some pointers really are not addresses but values, like the char R or the Fixnum 10. These will be eq? since the "address" is an embedded type+value. Some implementations also reuse immutable constants. (eq? '(1 2 3) '(1 2 3)) might be #f when interpreted but #t when compiled since it might get the same address. (Like the constant String pool in Java). Because of this, many expresions involving eq? are unspecified, thus wether it evaluates to #t or #f is implementation dependent.
eqv? are #t for the same things as eq?. It is also #t if it's a number or character and it's value is the same, even when the data is too big to fit into a pointer. Thus for those eqv? does the extra work of checking that type is one of the supported, that both are the same type and it's target objects have the same data value.
equal? is #t for the same things as eqv? and if it's a compound type like pair, vector,
string, and bytevector it recursively does equal? with the parts. In practice it will return #t if the two objects looks the same. Prior to R6RS, it's unsafe to use equal? on circular structures.
= is like eqv? but it only works for numeric types. It might be more efficient.
string=? is like equal?, but it only works for strings. It might be more efficient.
equal? recursively compares two objects (of any type) for equality.
Note this could be expensive for a large data structure since potentially the entire list, string, vector, etc must be traversed.
If the object just contains a single element (EG: number, character, etc), this is the same as eqv?.
eqv? tests two objects to determine if both are "normally regarded as the same object".
eqv? and eq? are very similar operations, and the differences between them are going to be somewhat implementation specific.
eq? is the same as eqv? but may be able to discern finer distinctions, and may be implemented more efficiently.
According to the spec, this might be implemented as a fast and efficient pointer comparison, as opposed to a more complicated operation for eqv?.
= compares numbers for numerical equality.
Note that more than two numbers can be provided, eg: (= 1 1.0 1/1 2/2)
You don't mention a scheme implementation, but in Racket, eq? only returns true if the arguments refer to the same object. Your second example is yielding #f because the system is creating a new floating point number for each argument; they're not the same object.
equal? and = are checking for value equivalence, but = is only applicable to numbers.
If you're using Racket, check here for more information. Otherwise, check the documentation of your scheme implementation.
Think of eq? as pointer equality. The authors of the Report want it to be as general as possible so they don't say this outright because it's implementation-dependent, and to say it, would favor the pointer-based implementations. But they do say
It will usually be possible to implement eq? much more efficiently than eqv?, for example, as a simple pointer comparison
Here's what I mean. (eqv? 2 2) is guaranteed to return #t but (eq? 2 2) is unspecified. Now imagine a pointer-based implementation. In it eq? is just pointer comparison. Since (eq? 2 2) is unspecified, it means that this implementation is free to just create new memory object representation of each new number it reads from the source code. eqv? must actually inspect its arguments.
OTOH (eq 'a 'a) is #t. This means that such implementation must recognize symbols with duplicate names and use the same one representation object in memory for all of them.
Suppose an implementation is not pointer-based. As long as it adheres to the Report, it doesn't matter. The authors just don't want to be seen as dictating the specifics of implementations to the implementors, so they choose their wording carefully.
This is my guess anyway.
So very coarsely, eq? is pointer equality, eqv? is (atomic-)values-aware, equal? is also structure-aware (checks into its arguments recursively, so that finally (equal? '(a) '(a)) is required to be #t), = is for numbers, string=? is for strings, and the details are in the Report.
Apart from the previous answers, I will add some comments.
All these predicates want to define the abstract function of identity for an object but in different contextes.
EQ? is implementation-dependent and it does not answer the question are 2 objects the same? only in limited use. From implementation point of view, this predicate just compares 2 numbers (pointer to objects), it does not look at the content of the objects. So, for example, if your implementation does not uniquely keep the strings inside but allocates different memory for each string, then (eq? "a" "a") will be false.
EQV? -- this looks inside the objects, but with limited use. It is implementation-dependent if it returns true for (eqv? (lambda(x) x) (lambda(x) x)). Here it's a full philosophy how to define this predicate, as we know nowadays that there are some fast methods to compare the functionality of some functions, with limited use. But eqv? provides coherent answer for big numbers, strings, etc.
