[julia-users] Re: Unexpected type behaviour in arithmetic operations
El viernes, 28 de febrero de 2014 07:49:09 UTC-6, andrew cooke escribió:
the current behaviour is explained at
http://julia.readthedocs.org/en/latest/manual/conversion-and-promotion/
But I don't find it evident from that document that Int32+Int32 gives Int64
on a 64-bit machine!
defining
Base.promote_rule(::Type{Int32}, ::Type{Int32}) = Int32
doesn't help either, and i'm not sure why.
andrew
On Friday, 28 February 2014 10:46:16 UTC-3, Tobias Knopp wrote:
There was quite some discussion on this topic in
https://groups.google.com/forum/?fromgroups=#!searchin/julia-users/Int32/julia-users/Rte_I6htLRc/VJG5DWVcZbQJ
I had the feeling that we almost convinced Stefan :-)
The good thing is that Array{Int32} operations do not promote to
Array{Int64}. But I still vote for promoting Int32 + Int32 to Int32.
[julia-users] Interesting behaviour in IntSet
I am investigating possible data structures for an application. Here is an interesting behaviour in IntSet, which is no doubt to do with the implementation. Maybe it should just throw an exception if someone tries to add a really large integer like this! julia s = IntSet() IntSet() julia push!(s, 10) IntSet(10) julia sizeof(s) 24 julia push!(s, 100) IntSet(10, 100) julia sizeof(s) 24 julia push!(s, 1000) IntSet(10, 100, 1000) julia push!(s, 1) IntSet(10, 100, 1000, 1) julia push!(s, 10) IntSet(10, 100, 1000, 1, 10) julia sizeof(s) 24 julia push!(s, 100) IntSet(10, 100, 1000, 1, 10, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408^CEvaluation succeeded, but an error occurred while showing value of type IntSet: ERROR: interrupt in show at intset.jl:172 in anonymous at show.jl:973 in showlimited at show.jl:972 in writemime at repl.jl:2 in display at multimedia.jl:117 in display at multimedia.jl:119 in display at multimedia.jl:151
[julia-users] Re: Interesting behaviour in IntSet
El viernes, 28 de febrero de 2014 08:41:37 UTC-6, Ivar Nesje escribió: The documentation states very clear that IntSethttp://docs.julialang.org/en/latest/stdlib/base/#Base.IntSet should only be used for dense collections, and that Sethttp://docs.julialang.org/en/latest/stdlib/base/#Base.Set, should be used for sparse collections. Agreed. Of course, this was just a toy example to test the limits of IntSet. In the real application that I am working towards, I want to think about systems of size at least 10^5 x 10^5. Mapping pairs (x,y) in this system to a single number gives up to 10^10, which is what I was testing. Construct a sorted set of the integers generated by the given iterable object, or an empty set. Implemented as a bit string, and therefore designed for dense integer sets. If the set will be sparse (for example holding a single very large integer), use Set instead. Do you happen to know a nice limit to how much memory IntSet should be allowed to use? In this case, it seems to be using more memory than I have available on my machine (4GB on my laptop). I guess my point is that normally I would expect that to give me an out-of-memory error, rather than enter an infinite loop producing garbage. On my laptop 100 MB would be more than I can afford Not sure what you mean by that -- doesn't it rather depend on the application? If I am doing a heavy computation on my laptop over night, I am happy for it to use all available memory. , but that would make IntSet unusable for bigger calculations on bigger systems, so it should be no smaller than 10 GB. I have another machine with a lot of memory (128 GB), so I certainly do not want to impose an arbitrary restriction. David. Ivar kl. 15:16:33 UTC+1 fredag 28. februar 2014 skrev David P. Sanders følgende: I am investigating possible data structures for an application. Here is an interesting behaviour in IntSet, which is no doubt to do with the implementation. Maybe it should just throw an exception if someone tries to add a really large integer like this! julia s = IntSet() IntSet() julia push!(s, 10) IntSet(10) julia sizeof(s) 24 julia push!(s, 100) IntSet(10, 100) julia sizeof(s) 24 julia push!(s, 1000) IntSet(10, 100, 1000) julia push!(s, 1) IntSet(10, 100, 1000, 1) julia push!(s, 10) IntSet(10, 100, 1000, 1, 10) julia sizeof(s) 24 julia push!(s, 100) IntSet(10, 100, 1000, 1, 10, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408, 1410065408^CEvaluation succeeded, but an error occurred while showing value of type IntSet: ERROR: interrupt in show at intset.jl:172 in anonymous at show.jl:973 in showlimited at show.jl:972 in writemime at repl.jl:2 in display at multimedia.jl:117 in display at multimedia.jl:119 in display at multimedia.jl:151
[julia-users] Re: Iteration over a Set where elements may be deleted in the process -- bug?
