I am looking for a simple way to extract the extension (.txt, .py, .jl, etc) from a file in Julia. I looked through the Julia docs but didn't see anything specifically built for this.
Use splitext from Filesystem
julia> splitext("/home/myuser/example.jl")
("/home/myuser/example", ".jl")
julia> splitext("/home/myuser/example")
("/home/myuser/example", "")
While there is no built in Julia function to do this, what you can do is given the filename, split the string on a period "." as follows:
julia> filePath = "some/path/my_program.jl"
"some/path/my_program.jl"
julia> split(filePath, ".")[2]
"jl"
This is a faster alternative, if that is of any importance:
file_extension(file::String) = file[findlast(==('.'), file)+1:end]
Related
The result variable is a json-type string, which is very long. What is the option in Julia REPL that allows only a limited output when the variable is this long? DataFrame is originally only partially output. I hope that the general variables will also be output like that.
You can overwrite the display method for AbstractStrings:
import Main.display
display(x::AbstractString) =
show(length(x)<=50 ? x : SubString(x,1,50)*"…")
Let us test it:
julia> str = join(rand('a':'z', 200))
"wcbifwzglgqyenrcdgdxagohlwdoxrrumoaltklkjauptwzrmi…"
What is the sane way to go from a Module object to a path to the file in which it was declared?
To be precise, I am looking for the file where the keyword module occurs.
The indirect method is to find the location of the automatically defined eval method in each module.
moduleloc(mm::Module) = first(functionloc(mm.eval, (Symbol,)))
for example
moduleloc(mm::Module) = first(functionloc(mm.eval, (Symbol,)))
using DataStructures
moduleloc(DataStructures)
Outputs:
/home/oxinabox/.julia/v0.6/DataStructures/src/DataStructures.jl
This indirect method works, but it feels like a bit of a kludge.
Have I missed some inbuilt function to do this?
I will remind answered that Modules are not the same thing as packages.
Consider the existence of submodules, or even modules that are being loaded via includeing some abolute path that is outside the package directory or loadpath.
Modules simply do not store the file location where they were defined. You can see that for yourself in their definition in C. Your only hope is to look through the bindings they hold.
Methods, on the other hand, do store their file location. And eval is the one function that is defined in every single module (although not baremodules). Slightly more correct might be:
moduleloc(mm::Module) = first(functionloc(mm.eval, (Any,)))
as that more precisely mirrors the auto-defined eval method.
If you aren't looking for a programmatic way of doing it you can use the methods function.
using DataFrames
locations = methods(DataFrames.readtable).ms
It's for all methods but it's hardly difficult to find the right one unless you have an enormous number of methods that differ only in small ways.
There is now pathof:
using DataStructures
pathof(DataStructures)
"/home/ederag/.julia/packages/DataStructures/59MD0/src/DataStructures.jl"
See also: pkgdir.
pkgdir(DataStructures)
"/home/ederag/.julia/packages/DataStructures/59MD0"
Tested with julia-1.7.3
require obviously needs to perform that operation. Looking into loading.jl, I found that finding the module path has changed a bit recently: in v0.6.0, there is a function
load_hook(prefix::String, name::String, ::Void)
which you can call "manually":
julia> Base.load_hook(Pkg.dir(), "DataFrames", nothing)
"/home/philipp/.julia/v0.6/DataFrames/src/DataFrames.jl"
However, this has changed to the better in the current master; there's now a function find_package, which we can copy:
macro return_if_file(path)
quote
path = $(esc(path))
isfile(path) && return path
end
end
function find_package(name::String)
endswith(name, ".jl") && (name = chop(name, 0, 3))
for dir in [Pkg.dir(); LOAD_PATH]
dir = abspath(dir)
#return_if_file joinpath(dir, "$name.jl")
#return_if_file joinpath(dir, "$name.jl", "src", "$name.jl")
#return_if_file joinpath(dir, name, "src", "$name.jl")
end
return nothing
end
and add a little helper:
find_package(m::Module) = find_package(string(module_name(m)))
Basically, this takes Pkg.dir() and looks in the "usual locations".
Additionally, chop in v0.6.0 doesn't take these additional arguments, which we can fix by adding
chop(s::AbstractString, m, n) = SubString(s, m, endof(s)-n)
Also, if you're not on Unix, you might want to care about the definitions of isfile_casesensitive above the linked code.