Practically, some of these predicates tries to use the abstract definition of an object (mathematically), while others use the representation of an object (how it's implemented on a real machine). The mathematical definition of identity comes from Leibniz and it says:
X = Y iff for any P, P(X) = P(Y)
X, Y being objects and
P being any property associated with object X and Y.
Ideally it would be to be able to implement this very definition on computer but for reasons of indecidability and/or speed it is not implemented literally. This is why there are lots of operators that try each one to focus on different viewpoints around this definition.
Try to imagine the abstract definition of an identity for a continuation. Even if you can provide a definition of a subset of functions (sigma-recursive class of functions), the language does not impose any predicate to be true or false. It would complicate a lot both the definition of the language and much more the implementation.
The context for the other predicates is easier to analyze.
Is there way to perform arbitrary precision exponentiation in Clojure? I've tried Math/pow and the expt function from clojure.math.numeric-tower, but both will only return limited precision. For example:
(with-precision 100 (expt 2 1/2))
=> 1.4142135623730951
How do I get more digits?
Apfloat for Java provides fast arbitrary precision arithmetic. You can easily use it by adding the following dependency information to your project.clj file, if your project comes with Leiningen.
[org.apfloat/apfloat "1.6.3"]
You can perform arbitrary precision exponentiation in Clojure using Apfloat. For example:
user> (import '(org.apfloat Apfloat ApfloatMath))
org.apfloat.ApfloatMath
user> (-> (Apfloat. 2M 100) (ApfloatMath/pow (Apfloat. 0.5M 100)))
1.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727
math/expt is likely not the function you are looking for as it returns a double instead of a BigDecimal in this context, and hence ignores your with-precision statement:
Returns an exact number if the base is an exact number and the power is an integer, otherwise returns a double.
user> (type (with-precision 100 (math/expt 2M 1/2)))
java.lang.Double
the answer to this question seems to cover how to get arbitrary precision out of BigDecimal exponentiation. BigDecimal seems not to provide this "out of the box"
In Haskell, like in many other functional languages, the function foldl is defined such that, for example, foldl (-) 0 [1,2,3,4] = -10.
This is OK, because foldl (-) 0 [1, 2,3,4] is, by definition, ((((0 - 1) - 2) - 3) - 4).
But, in Racket, (foldl - 0 '(1 2 3 4)) is 2, because Racket "intelligently" calculates like this: (4 - (3 - (2 - (1 - 0)))), which indeed is 2.
Of course, if we define auxiliary function flip, like this:
(define (flip bin-fn)
(lambda (x y)
(bin-fn y x)))
then we could in Racket achieve the same behavior as in Haskell: instead of (foldl - 0 '(1 2 3 4)) we can write: (foldl (flip -) 0 '(1 2 3 4))
The question is: Why is foldl in racket defined in such an odd (nonstandard and nonintuitive) way, differently than in any other language?
The Haskell definition is not uniform. In Racket, the function to both folds have the same order of inputs, and therefore you can just replace foldl by foldr and get the same result. If you do that with the Haskell version you'd get a different result (usually) — and you can see this in the different types of the two.
(In fact, I think that in order to do a proper comparison you should avoid these toy numeric examples where both of the type variables are integers.)
This has the nice byproduct where you're encouraged to choose either foldl or foldr according to their semantic differences. My guess is that with Haskell's order you're likely to choose according to the operation. You have a good example for this: you've used foldl because you want to subtract each number — and that's such an "obvious" choice that it's easy to overlook the fact that foldl is usually a bad choice in a lazy language.
Another difference is that the Haskell version is more limited than the Racket version in the usual way: it operates on exactly one input list, whereas Racket can accept any number of lists. This makes it more important to have a uniform argument order for the input function).
Finally, it is wrong to assume that Racket diverged from "many other functional languages", since folding is far from a new trick, and Racket has roots that are far older than Haskell (or these other languages). The question could therefore go the other way: why is Haskell's foldl defined in a strange way? (And no, (-) is not a good excuse.)
