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To split a number into digits in a given base, Julia has the digits() function:
julia> digits(36, base = 4)
3-element Array{Int64,1}:
0
1
2
What's the reverse operation? If you have an array of digits and the base, is there a built-in way to convert that to a number? I could print the array to a string and use parse(), but that sounds inefficient, and also wouldn't work for bases > 10.
The previous answers are correct, but there is also the matter of efficiency:
sum([x[k]*base^(k-1) for k=1:length(x)])
collects the numbers into an array before summing, which causes unnecessary allocations. Skip the brackets to get better performance:
sum(x[k]*base^(k-1) for k in 1:length(x))
This also allocates an array before summing: sum(d.*4 .^(0:(length(d)-1)))
If you really want good performance, though, write a loop and avoid repeated exponentiation:
function undigit(d; base=10)
s = zero(eltype(d))
mult = one(eltype(d))
for val in d
s += val * mult
mult *= base
end
return s
end
This has one extra unnecessary multiplication, you could try to figure out some way of skipping that. But the performance is 10-15x better than the other approaches in my tests, and has zero allocations.
Edit: There's actually a slight risk to the type handling above. If the input vector and base have different integer types, you can get a type instability. This code should behave better:
function undigits(d; base=10)
(s, b) = promote(zero(eltype(d)), base)
mult = one(s)
for val in d
s += val * mult
mult *= b
end
return s
end
The answer seems to be written directly within the documentation of digits:
help?> digits
search: digits digits! ndigits isdigit isxdigit disable_sigint
digits([T<:Integer], n::Integer; base::T = 10, pad::Integer = 1)
Return an array with element type T (default Int) of the digits of n in the given base,
optionally padded with zeros to a specified size. More significant digits are at higher
indices, such that n == sum([digits[k]*base^(k-1) for k=1:length(digits)]).
So for your case this will work:
julia> d = digits(36, base = 4);
julia> sum([d[k]*4^(k-1) for k=1:length(d)])
36
And the above code can be shortened with the dot operator:
julia> sum(d.*4 .^(0:(length(d)-1)))
36
Using foldr and muladd for maximum conciseness and efficiency
undigits(d; base = 10) = foldr((a, b) -> muladd(base, b, a), d, init=0)
I want to infinitely repeat T elements in a Sequence<T>. This can't be done using kotlin.collections.asSequence. For example:
val intArray = intArrayOf(1, 2, 3)
val finiteIntSequence = intArray.asSequence()
val many = 10
finiteIntSequence.take(many).forEach(::print)
// 123
This is not what I want. I expected some kind of kotlin.collections.repeat function to exist, but there isn't, so I implemented one myself (e.g. for this IntArray):
var i = 0
val infiniteIntSequence = generateSequence { intArray[i++ % intArray.size] }
infiniteIntSequence.take(many).forEach(::print)
// 1231231231
This is quite imperative, so I feel there must be a more functional and less verbose way to do this. If it exists, what is/are Kotlin's standard way(s) to repeat collections / arrays a(n) (in)finite amount of times?
Update: coroutines are no longer experimental as of Kotlin 1.3! Use them as much as you like :)
If you allow the use of coroutines you can do this in a pretty clean way using sequence:
an infinite amount of times
fun <T> Sequence<T>.repeat() = sequence { while (true) yieldAll(this#repeat) }
Note the use of a qualified this expression this#repeat - simply using this would refer to the lambda's receiver, a SequenceScope.
then you can do
val intArray = intArrayOf(1, 2, 3)
val finiteIntSequence = intArray.asSequence()
val infiniteIntSequence = finiteIntSequence.repeat()
println(infiniteIntSequence.take(10).toList())
// ^ [1, 2, 3, 1, 2, 3, 1, 2, 3, 1]
a finite amount of times
fun <T> Sequence<T>.repeat(n: Int) = sequence { repeat(n) { yieldAll(this#repeat) } }
To avoid using the experimental coroutines, use:
generateSequence { setOf("foo", 'b', 'a', 'r') }
.flatten() // Put the Iterables' contents into one Sequence
.take(5) // Take 5 elements
.joinToString(", ")
// Result: "foo, b, a, r, foo"
or alternatively, if you want to repeat the entire collection a number of times, just take before flattening:
generateSequence { setOf("foo", 'b', 'a', 'r') }
.take(5) // Take the entire Iterable 5 times
.flatten() // Put the Iterables' contents into one Sequence
.joinToString(", ")
// Result: "foo, b, a, r, foo, b, a, r, foo, b, a, r, foo, b, a, r, foo, b, a, r"
For the original question's IntArray, the array first must be converted to an Iterable<Int> (otherwise flatten() isn't available):
val intArray = intArrayOf(1, 2, 3)
generateSequence { intArray.asIterable() }
.flatten()
.take(10)
.joinToString(", ")
// Result: "1, 2, 3, 1, 2, 3, 1, 2, 3, 1"
Furthermore, other types of Array, e.g. ByteArray or LongArray, as well as Map are not Iterable, but they all implement the asIterable() method like IntArray in the example above.