And if you're not so concerned about corner cases, maybe this is enough or can serve as a basis:
function modulepath(m::Module)
name = string(module_name(m))
Pkg.dir(name, "src", "$name.jl")
end
julia> Pkg.dir("DataStructures")
"/home/liso/.julia/v0.7/DataStructures"
Edit: I now realized that you want to use Module object!
julia> m = DataStructures
julia> Pkg.dir(repr(m))
"/home/liso/.julia/v0.7/DataStructures"
Edit2: I am not sure if you are trying to find path to module or to object defined in module (I hope that parsing path from next result is easy):
julia> repr(which(DataStructures.eval, (String,)))
"eval(x) in DataStructures at /home/liso/.julia/v0.7/DataStructures/src/DataStructures.jl:3"
I would like to retrieve the path of the currently running Julia interpreter from Julia. In Python, this can be achieved with sys.executable.
Base.julia_cmd() is probably what you need. It returns the full command line that was used to invoke the current julia process, with the default options spelled out. Base.julia_exename() returns the name of the executable.
julia> Base.julia_cmd()
/Users/aviks/dev/julia/julia5/usr/bin/julia -Cnative -J/usr/lib/julia/sys.dylib --compile=yes --depwarn=yes
julia> Base.julia_exename()
"julia"
If you just want the location of the julia executable, try one of these:
julia> julia_bin_exe = joinpath(Base.Sys.BINDIR,Base.julia_exename())
"/home/mkitti/src/julia/usr/bin/julia"
julia> Base.julia_cmd()
`/home/mkitti/src/julia/usr/bin/julia -Cnative -J/home/mkitti/src/julia/usr/lib/julia/sys.so -g1`
julia> typeof(Base.julia_cmd())
Cmd
julia> Base.julia_cmd()[1]
"/home/mkitti/src/julia/usr/bin/julia"
julia> julia_bin_exe == Base.julia_cmd()[1]
true
Given a function object f, how do I find:
Function's name.
Module(s) of its method(s)?
In Julia 0.4 I was able to find name using f.env.name, but no tips for module. For Julia 0.5 I wasn't able to find any of two.
Name is easy: Symbol(f) or string(f) if you want a string
Module is, as you know going to be per method (i.e per type signature).
methods(f) with return a method table that prints out all the methods and where they are, in terms of files.
You can do [meth.module for meth in methods(f)] to get there modules
So to use an example, the collect function.
julia> using DataStructures #so we have some non-Base definitions
julia> Symbol(collect)
:collect
julia> methods(collect)
# 5 methods for generic function "collect":
collect(r::Range) at range.jl:813
collect{T}(::Type{T}, itr) at array.jl:211
collect(itr::Base.Generator) at array.jl:265
collect{T}(q::DataStructures.Deque{T}) at /home/ubuntu/.julia/v0.5/DataStructures/src/deque.jl:170
collect(itr) at array.jl:236
julia> [meth.module for meth in methods(collect)]
5-element Array{Module,1}:
Base
Base
Base
DataStructures
Base
julia> first(methods(collect, (Deque,))).module
DataStructures
#oxinabox's answer is correct. To add, typeof(f).name.mt.name is the v0.5 replacement for f.env.name. That can be useful to avoid the . that occurs when just applying string to a function introduced in a non-stdlib module. There also exists Base.function_name(f) which is probably less likely to break when the Julia version changes.
To get the module that a function (type) is introduced in, rather than the modules of individual methods, there's typeof(f).name.module, or the probably-better version, Base.function_module(f). The module of the method table is probably the same; that can be obtained through typeof(f).name.mt.module.
Note that f.env in v0.4 is a direct equivalent of typeof(f).name.mt, so on v0.4 the same f.env.name and f.env.module apply.
In Julia 1.x the commands are:
Base.nameof
Base.parentmodule
Since this two methods are exported, also nameof and parentmodule works.
Not exactly the answer but closely related: you can find the file name and line number of f using functionloc(f)
Problem
I read in an array of strings from a file.
julia> file = open("word-pairs.txt");
julia> lines = readlines(file);
But Julia doesn't know that they're strings.
julia> typeof(lines)
Array{Any,1}
Question
Can I tell Julia this somehow?
Is it possible to insert type information onto a computed result?
It would be helpful to know the context where this is an issue, because there might be a better way to express what you need - or there could be a subtle bug somewhere.
Can I tell Julia this somehow?
No, because the readlines function explicitly creates an Any array (a = {}): https://github.com/JuliaLang/julia/blob/master/base/io.jl#L230
Is it possible to insert type information onto a computed result?
You can convert the array:
r = convert(Array{ASCIIString,1}, w)
Or, create your own readstrings function based on the link above, but using ASCIIString[] for the collection array instead of {}.
Isaiah is right about the limits of readlines. More generally, often you can say
n = length(A)::Int
when generic type inference fails but you can guarantee the type in your particular case.
As of 0.3.4:
julia> typeof(lines)
Array{Union(ASCIIString,UTF8String),1}
I just wanted to warn against:
convert(Array{ASCIIString,1}, lines)
that can fail (for non-ASCII) while I guess, in this case nothing needs to be done, this should work:
convert(Array{UTF8String,1}, lines)