Historical update:
Since this seems to bother people again and again, I did a little bit of legwork. This is not definitive in any way, just my second-hand guessing. Feel free to edit this if you know more, or even better, email the relevant people and ask. Specifically, I don't know the dates where these decisions were made, so the following list is in rough order.
First there was Lisp, and no mention of "fold"ing of any kind. Instead, Lisp has reduce which is very non-uniform, especially if you consider its type. For example, :from-end is a keyword argument that determines whether it's a left or a right scan and it uses different accumulator functions which means that the accumulator type depends on that keyword. This is in addition to other hacks: usually the first value is taken from the list (unless you specify an :initial-value). Finally, if you don't specify an :initial-value, and the list is empty, it will actually apply the function on zero arguments to get a result.
All of this means that reduce is usually used for what its name suggests: reducing a list of values into a single value, where the two types are usually the same. The conclusion here is that it's serving a kind of a similar purpose to folding, but it's not nearly as useful as the generic list iteration construct that you get with folding. I'm guessing that this means that there's no strong relation between reduce and the later fold operations.
The first relevant language that follows Lisp and has a proper fold is ML. The choice that was made there, as noted in newacct's answer below, was to go with the uniform types version (ie, what Racket uses).
The next reference is Bird & Wadler's ItFP (1988), which uses different types (as in Haskell). However, they note in the appendix that Miranda has the same type (as in Racket).
Miranda later on switched the argument order (ie, moved from the Racket order to the Haskell one). Specifically, that text says:
WARNING - this definition of foldl differs from that in older versions of Miranda. The one here is the same as that in Bird and Wadler (1988). The old definition had the two args of `op' reversed.
Haskell took a lot of stuff from Miranda, including the different types. (But of course I don't know the dates so maybe the Miranda change was due to Haskell.) In any case, it's clear at this point that there was no consensus, hence the reversed question above holds.
OCaml went with the Haskell direction and uses different types
I'm guessing that "How to Design Programs" (aka HtDP) was written at roughly the same period, and they chose the same type. There is, however, no motivation or explanation — and in fact, after that exercise it's simply mentioned as one of the built-in functions.
Racket's implementation of the fold operations was, of course, the "built-ins" that are mentioned here.
Then came SRFI-1, and the choice was to use the same-type version (as Racket). This decision was question by John David Stone, who points at a comment in the SRFI that says
Note: MIT Scheme and Haskell flip F's arg order for their reduce and fold functions.
Olin later addressed this: all he said was:
Good point, but I want consistency between the two functions.
state-value first: srfi-1, SML
state-value last: Haskell
Note in particular his use of state-value, which suggests a view where consistent types are a possibly more important point than operator order.
"differently than in any other language"
As a counter-example, Standard ML (ML is a very old and influential functional language)'s foldl also works this way: http://www.standardml.org/Basis/list.html#SIG:LIST.foldl:VAL
Racket's foldl and foldr (and also SRFI-1's fold and fold-right) have the property that
(foldr cons null lst) = lst
(foldl cons null lst) = (reverse lst)
I speculate the argument order was chosen for that reason.
From the Racket documentation, the description of foldl:
(foldl proc init lst ...+) → any/c
Two points of interest for your question are mentioned:
the input lsts are traversed from left to right
And
foldl processes the lsts in constant space
I'm gonna speculate on how the implementation for that might look like, with a single list for simplicity's sake:
(define (my-foldl proc init lst)
(define (iter lst acc)
(if (null? lst)
acc
(iter (cdr lst) (proc (car lst) acc))))
(iter lst init))
As you can see, the requirements of left-to-right traversal and constant space are met (notice the tail recursion in iter), but the order of the arguments for proc was never specified in the description. Hence, the result of calling the above code would be:
(my-foldl - 0 '(1 2 3 4))
> 2
If we had specified the order of the arguments for proc in this way:
(proc acc (car lst))
Then the result would be:
(my-foldl - 0 '(1 2 3 4))
> -10
My point is, the documentation for foldl doesn't make any assumptions on the evaluation order of the arguments for proc, it only has to guarantee that constant space is used and that the elements in the list are evaluated from left to right.