El viernes, 28 de febrero de 2014 16:41:20 UTC-6, Pierre-Yves Gérardy
escribió:
To realize what you want to do, you can add the elements to be removed to
another set.
On each iteration, verify that the current element isn't already in the
set of elements to be removed, and after the loop, clean the first set
using the elements of the second one.
Ah, that's a neat solution, thanks!
David.
—Pierre-Yves
On Thursday, February 27, 2014 4:54:58 PM UTC+1, David P. Sanders wrote:
I need to iterate over a Set whose elements will be deleted in the
process.
However, some deleted elements are included in the iteration, e.g. the
element 4 in the code below.
Is this a bug, or am I misunderstanding?
Thanks.
julia s = Set(5, 3, 4, 6)
Set{Int64}({5,4,6,3})
julia for i in s
println(i=$i s=$s)
delete!(s, i-1)
println(now s=$s\n)
end
i=5 s=Set{Int64}({5,4,6,3})
now s=Set{Int64}({5,6,3})
i=4 s=Set{Int64}({5,6,3})
now s=Set{Int64}({5,6})
i=6 s=Set{Int64}({5,6})
now s=Set{Int64}({6})
[julia-users] Iteration over a Set where elements may be deleted in the process -- bug?
I need to iterate over a Set whose elements will be deleted in the process.
However, some deleted elements are included in the iteration, e.g. the
element 4 in the code below.
Is this a bug, or am I misunderstanding?
Thanks.
julia s = Set(5, 3, 4, 6)
Set{Int64}({5,4,6,3})
julia for i in s
println(i=$i s=$s)
delete!(s, i-1)
println(now s=$s\n)
end
i=5 s=Set{Int64}({5,4,6,3})
now s=Set{Int64}({5,6,3})
i=4 s=Set{Int64}({5,6,3})
now s=Set{Int64}({5,6})
i=6 s=Set{Int64}({5,6})
now s=Set{Int64}({6})
[julia-users] Re: Iteration over a Set where elements may be deleted in the process -- bug?
El jueves, 27 de febrero de 2014 09:54:58 UTC-6, David P. Sanders escribió:
I need to iterate over a Set whose elements will be deleted in the process.
However, some deleted elements are included in the iteration, e.g. the
element 4 in the code below.
According to the following, it is enough to check if the element is `in`
the Set.
It seems wrong that this should be necessary, though!
julia for i in s
println(i=$i s=$s $(i in s))
delete!(s, i-1)
println(now s=$s\n)
end
i=5 s=Set{Int64}({5,4,6,3}) true
now s=Set{Int64}({5,6,3})
i=4 s=Set{Int64}({5,6,3}) false
now s=Set{Int64}({5,6})
i=6 s=Set{Int64}({5,6}) true
now s=Set{Int64}({6})
Is this a bug, or am I misunderstanding?
