How to retrieve both #elapsed and #allocated as floats simultaneously in julia? - profiling

I would like to use retrieve both the bytes allocated and seconds used for my function in julia. Right now I can get each number separately, but I would like to do something like
#elapsed, #allocated f(1)
to get both numbers for the same call on my function.

Why not use #timed?
julia> #timed(f(1))
(0.8414709848078965, 0.012771058, 76527, 0.0, Base.GC_Diff(76527, 0, 0, 1351, 0, 0, 0, 0, 0))
# the first three values are result, elapsed and allocated, so
julia> (x, elapsed, alloced) = #timed(f(1))[1:3]
(0.8414709848078965, 0.012771058, 76527)

Here is some quick fix with a macro using #elapsed and #allocated:
macro timemem(ex)
return quote begin
local t,m
m = #allocated t = #elapsed $(esc(ex))
t,m-2*sizeof(Int)
end end
end
This is not the best test case, but it gives the same answer as the macros from Base:
julia> #timemem sleep(1.0)
(1.002348175, 192)

Related

Scala: decrement for-loop

I noticed that the following two for-loop cases behave differently sometimes while most of the time they are the same. I couldn't figure out the pattern, does anyone have any idea? Thanks!
case 1:
for (i <- myList.length - 1 to 0 by -1) { ... }
case 2:
for (i <- myList.length - 1 to 0) { ...}
Well, they definitely don't do the same things. n to 0 by -1 means "start at n and go to 0, counting backwards by 1. So:
5 to 0 by -1
// res0: scala.collection.immutable.Range = Range(5, 4, 3, 2, 1, 0)
Whereas n to 0 means "start at n and got to 0 counting forward by 1". But you'll notice that if n > 0, then there will be nothing in that list, since there is no way to count forward to 0 from anything greater than zero.
5 to 0
// res1: scala.collection.immutable.Range.Inclusive = Range()
The only way that they would produce the same result is if n=0 since counting from 0 to 0 is the same forwards and backwards:
0 to 0 by -1 // Range(0)
0 to 0 // Range(0)
In your case, since you're starting at myList.length - 1, they will produce the same result when the length of myList is 1.
In summary, the first version makes sense, because you want to count down to 0 by counting backward (by -1). And the second version doesn't make sense because you're not going to want to count forward to 0 from a length (which is necessarily non-negative).
First, we need to learn more about how value members to and by works.
to - Click Here for API documentation
to is a value member that appears in classes like int, double etc.
scala> 1 to 3
res35: scala.collection.immutable.Range.Inclusive = Range(1, 2, 3)
Honestly, you don't have to use start to end by step and start.to(end, step) will also work if you are more comfortable working with in this world. Basically, to will return you a Range.Inclusive object if we are talking about integer inputs.
by - Click Here for API documentation
Create a new range with the start and end values of this range and a new step
scala> Range(1,8) by 3
res54: scala.collection.immutable.Range = Range(1, 4, 7)
scala> Range(1,8).by(3)
res55: scala.collection.immutable.Range = Range(1, 4, 7)
In the end, lets spend some time looking at what happens when the step is on a different direction from start to end. Like 1 to 3 by -1
Here is the source code of the Range class and it is actually pretty straightforward to read:
def by(step: Int): Range = copy(start, end, step)
So by is actually calling a function copy, so what is copy?
protected def copy(start: Int, end: Int, step: Int): Range = new Range(start, end, step)
So copy is literally recreate a new range with different step, then lets look at the constructor or Range itself.
Reading this paragraph of code
override final val isEmpty = (
(start > end && step > 0)
|| (start < end && step < 0)
|| (start == end && !isInclusive)
)
These cases will trigger the exception and your result will be a empty Range in cases like 1 to 3 by -1..etc.
Sorry the length of my post is getting out of control since I am also learning Scala now.
Why don't you just read the source code of Range, it is written by Martin Odersky and it is only 500 lines including comments :)

Getting all combinations of splitting an array into two equally sized groups in Julia

