Is there a way to pass a vector variable as a command line argument in Julia? In my case I have two arguments: an integer and a vector of integers. While I can easily parse the first argument, I didn't find any pleasant way to parse a vector. For now I simply set the vector to be v = parse.(Int, ARGS[2:end]) but it is quite confusing since the items of the vector are treated as arguments. Is there a some special syntax to treat such cases?
I think your current solution is fine as it is, and in line with the way many command-line tools do things.
If you really want to pass your whole array as one command-line argument, you'll have to:
somehow make sure it is correctly parsed as one argument by the shell
parse it as an array within Julia
Both steps vary depending on the syntax you want to use.
Two examples:
example 1: Julia-like syntax
shell$ julia myscript.jl 42 "[1,2,3]"
i = 42
v = [1, 2, 3]
We can take advantage of the Julia parser being able to parse such arrays (but let's be cautious about not evaluating arbitrary julia code input by the user):
# First argument: an integer
i = parse(Int, ARGS[1])
# Second argument, a Vector{Int} in Julia-like format: "[1, 2, 3]"
v = let expr = Meta.parse(ARGS[2])
#assert expr.head == :vect
Int.(expr.args)
end
#show i
#show v
Example 2: space- or comma-separated values
shell$ julia myscript.jl 42 "1,2,3"
i = 42
v = [1, 2, 3]
Here, we can use DelimitedFiles to parse the array (change the delimiter to whatever you like):
# First argument: an integer
i = parse(Int, ARGS[1])
# Second argument, a Vector{Int} as comma-separated values
using DelimitedFiles
v = reshape(readdlm(IOBuffer(ARGS[2]), ',', Int), :)
# set the delimiter here ^
#show i
#show v
Related
In Julia it is possible to create arrays of any size using the functions zeros(.) or ones(.). Is there a similar function to create an array that is filled with nothing at initialization but also accepts floats? I mean a function like in this example:
a = array_of_nothing(3)
# a = [nothing,nothing,nothing]
a[1] = 3.14
# a = [3.14,nothing,nothing]
I tried to find information on internet, but without success... Sorry, I am a beginner in Julia.
The fill function can be used to create arrays of arbitrary values, but it's not so easy to use here, since you want a Vector{Union{Float64, Nothing}}. Two options come to mind:
A comprehension:
a = Union{Float64, Nothing}[nothing for _ in 1:3];
a[2] = 3.14;
>> a
3-element Array{Union{Nothing, Float64},1}:
nothing
3.14
nothing
Or ordinary array initialization:
a = Vector{Union{Float64, Nothing}}(undef, 3)
fill!(a, nothing)
a[2] = 3.14
It seems that when you do Vector{Union{Float64, Nothing}}(undef, 3) the vector automatically contains nothing, but I wouldn't rely on that, so fill! may be necessary.
I think you are looking for the Base.fill — Function.
fill(x, dims)
This creates an array filled with value x.
println(fill("nothing", (1,3)))
You can also pass a function Foo() like fill(Foo(), dims) which will return an array filled with the result of evaluating Foo() once.
I have the following code:
# install.packages("jsonlite")
require("jsonlite")
x = list(
test = "my_test",
data = c(1, 2, 3)
)
toJSON(x)
This prints:
{"test":["my_test"],"data":[1,2,3]}
I was expecting:
{"test":"my_test","data":[1,2,3]}
I've tried using some of the parameters from the documentation, but can't seem to get it right.
The argument auto_unbox=TRUE did the trick:
automatically unbox all atomic vectors of length 1. It is usually safer to avoid this and instead use the unbox function to unbox individual elements. An exception
is that objects of class AsIs (i.e. wrapped in I()) are not automatically unboxed. This is a way to mark single values as length-1 arrays.
I.e., the solution was toJSON(x, auto_unbox=TRUE), which returns what I expected:
{"test":"my_test","data":[1,2,3]}
I would like to create an empty vector and append to it an array in Julia. How do I do that?
x = Vector{Float64}
append!(x, rand(10))
results in
`append!` has no method matching append!(::Type{Array{Float64,1}}, ::Array{Float64,1})
Thanks.