Thanks.
julia s = Set(5, 3, 4, 6)
Set{Int64}({5,4,6,3})
julia for i in s
println(i=$i s=$s)
delete!(s, i-1)
println(now s=$s\n)
end
i=5 s=Set{Int64}({5,4,6,3})
now s=Set{Int64}({5,6,3})
i=4 s=Set{Int64}({5,6,3})
now s=Set{Int64}({5,6})
i=6 s=Set{Int64}({5,6})
now s=Set{Int64}({6})
Re: [julia-users] Re: Best data structure to remove elements one by one
El domingo, 23 de febrero de 2014 18:09:36 UTC-6, Jeff Bezanson escribió: This was just changed on master so that you'd do Set([i*2 for i in 1:5]). OK, that seems reasonable, thanks. On Sun, Feb 23, 2014 at 12:45 AM, David P. Sanders dpsa...@gmail.comjavascript: wrote: El viernes, 21 de febrero de 2014 16:03:19 UTC-6, Steven G. Johnson escribió: On Friday, February 21, 2014 9:02:48 AM UTC-5, David P. Sanders wrote: OK, I think I have answered my own question: a Set is the good structure. And to create a set from an array I can do something like s = Set([3, 4, 5]...) or s = Set([i*2 for i in 1:5]...) which would be the closest thing to a set comprehension? The closest thing that avoids the overhead of creating an array first would currently be a loop: s = Set(Int) for i in 1:5 push!(s, i*2) end OK, this is the conclusion I had reached, thanks. At some point, if (when?) https://github.com/JuliaLang/julia/issues/4470 gets added, it will be easy to add support for syntax like Set(Int, i*2 for i in 1:5), where the second argument implicitly creates an iterable type. Sounds good!
Re: [julia-users] List of indices of matrix satisfying certain criterion
El domingo, 23 de febrero de 2014 02:36:57 UTC-6, Mauro escribió:
On Sun, 2014-02-23 at 05:51, dpsa...@gmail.com javascript: wrote:
El viernes, 21 de febrero de 2014 08:25:00 UTC-6, Mauro escribió:
Given a matrix, which will be large (say 10^5 x 10^5), I need to
extract
the list of indices (i.e. the pairs (x,y) of positions) of those
places
in
the matrix
where the value stored satisfies a certain condition. For a minimal
example, the condition can just be that the value is greater than
0.5.
Give you a logical array which you can use for indexing:
r..5
and to get the linear indices
find(r..5)
Great, didn't know about find.
Maybe I should just rewrite everything thinking in terms of 1D arrays.
(This is something I have taught in several courses,
but for some reason it didn't occur to me!)
No need to rewrite in terms of 1D arrays. You can index both with a
linear index or with a 2D index (or higher D, depending on the array):
julia a = [1 2; 3 4]
2x2 Array{Int64,2}:
1 2
3 4
julia a[2]
3
julia a[2,1]
3
If you want to translate between the two use ind2sub and sub2ind:
julia ind = ind2sub(size(a), 2)
(2,1)
julia a[ind...]
3
Excellent, thanks!
[julia-users] Re: Best data structure to remove elements one by one
El viernes, 21 de febrero de 2014 16:03:19 UTC-6, Steven G. Johnson escribió: On Friday, February 21, 2014 9:02:48 AM UTC-5, David P. Sanders wrote: OK, I think I have answered my own question: a Set is the good structure. And to create a set from an array I can do something like s = Set([3, 4, 5]...) or s = Set([i*2 for i in 1:5]...) which would be the closest thing to a set comprehension? The closest thing that avoids the overhead of creating an array first would currently be a loop: s = Set(Int) for i in 1:5 push!(s, i*2) end OK, this is the conclusion I had reached, thanks. At some point, if (when?) https://github.com/JuliaLang/julia/issues/4470 gets added, it will be easy to add support for syntax like Set(Int, i*2 for i in 1:5), where the second argument implicitly creates an iterable type. Sounds good!