I think this is pretty clear:
generateSequence(0) { (it + 1) % intArray.size }
.map { intArray[it] }
.forEach { println(it) }
A generic solution would be to reuse the proposal from this answer with extension functions:
fun <T> Array<T>.asRepeatedSequence() =
generateSequence(0) {
(it + 1) % this.size
}.map(::get)
fun <T> List<T>.asRepeatedSequence() =
generateSequence(0) {
(it + 1) % this.size
}.map(::get)
Called like this:
intArray.asRepeatedSequence().forEach(::println)
I'm unsure if this is due to API changes in Kotlin, but it's possible to do the following:
fun <T> Sequence<T>.repeatForever() =
generateSequence(this) { it }.flatten()
Live example: https://pl.kotl.in/W-h1dnCFx
If you happen to have Guava on your classpath, you can do the following:
val intArray = intArrayOf(1, 2, 3)
val cyclingSequence = Iterators.cycle(intArray.asList()).asSequence()
// prints 1,2,3,1,2,3,1,2,3,1
println(cyclingSequence.take(10).joinToString(","))
I'd like to write an iterator that behaves exactly like ipairs, except which takes a second argument. The second argument would be a table of the indices that ipairs should loop over.
I'm wondering if my current approach is inefficient, and how I could improve it with closures.
I'm also open to other methods of accomplishing the same thing. But I like iterators because they're easy to use and debug.
I'll be making references to and using some of the terminology from Programming in Lua (PiL), especially the chapter on closures (chapter 7 in the link).
So I'd like to have this,
ary = {10,20,30,40}
for i,v in selpairs(ary, {1,3}) do
ary[i] = v+5
print(string.format("ary[%d] is now = %g", i, ary[i]))
end
which would output this:
ary[1] is now = 15
ary[3] is now = 35
My current approach is this : (in order: iterator, factory, then generic for)
iter = function (t, s)
s = s + 1
local i = t.sel[s]
local v = t.ary[i]
if v then
return s, i, v
end
end
function selpairs (ary, sel)
local t = {}
t.ary = ary
t.sel = sel
return iter, t, 0
end
ary = {10,20,30,40}
for _,i,v in selpairs(ary, {1,3}) do
ary[i] = v+5
print(string.format("ary[%d] is now = %g", i, ary[i]))
end
-- same output as before
It works. sel is the array of 'selected' indices. ary is the array you want to perform the loop on. Inside iter, s indexes sel, and i indexes ary.
But there are a few glaring problems.
I must always discard the first returned argument s (_ in the for loop). I never need s, but it has to be returned as the first argument since it is the "control variable".
The "invariant state" is actually two invariant states (ary and sel) packed into a single table. Pil says that this is more expensive, and recommends using closures. (Hence my writing this question).
The rest can of this can be ignored. I'm just providing more context for what I'm wanting to use selpairs for.
I'm mostly concerned with the second problem. I'm writing this for a library I'm making for generating music. Doing simple stuff like ary[i] = v+5 won't really be a problem. But when I do stuff like accessing object properties and checking bounds, then I get concerned that the 'invariant state as a table' approach may be creating unnecessary overhead. Should I be concerned about this?
If anything, I'd like to know how to write this with closures just for the knowledge.
Of course, I've tried using closures, but I'm failing to understand the scope of "locals in enclosing functions" and how it relates to a for loop calling an iterator.
As for the first problem, I imagine I could make the control variable a table of s, i, and v. And at the return in iter, unpack the table in the desired order.
But I'm guessing that this is inefficient too.
Eventually, I'd like to write an iterator which does this, except nested into itself. My main data structure is arrays of arrays, so I'd hope to make something like this:
ary_of_arys = {
{10, 20, 30, 40},
{5, 6, 7, 8},
{0.9, 1, 1.1, 1.2},
}
for aoa,i,v in selpairs_inarrays(ary_of_arys, {1,3}, {2,3,4}) do
ary_of_arys[aoa][i] = v+5
end
And this too, could use the table approach, but it'd be nice to know how to take advantage of closures.
I've actually done something similar: A function that basically does the same thing by taking a function as it's fourth and final argument. It works just fine, but would this be less inefficient than an iterator?
You can hide "control variable" in an upvalue:
local function selpairs(ary, sel)
local s = 0
return
function()
s = s + 1
local i = sel[s]
local v = ary[i]
if v then
return i, v
end
end
end
Usage:
local ary = {10,20,30,40}
for i, v in selpairs(ary, {1,3}) do
ary[i] = v+5
print(string.format("ary[%d] is now = %g", i, ary[i]))
end
Nested usage:
local ary_of_arys = {
{10, 20, 30, 40},
{5, 6, 7, 8},
{0.9, 1, 1.1, 1.2},
}
local outer_indices = {1,3}
local inner_indices = {2,3,4}
for aoa, ary in selpairs(ary_of_arys, outer_indices) do
for i, v in selpairs(ary, inner_indices) do
ary[i] = v+5 -- This is the same as ary_of_arys[aoa][i] = v+5
end
end
Not sure if I understand what you want to achive but why not simply write
local values = {"a", "b", "c", "d"}
for i,key in ipairs {3,4,1} do
print(values[key])
end
and so forth, instead of implementing all that interator stuff? I mean your use case is rather simple. It can be easily extended to more dimensions.