Your variable x does not contain an array but a type.
x = Vector{Float64}
typeof(x) # DataType
You can create an array as Array(Float64, n)
(but beware, it is uninitialized: it contains arbitrary values) or zeros(Float64, n),
where n is the desired size.
Since Float64 is the default, we can leave it out.
Your example becomes:
x = zeros(0)
append!( x, rand(10) )
I am somewhat new to Julia and came across this question after getting a similar error. To answer the original question for Julia version 1.2.0, all that is missing are ():
x = Vector{Float64}()
append!(x, rand(10))
This solution (unlike x=zeros(0)) works for other data types, too. For example, to create an empty vector to store dictionaries use:
d = Vector{Dict}()
push!(d, Dict("a"=>1, "b"=>2))
A note regarding use of push! and append!:
According to the Julia help, push! is used to add individual items to a collection, while append! adds an collection of items to a collection. So, the following pieces of code create the same array:
Push individual items:
a = Vector{Float64}()
push!(a, 1.0)
push!(a, 2.0)
Append items contained in an array:
a = Vector{Float64}()
append!(a, [1.0, 2.0])
You can initialize an empty Vector of any type by typing the type in front of []. Like:
Float64[] # Returns what you want
Array{Float64, 2}[] # Vector of Array{Float64,2}
Any[] # Can contain anything
New answer, for Julia 1. append! is deprecated, you now need to use push!(array, element) to add elements to an array
my_stuff = zeros()
push!(my_stuff, "new element")
I have:
z = data.frame(x1=a, x2=b, x3=c, etc)
I am trying to do:
for (i in 1:10)
{
paste(c('N'),i,sep="") -> paste(c('z$x'),i,sep="")
}
Problems:
paste(c('z$x'),i,sep="") yields "z$x1", "z$x1" instead of calling the actual values. I need the expression to be evaluated. I tried as.numeric, eval. Neither seemed to work.
paste(c('N'),i,sep="") yields "N1", "N2". I need the expression to be merely used as name. If I try to assign it a value such as paste(c('N'),5,sep="") -> 5, ie "N5" -> 5 instead of N5 -> 5, I get target of assignment expands to non-language object.
This task is pretty trivial since I can simply do:
N1 = x1...
N2 = x2...
etc, but I want to learn something new
I'd suggest using something like for( i in 1:10 ) z[,i] <- N[,i]...
BUT, since you said you want to learn something new, you can play around with parse and substitute.
NOTE: these little tools are funny, but experienced users (not me) avoid them.
This is called "computing on the language". It's very interesting, and it helps understanding the way R works. Let me try to give an intro:
The basic language construct is a constant, like a numeric or character vector. It is trivial because it is not different from its "unevaluated" version, but it is one of the building blocks for more complicated expressions.
The (officially) basic language object is the symbol, also known as a name. It's nothing but a pointer to another object, i.e., a token that identifies another object which may or may not exist. For instance, if you run x <- 10, then x is a symbol that refers to the value 10. In other words, evaluating the symbol x yields the numeric vector 10. Evaluating a non-existant symbol yields an error.
A symbol looks like a character string, but it is not. You can turn a string into a symbol with as.symbol("x").
The next language object is the call. This is a recursive object, implemented as a list whose elements are either constants, symbols, or another calls. The first element must not be a constant, because it must evaluate to the real function that will be called. The other elements are the arguments to this function.
If the first argument does not evaluate to an existing function, R will throw either Error: attempt to apply non-function or Error: could not find function "x" (if the first argument is a symbol that is undefined or points to something other than a function).
Example: the code line f(x, y+z, 2) will be parsed as a list of 4 elements, the first being f (as a symbol), the second being x (another symbol), the third another call, and the fourth a numeric constant. The third element y+z, is just a function with two arguments, so it parses as a list of three names: '+', y and z.