As a side note, you can get the desired evaluation order for your expression by simply writing this:
(- 0 1 2 3 4)
> -10
can some one please explain why
(some #(= 3 %) (range))
returns true but
(some #(= 4/3 %) (range))
never returns?
(range) produces an infinite sequence, so it'll just keep looking until it finds a natural number which is equal to 4/3, which isn't terribly likely.
Fold (aka reduce) is considered a very important higher order function. Map can be expressed in terms of fold (see here). But it sounds more academical than practical to me. A typical use could be to get the sum, or product, or maximum of numbers, but these functions usually accept any number of arguments. So why write (fold + 0 '(2 3 5)) when (+ 2 3 5) works fine. My question is, in what situation is it easiest or most natural to use fold?
The point of fold is that it's more abstract. It's not that you can do things that you couldn't before, it's that you can do them more easily.
Using a fold, you can generalize any function that is defined on two elements to apply to an arbitrary number of elements. This is a win because it's usually much easier to write, test, maintain and modify a single function that applies two arguments than to a list. And it's always easier to write, test, maintain, etc. one simple function instead of two with similar-but-not-quite functionality.
Since fold (and for that matter, map, filter, and friends) have well-defined behaviour, it's often much easier to understand code using these functions than explicit recursion.
Basically, once you have the one version, you get the other "for free". Ultimately, you end up doing less work to get the same result.
Here are a few simple examples where reduce works really well.
Find the sum of the maximum values of each sub-list
Clojure:
user=> (def x '((1 2 3) (4 5) (0 9 1)))
#'user/x
user=> (reduce #(+ %1 (apply max %2)) 0 x)
17
Racket:
> (define x '((1 2 3) (4 5) (0 9 1)))
> (foldl (lambda (a b) (+ b (apply max a))) 0 x)
17
Construct a map from a list
Clojure:
user=> (def y '(("dog" "bark") ("cat" "meow") ("pig" "oink")))
#'user/y
user=> (def z (reduce #(assoc %1 (first %2) (second %2)) {} y))
#'user/z
user=> (z "pig")
"oink"
For a more complicated clojure example featuring reduce, check out my solution to Project Euler problems 18 & 67.
See also: reduce vs. apply
In Common Lisp functions don't accept any number of arguments.
There is a constant defined in every Common Lisp implementation CALL-ARGUMENTS-LIMIT, which must be 50 or larger.
This means that any such portably written function should accept at least 50 arguments. But it could be just 50.
This limit exists to allow compilers to possibly use optimized calling schemes and to not provide the general case, where an unlimited number of arguments could be passed.
Thus to really process large (larger than 50 elements) lists or vectors in portable Common Lisp code, it is necessary to use iteration constructs, reduce, map, and similar. Thus it is also necessary to not use (apply '+ large-list) but use (reduce '+ large-list).
Code using fold is usually awkward to read. That's why people prefer map, filter, exists, sum, and so on—when available. These days I'm primarily writing compilers and interpreters; here's some ways I use fold:
Compute the set of free variables for a function, expression, or type
Add a function's parameters to the symbol table, e.g., for type checking
Accumulate the collection of all sensible error messages generated from a sequence of definitions
Add all the predefined classes to a Smalltalk interpreter at boot time
What all these uses have in common is that they're accumulating information about a sequence into some kind of set or dictionary. Eminently practical.
Your example (+ 2 3 4) only works because you know the number of arguments beforehand. Folds work on lists the size of which can vary.
fold/reduce is the general version of the "cdr-ing down a list" pattern. Each algorithm that's about processing every element of a sequence in order and computing some return value from that can be expressed with it. It's basically the functional version of the foreach loop.
Here's an example that nobody else mentioned yet.
By using a function with a small, well-defined interface like "fold", you can replace that implementation without breaking the programs that use it. You could, for example, make a distributed version that runs on thousands of PCs, so a sorting algorithm that used it would become a distributed sort, and so on. Your programs become more robust, simpler, and faster.
Your example is a trivial one: + already takes any number of arguments, runs quickly in little memory, and has already been written and debugged by whoever wrote your compiler. Those properties are not often true of algorithms I need to run.