And here's a co-routine based possibility:
function selpairs(t,selected)
return coroutine.wrap(function()
for _,k in ipairs(selected) do
coroutine.yield(k,t[k])
end
end)
end
Right now I have an SML function:
method([1,1,1,1,2,2,2,3,3,3]);
returns:
val it = [[2,2,2],[3,3,3]] : int list list
but I need it to return:
val it = [[1,1,1,1],[2,2,2],[3,3,3]] : int list list
This is my current code:
- fun method2(L: int list) =
= if tl(L) = [] then [hd(L)] else
= if hd(tl(L)) = hd(L) then hd(L)::method(tl(L)) else [hd(L)];
- fun method(L: int list) =
= if tl(L) = [] then [] else
= if hd(tl(L)) = hd(L) then method(tl(L)) else
= method2(tl(L))::method(tl(L));
As you can see it misses the first method2 call. Any ideas on how I can fix this? I am completely stumped.
Your problem is here if hd(tl(L)) = hd(L) then method(tl(L)) else. This is saying if the head of the tail is equal to the head, then continue processing, but don't add it to the result list. this will skip the first contiguous chunk of equal values. I would suggest separating the duties of these functions a bit more. The way to do this is to have method2 strip off the next contiguous chunk of values, and return a pair, where the first element will have the contiguous chunk removed, and the second element will have the remaining list. For example, method2([1, 1, 1, 2, 2, 3, 3]) = ([1, 1, 1], [2, 2, 3, 3]) and method2([2, 2, 3, 3]) = ([2, 2], [3, 3]). Now, you can just keep calling method2 until the second part of the pair is nil.
I'm not quite sure what you are trying to do with your code. I would recommend creating a tail recursive helper function which is passed three arguments:
1) The list of lists you are trying to build up
2) The current list you are building up
3) The list you are processing
In your example, a typical call somewhere in the middle of the computation would look like:
helper([[1,1,1,1]], [2,2],[2,3,3,3])
The recursion would work by looking at the head of the last argument ([2,3,3,3]) as well as the head of the list which is currently being built up ([2,2]) and, since they are the same -- the 2 at the end of the last argument is shunted to the list being built up:
helper([[1,1,1,1]], [2,2,2],[3,3,3])
in the next step in the recursion the heads are then compared and found to be different (2 != 3), so the helper function will put the middle list at the front of the list of lists:
helper([[2,2,2], [1,1,1,1]], [3],[3,3])
the middle list is re-initialized to [3] so it will start growing
eventually you reach something like this:
helper([[2,2,2], [1,1,1,1]], [3,3,3],[])
the [3,3,3] is then tacked onto the list of lists and the reverse of this list is returned.
Once such a helper function is defined, the main method checks for an empty list and, if not empty, initializes the first call to the helper function. The following code fleshes out theses ideas -- using pattern-matching style rather than hd and tl (I am not a big fan of using those functions explicitly -- it makes the code too Lisp-like). If this is homework then you should probably thoroughly understand how it works and then translate it to code involving hd and tl since your professor would regard it as plagiarized if you use things you haven't yet studied and haven't made it your own work:
fun helper (xs, ys, []) = rev (ys::xs)
| helper (xs, y::ys, z::zs) =
if y = z
then helper(xs, z :: y :: ys, zs)
else helper((y::ys)::xs,[z],zs);
fun method [] = []
| method (x::xs) = helper([],[x],xs);
I have a weird question..
Essentially, I have a function which takes a data frame of dimension Nx(2k) and transforms it into an array of dimension Nx2xk. I then further use that array in various locations in the function.
My issue is this, when k == 2, I'm left with a matrix of degree Nx2, and even worse, if N = 1, I'm stuck with a matrix of degree 1x2.
I would like to write myArray[thisRow,,] to select that slice of the array, but this falls short for the N = 1, k = 2 case. I tried myArray[thisRow,,,drop = FALSE] but that gives an 'incorrect number of dimensions' error. This same issue arrises for the Nx2 case.
Is there a work around for this issue, or do I need to break my code into cases?
Sample Code Shown Below:
thisFunction <- function(myDF)
{
nGroups = NCOL(myDF)/2
afMyArray = myDF
if(nGroups > 1)
{
afMyArray = abind(lapply(1:nGroups, function(g){myDF[,2*(g-1) + 1:2]}),
along = 3)
}
sapply(1:NROW(myDF),
function(r)
{
thisSlice = afMyArray[r,,]
*some operation on thisSlice*
})
}
Thanks,
James