Finally, there is also the expression object, that is a list of calls/symbols/constants, that are meant to be evaluated one by one.
You'll find lots of information here:
https://github.com/hadley/devtools/wiki/Computing-on-the-language
OK, now let's get back to your question :-)
What you have tried does not work because the output of paste is a character string, and the assignment function expects as its first argument something that evaluates to a symbol, to be either created or modified. Alternativelly, the first argument can also evaluate to a call associated with a replacement function. These are a little trickier, but they are handled by the assignment function itself, not by the parser.
The error message you see, target of assignment expands to non-language object, is triggered by the assignment function, precisely because your target evaluates to a string.
We can fix that building up a call that has the symbols you want in the right places. The most "brute force" method is to put everything inside a string and use parse:
parse(text=paste('N',i," -> ",'z$x',i,sep=""))
Another way to get there is to use substitute:
substitute(x -> y, list(x=as.symbol(paste("N",i,sep="")), y=substitute(z$w, list(w=paste("x",i,sep="")))))
the inner substitute creates the calls z$x1, z$x2 etc. The outer substitute puts this call as the taget of the assignment, and the symbols N1, N2 etc as the values.
parse results in an expression, and substitute in a call. Both can be passed to eval to get the same result.
Just one final note: I repeat that all this is intended as a didactic example, to help understanding the inner workings of the language, but it is far from good programming practice to use parse and substitute, except when there is really no alternative.
A data.frame is a named list. It usually good practice, and idiomatically R-ish not to have lots of objects in the global environment, but to have related (or similar) objects in lists and to use lapply etc.
You could use list2env to multiassign the named elements of your list (the columns in your data.frame) to the global environment
DD <- data.frame(x = 1:3, y = letters[1:3], z = 3:1)
list2env(DD, envir = parent.frame())
## <environment: R_GlobalEnv>
## ta da, x, y and z now exist within the global environment
x
## [1] 1 2 3
y
## [1] a b c
## Levels: a b c
z
## [1] 3 2 1
I am not exactly sure what you are trying to accomplish. But here is a guess:
### Create a data.frame using the alphabet
data <- data.frame(x = 'a', y = 'b', z = 'c')
### Create a numerical index corresponding to the letter position in the alphabet
index <- which(tolower(letters[1:26]) == data[1, ])
### Use an 'lapply' to apply a function to every element in 'index'; creates a list
val <- lapply(index, function(x) {
paste('N', x, sep = '')
})
### Assign names to our list
names(val) <- names(data)
### Observe the result
val$x
What I want to do is simple:
collection = {'a':[], 'b':[], 'c':[]}
values = [1,2,3]
I want to make a function that produces the following: (append the values into the list element of the dictionary, the dic and the list are the same length)
{'a':[1], 'b':[2], 'c':[3]}
This is simple enough and I can do it using, a couple of for x in. But I want to do this in one line. (using two loops in the same line) and I can not get the syntax to work.
I have tried some things similar to this, but they all result in syntax error:
collection[c].append(value), for c in d.iteritems(), for value in values
You can't do what you want to do on one line. You can create a new dictionary on one line though:
{k: collection[k] + [v] for k, v in zip(collection.keys(), values)}
Result:
>>> collection = {'a':[], 'b':[], 'c':[]}
>>> values = [1,2,3]
>>> {k: collection[k] + [v] for k, v in zip(collection.keys(), values)}
{'a': [1], 'c': [2], 'b': [3]}
This is called a dict comprehension. Like a list comprehension and a generator expression, you can use multiple loops in one of those, but you don't need one here. zip() will pair up the keys from collection with the integers from values.
To modify a dict in-place, you'll have to use 2 lines at least:
for k, v in zip(collection.keys(), values):
collection[k].append(v)
Python does accept that on one line, but that goes against just about every styleguide I can look up for you:
for k, v in zip(collection.keys(), values): collection[k].append(v)
Python throws a syntax error because it interprets your line as a tuple of expressions (the commas make it a tuple), and two of your expressions are for statements, which cannot be used in an expression.