Given an array of 20 numbers, I would like to extract all possible combinations of two groups, with ten numbers in each, order is not important.
combinations([1, 2, 3], 2)
in Julia will give me all possible combinations of two numbers drawn from the array, but I also need the ones that were not drawn...
You can use setdiff to determine the items missing from any vector, e.g.,
y = setdiff(1:5, [2,4])
yields [1,3,5].
After playing around for a bit, I came up with this code, which seems to work. I'm sure it could be written much more elegantly, etc.
function removeall!(remove::Array, a::Array)
for i in remove
if in(i, a)
splice!(a, indexin([i], a)[1])
end
end
end
function combinationgroups(a::Array, count::Integer)
result = {}
for i in combinations(a, count)
all = copy(a)
removeall!(i, all)
push!(result, { i; all } )
end
result
end
combinationgroups([1,2,3,4],2)
6-element Array{Any,1}:
{[1,2],[3,4]}
{[1,3],[2,4]}
{[1,4],[2,3]}
{[2,3],[1,4]}
{[2,4],[1,3]}
{[3,4],[1,2]}
Based on #tholy's comment about instead of using the actual numbers, I could use positions (to avoid problems with numbers not being unique) and setdiff to get the "other group" (the non-selected numbers), I came up with the following. The first function grabs values out of an array based on indices (ie. arraybyindex([11,12,13,14,15], [2,4]) => [12,14]). This seems like it could be part of the standard library (I did look for it, but might have missed it).
The second function does what combinationgroups was doing above, creating all groups of a certain size, and their complements. It can be called by itself, or through the third function, which extracts groups of all possible sizes. It's possible that this could all be written much faster, and more idiomatical.
function arraybyindex(a::Array, indx::Array)
res = {}
for e in indx
push!(res, a[e])
end
res
end
function combinationsbypos(a::Array, n::Integer)
res = {}
positions = 1:length(a)
for e in combinations(positions, n)
push!(res, { arraybyindex(a, e) ; arraybyindex(a, setdiff(positions, e)) })
end
res
end
function allcombinationgroups(a::Array)
maxsplit = floor(length(a) / 2)
res = {}
for e in 1:5
println("Calculating for $e, so far $(length(res)) groups calculated")
push!(res, combinationsbypos(a, e))
end
res
end
Running this in IJulia on a 3 year old MacBook pro gives
#time c=allcombinationgroups([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20])
println(length(c))
c
Calculating for 1, so far 0 groups calculated
Calculating for 2, so far 20 groups calculated
Calculating for 3, so far 210 groups calculated
Calculating for 4, so far 1350 groups calculated
Calculating for 5, so far 6195 groups calculated
Calculating for 6, so far 21699 groups calculated
Calculating for 7, so far 60459 groups calculated
Calculating for 8, so far 137979 groups calculated
Calculating for 9, so far 263949 groups calculated
Calculating for 10, so far 431909 groups calculated
elapsed time: 11.565218719 seconds (1894698956 bytes allocated)
Out[49]:
616665
616665-element Array{Any,1}:
{{1},{2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}}
{{2},{1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}}
⋮
{{10,12,13,14,15,16,17,18,19,20},{1,2,3,4,5,6,7,8,9,11}}
{{11,12,13,14,15,16,17,18,19,20},{1,2,3,4,5,6,7,8,9,10}}
ie. 53,334 groups calculated per second.
As a contrast, using the same outer allcombinationgroups function, but replacing the call to combinationsbypos with a call to combinationgroups (see previous answer), is 10x slower.
I then rewrote the array by index group using true or false flags as suggested by #tholy (I couldn't figure out how to get it work using [], so I used setindex! explicitly, and moved it into one function. Another 10x speedup! 616,665 groups in 1 second!
Final code (so far):
function combinationsbypos(a::Array, n::Integer)
res = {}
positions = 1:length(a)
emptyflags = falses(length(a))
for e in combinations(positions, n)
flag = copy(emptyflags)
setindex!(flag, true, e)
push!(res, {a[flag] ; a[!flag]} )
end
res
end
function allcombinationgroups(a::Array)
maxsplit = floor(length(a) / 2)
res = {}
for e in 1:maxsplit
res = vcat(res, combinationsbypos(a, e))
end
res
end