Re: [julia-users] Best data structure for a dynamic array
El viernes, 21 de febrero de 2014 09:57:10 UTC-6, Kevin Squire escribió: It's hard to answer without more information. Is this related to your other, self-answered post, where you mention adding and deleting elements? Yes, it's a simulation I'm working on, which is a particular kind of dynamics in a random environment. I have a version in C++ but am converting it to Julia since it is so much more quick and flexible to write. If so, a Set is the better choice for anything but small arrays, as deleting from the middle of a vector is inefficient for large vectors. If insertion order is important, an OrderedSet (in the DataStructures package) might be useful. It will be slightly slower than a Set, but more efficient than deleting from an array. Great, thanks -- didn't know about the DataStructures package. It's often pretty hard to find my way around the julia ecosystem! Thanks, David But without more information, these might all be wrong (or inefficient) for your problem. :-) Cheers, Kevin On Friday, February 21, 2014, David P. Sanders dpsa...@gmail.comjavascript: wrote: If I were to use instead a Set, instead of a Vector, how would this impact performance?
[julia-users] Re: Best data structure to remove elements one by one
El viernes, 21 de febrero de 2014 07:49:27 UTC-6, David P. Sanders escribió: Hi, Suppose I have a collection of elements that I will create once at the beginining, visit in an unknown order and delete one by one when they are visited. What is the most efficient data structure for this in julia? I believe this would be an unordered_set (C++) or HashSet (java?), i.e. a Set implemented via a dictionary. (But I am not a computer scientist, so please correct me if I am wrong!) Should I in fact, then, just use a Set in julia? If, instead, I can add *and* remove elements in a random way, is the answer the same? OK, I think I have answered my own question: a Set is the good structure. And to create a set from an array I can do something like s = Set([3, 4, 5]...) or s = Set([i*2 for i in 1:5]...) which would be the closest thing to a set comprehension? (The ellipsis is necessary, otherwise a set with a single array object inside is created.) Thanks, David. Thanks, David.
[julia-users] Re: Best data structure for a dynamic array
El viernes, 21 de febrero de 2014 08:23:56 UTC-6, David P. Sanders escribió:
Hi,
I am having trouble finding the best equivalent of a list in Python, and
the Pythonic idiom of creating an empty list and then adding elements one
by one with .append().
For example, how would I do this for a collection of ordered pairs (Int64,
Int64)? I don't seem able even to create the correct type:
julia v = Vector{(Int64, Int64)}()
ERROR: type cannot be constructed
Then I want to do something like
push!(v, (1, 2))
Thanks,
David.
I found a similar question which I think gives the (or at least an) answer:
julia v = (Int64, Int64)[]
0-element Array{(Int64,Int64),1}
julia push!(v, (1,2))
1-element Array{(Int64,Int64),1}:
(1,2)
Sorry for the noise.
[julia-users] Re: Best data structure for a dynamic array
El viernes, 21 de febrero de 2014 08:26:52 UTC-6, Mauro escribió:
Yes, Julia vectors are basically like python lists.
OK, great, thanks!
David.
julia v=(Int,Int)[]
0-element Array{(Int64,Int64),1}
julia push!(v,(1,2))
1-element Array{(Int64,Int64),1}:
(1,2)
On Friday, February 21, 2014 2:23:56 PM UTC, David P. Sanders wrote:
Hi,
I am having trouble finding the best equivalent of a list in Python, and
the Pythonic idiom of creating an empty list and then adding elements one
by one with .append().
For example, how would I do this for a collection of ordered pairs
(Int64, Int64)? I don't seem able even to create the correct type:
julia v = Vector{(Int64, Int64)}()
ERROR: type cannot be constructed
Then I want to do something like
push!(v, (1, 2))
Thanks,
David.
[julia-users] Re: Best data structure for a dynamic array
If I were to use instead a Set, instead of a Vector, how would this impact performance?