Mathematica - Functions with Lists

I got this error in Mathematica today:
Set::shape: "Lists {0,0,0,0,0,0,0,0,0,0} and {0,0,0,0,0,0,0,0,0,0,{1}} are not the same shape" >>
And after 3 of those :
General::stop : Further output of Set::shape will be suppressed during this calculation. >>
I am confused as to why I cannot append a "1" to my list of zeros. Is this because I cannot edit the list that is passed into the function? If so, how could I edit that list and somehow return or print it?
Here is my full code:
notFunctioningFunction[list_] := (For[i = 1, i < 10, i++, list = Append[list, {1}]];
Print[list])
list = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
notFunctioningFunction[list]
The reason why I am appending a "{1}" is because in my function, I am solving an equation, and getting the value of the variable which outputs {1}. Here is my code for that :
varName / . Solve[ function1 == function2 ]
Obviously I am a beginner with Mathematica so please be patient :)
Thanks,
Bucco
Append needs to take one list and one element. Like so:
Append[{1,2,3,4},5]
If you have two lists, you can use Join. Like so:
Join[{1,2,3,4},{5}]
Both of these will yield the same result: {1,2,3,4,5}.
Dear Mathematica beginner.
First, when you use something like
{a,b} = {c,d,e};
in Mathematica, between two lists, the program has a difficulty because this is a construct used to assign values to variables, and it requires (among other things) the two lists to be equal.
If what you want is just to add a "1" to an existing and named list, one at a time, the best construct is:
AppendTo[list, 1];
(this construct will modify the variable 'list')
or
list = Join[list, {1}];
Second: about the error messages, they are printed 3 times by default in an evaluation, then muted so that a long list of identical error messages does not clutter your display.
Third, if what you need is adding 10 1s to a list, there is no need to construct that in a loop. You can do that in one pass:
list = Join[list, Table[1, {10}]]
or, more cryptic for beginners
list = Join[list, Array[1&, 10]]

Assigning list elements to variables

The output from Mathematica with the following operation FactorInteger[28851680048402838857] is as follows:
{{3897424303, 1}, {7402755719, 1}}
My question is: how could I go about extracting the two prime numbers (without the exponents) and assign them to an arbitrary variable?
I basically want to retrieve two primes, whatever they may be, and assign them some variables.
Ex: x0 = 3897424303 and x1 = 7402755719
Thanks!
The output is a list and you can use list manipulating functions like Part ([[ ]]) to pick the pieces you want, e.g.,
{x0, x1} = FactorInteger[28851680048402838857][[All, 1]]
or, without Part:
{{x0,dummy}, {x1,dummy}} = FactorInteger[28851680048402838857];
Implicit in your question is the issue of handing parts of the expression that is returned as output from functions such as FactorInteger. Allow me to suggest alternatives.
1. Keep all of the values in a {list} and access each element with Part:
x = First /# FactorInteger[7813426]
{2, 31, 126023}
x[[1]]
x[[3]]
2
126023
2. Store factors as values of the function x, mimicking indexation of an array:
(This code uses MapIndexed, Function.)
Clear[x]
MapIndexed[
(x[First##2] = First##1) &,
FactorInteger[7813426]
];
x[1]
x[3]
2
126023
You can see all the values using ? or ?? (see Information):
?x
Global`x
x[1]=2
x[2]=31
x[3]=126023

Which language understands 'variable a = 0 , 20, ..., 300'?