Re: [julia-users] Re: Re: Use an array as the index for selecting an element from another array
On Tue, Feb 18, 2014 at 8:15 PM, Eric Davies iam...@gmail.com wrote: FYI: julia function test_index() a = rand(100,100); pos = [1,1]; s = 0.0 for i = 1:1 s += a[pos[1], pos[2]] end end test_index (generic function with 1 method) julia function test_tuple() a = rand(100,100); pos = (1,1); s = 0.0 for i = 1:1 s += a[pos...] end end test_tuple (generic function with 1 method) julia test_splat(); test_normal(); test_index(); test_tuple(); julia @time test_splat() elapsed time: 0.004122959 seconds (2406892 bytes allocated) julia @time test_normal() elapsed time: 3.1479e-5 seconds (80128 bytes allocated) julia @time test_index() elapsed time: 4.3435e-5 seconds (80192 bytes allocated) julia @time test_tuple() elapsed time: 3.5078e-5 seconds (80128 bytes allocated) So if you know the number of dimensions you're indexing, you can gain a lot. Alternatively, if you have the option of using a tuple you can save a lot there too. This may have to do with the fact that the type of the tuple describes its length. On Tuesday, 18 February 2014 12:31:52 UTC-6, Aditya Mahajan wrote: On Mon, 17 Feb 2014, Stefan Karpinski wrote: r[pos...] will do it. This is considerable slower than normal array access. ~~~ function test_splat() a = rand(100,100); pos = [1,1]; # Just a silly test s = 0.0 for i = 1:1 s += a[pos...] end end function test_normal() a = rand(100,100); # Just a silly test s = 0.0 for i = 1:1 s += a[1,1] end end # Run both functions once test_splat() test_normal() @time test_splat() @time test_normal() Gives elapsed time: 0.006395408 seconds (2406892 bytes allocated) elapsed time: 1.5434e-5 seconds (80128 bytes allocated) Aditya Many thanks, very interesting. I was indeed worried about the relative times. Best, David.
[julia-users] Use an array as the index for selecting an element from another array
Hi, Suppose I have the following: r = rand(5, 5) I can select a single element of the array r using r[3, 4] Now suppose that I have the position [3, 4] stored in a variable as pos = [3, 4] How can I use pos as the index for selecting the *single* element r[3, 4]? I would like to do r[pos], but this returns a *two*-element array, given by the elements numbered 3 and 4 in the linear ordering of the array r. This sounds like a silly question, but I can't find the answer! Thanks, David.
[julia-users] Re: Use an array as the index for selecting an element from another array
El lunes, 17 de febrero de 2014 18:57:10 UTC-6, David P. Sanders escribió: Hi, Suppose I have the following: r = rand(5, 5) I can select a single element of the array r using r[3, 4] Now suppose that I have the position [3, 4] stored in a variable as pos = [3, 4] How can I use pos as the index for selecting the *single* element r[3, 4]? Of course, one solution is r[pos[1], pos[2]] but this is not suitable for a multi-dimensional array. I would like to do r[pos], but this returns a *two*-element array, given by the elements numbered 3 and 4 in the linear ordering of the array r. This sounds like a silly question, but I can't find the answer! Thanks, David.
[julia-users] Re: Use an array as the index for selecting an element from another array
El lunes, 17 de febrero de 2014 19:03:55 UTC-6, Patrick O'Leary escribió: On Monday, February 17, 2014 6:57:10 PM UTC-6, David P. Sanders wrote: Hi, Suppose I have the following: r = rand(5, 5) I can select a single element of the array r using r[3, 4] Now suppose that I have the position [3, 4] stored in a variable as pos = [3, 4] How can I use pos as the index for selecting the *single* element r[3, 4]? I would like to do r[pos], but this returns a *two*-element array, given by the elements numbered 3 and 4 in the linear ordering of the array r. You're looking for r[pos...]. The ellipsis operator will splat the elements of pos as a comma-separated list of arguments. Many thanks for the 3 super-fast replies! r[pos...] is indeed exactly what I was looking for, thanks. Maybe this could be included in the array docs? Best, David.