Which language is smart so that it could understand variable a = 0 , 20, ..., 300 ? so you could easily create arrays with it giving step start var last var (or, better no last variable (a la infinite array)) and not only for numbers (but even complex numbers and custom structures like Sedenion's which you would probably define on your own as a class or whatever...)
Point is, find a language or algorithm usable in a language that can cach the law of how array of variables you've given (or params of that variables) change. And compose using that law a structure from which you would be able to get any variable(s).
To everyone - examples you provide are very helpful for all beginners out there. And at the same time are the basic knowledge required to build such 'Smart Array' class. So thank you wary much for your enthusiastic help.
As JeffSahol noticed
all possible rules might include some
that require evaluation of some/all
existing members to generate the nth
member.
So it is a hard Question. And I think language that would do it 'Naturally' would be great to play\work with, hopefully not only for mathematicians.
Haskell:
Prelude> let a=[0,20..300]
Prelude> a
[0,20,40,60,80,100,120,140,160,180,200,220,240,260,280,300]
btw: infinite lists are possible, too:
Prelude> let a=[0,20..]
Prelude> take 20 a
[0,20,40,60,80,100,120,140,160,180,200,220,240,260,280,300,320,340,360,380]
Excel:
Write 0 in A1
Write 20 in A2
Select A1:2
Drag the corner downwards
MatLab:
a = [0:20:300]
F#:
> let a = [|0..20..300|];;
val a : int [] =
[|0; 20; 40; 60; 80; 100; 120; 140; 160; 180; 200; 220; 240; 260; 280; 300|]
With complex numbers:
let c1 = Complex.Create( 0.0, 0.0)
let c2 = Complex.Create(10.0, 10.0)
let a = [|c1..c2|]
val a : Complex [] =
[|0r+0i; 1r+0i; 2r+0i; 3r+0i; 4r+0i; 5r+0i; 6r+0i; 7r+0i; 8r+0i; 9r+0i; 10r+0i|]
As you can see it increments only the real part.
If the step is a complex number too, it will increment the real part AND the imaginary part, till the last var real part has been reached:
let step = Complex.Create(2.0, 1.0)
let a = [|c1..step..c2|]
val a: Complex [] =
[|0r+0i; 2r+1i; 4r+2i; 6r+3i; 8r+4i; 10r+5i|]
Note that if this behavior doesn't match your needs you still can overload (..) and (.. ..) operators. E.g. you want that it increments the imaginary part instead of the real part:
let (..) (c1:Complex) (c2:Complex) =
seq {
for i in 0..int(c2.i-c1.i) do
yield Complex.Create(c1.r, c1.i + float i)
}
let a = [|c1..c2|]
val a : Complex [] =
[|0r+0i; 0r+1i; 0r+2i; 0r+3i; 0r+4i; 0r+5i; 0r+6i; 0r+7i; 0r+8i; 0r+9i; 0r+10i|]
And PHP:
$a = range(1,300,20);
Wait...
Python:
print range(0, 320, 20)
gives
[0, 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220, 240, 260, 280, 300]
Props to the comments (I knew there was a more succinct way :P)
Scala:
scala> val a = 0 to 100 by 20
a: scala.collection.immutable.Range = Range(0, 20, 40, 60, 80, 100)
scala> a foreach println
0
20
40
60
80
100
Infinite Lists:
scala> val b = Stream from 1
b: scala.collection.immutable.Stream[Int] = Stream(1, ?)
scala> b take 5 foreach println
1
2
3
4
5
In python you have
a = xrange(start, stop, step)
(or simply range in python 3)
This gives you an iterator from start to stop. It can be infinite since it is built lazily.
>>> a = xrange(0, 300, 20)
>>> for item in a: print item
...
0
20
40
60
80
100
120
140
160
180
200
220
240
260
280
And C++ too [use FC++ library]:
// List is different from STL list
List<int> integers = enumFrom(1); // Lazy list of all numbers starting from 1
// filter and ptr_to_fun definitions provided by FC++
// The idea is to _filter_ prime numbers in this case
// prime is user provided routine that checks if a number is prime
// So the end result is a list of infinite primes :)
List<int> filtered_nums = filter( ptr_to_fun(&prime), integers );
FC++ lazy list implementation: http://www.cc.gatech.edu/~yannis/fc++/New/new_list_implementation.html
More details: http://www.cc.gatech.edu/~yannis/fc++/
Arpan
Groovy,
assert [ 1, *3..5, 7, *9..<12 ] == [1,3,4,5,7,9,10,11]
The SWYM language, which appears to no longer be online, could infer arithmetic and geometric progressions from a few example items and generate an appropriate list.
I believe the syntax in perl6 is start ... *+increment_value, end
You should instead use math.
- (int) infiniteList: (int)x
{
return (x*20);
}
The "smart" arrays use this format since I seriously doubt Haskel could let you do this:
a[1] = 15
after defining a.
C# for example does implement Enumerable.Range(int start, int count), PHP offers the function range(mixed low, mixed high, number step), ... There are programming languages that are "smart" enough.
Beside that, an infinite array is pretty much useless - it's not infinite at all but all-memory-consuming.
You cannot do this enumerating simply with complex numbers as there is no direct successor or predecessor for a given number. Edit: This does not mean that you cannot compare complex numbers or create an array with a specified step!
I may be misunderstanding the question, but the answers that specify way to code the specific example you gave (counting by 20's) don't really meet the requirement that the array "cache" an arbitrary rule for generating array members...it seems that almost any complete solution would require a custom collection class that allows generation of the members with a delegated function/method, especially since all possible rules might include some that require evaluation of some/all existing members to generate the nth member.
Just about any program language can give you this sequence. The question is what syntax you want to use to express it. For example, in C# you can write:
Enumerable.Range(0, 300).Where(x => (x % 20) == 0)
or
for (int i = 0; i < 300; i += 20) yield return i;
or encapsulated in a class:
new ArithmaticSequence(0, 301, 20);
or in a method in a static class:
Enumerable2.ArithmaticSequence(0, 301, 20);
So, what is your criteria?
Assembly:
Assuming edi contains the address of the desired array:
xor eax, eax
loop_location:
mov [edi], eax
add edi, #4
add eax, #20
cmp eax, #300
jl loop_location
MATLAB
it is not a Programming language itself but its a tool but still u can use it like a programming language.
It is built for such Mathematics operations to easily arrays are a breeze there :)
a = 0:1:20;
creates an array from 0 to 20 with an increment of 1.
instead of the number 1 you can also provide any value/operation for the increment
Php always does things much simpler, and sometimes dangerously simple too :)
Well… Java is the only language I've ever seriously used that couldn't do that (although I believe using a Vector instead of an Array allowed that).