I am looking for a function that does the following rending:
f("2") = 2²
f("15") = 2¹⁵
I tried f(s) = "2\^($s)" but this doesn't seem to be a valid exponent as I can't TAB.
You can try e.g.:
julia> function f(s::AbstractString)
codes = Dict(collect("1234567890") .=> collect("¹²³⁴⁵⁶⁷⁸⁹⁰"))
return "2" * map(c -> codes[c], s)
end
f (generic function with 1 method)
julia> f("2")
"2²"
julia> f("15")
"2¹⁵"
(I have not optimized it for speed, but I hope this is fast enough with the benefit of being easy to read the code)
this should be a little faster, and uses replace:
function exp2text(x)
two = '2'
exponents = ('⁰', '¹', '²', '³', '⁴', '⁵', '⁶', '⁷', '⁸', '⁹')
#'⁰':'⁹' does not contain the ranges
exp = replace(x,'0':'9' =>i ->exponents[Int(i)-48+1])
#Int(i)-48+1 returns the number of the character if the character is a number
return two * exp
end
in this case, i used the fact that replace can accept a Pair{collection,function} that does:
if char in collection
replace(char,function(char))
end
Related
Suppose I have this function, implemented without StaticArrays
function example_svector_bad(G)
vector_list = [ randn(G) for q in 1:1000]
for i in size(vector_list)
for g in 1:G
vector_list[i][g] = vector_list[i][g] * g
end
end
return vector_list
end
I'm hoping to implement it using StaticArrays for speed gains. However, I don't know how to do it without losing the flexibility of specifying G. For example, I could do
function example_svector()
vector_list = [#SVector randn(3) for q in 1:1000]
for i in size(vector_list)
vector_list[i] = SVector(vector_list[i][1] * 1, vector_list[i][1] * 2,
vector_list[i][1] * 3)
end
return vector_list
end
if I knew that G = 3 and I had to write out SVector(vector_list[i][1] * 1, vector_list[i][1] * 2, vector_list[i][1] * 3).
Is there a way to implement this for any arbitrary number of G?
The size of a static vector or array must be known at the compile time.
At the compile time only types are known (rather than values).
Hence your function could look like this:
function myRandVec(::Val{G}) where G
SVector{G}(rand(G))
end
Note that G is passed as type rather than as value and hence can be used to create a static vector.
This function could be used as:
julia> myRandVec(Val{2}())
2-element SVector{2, Float64} with indices SOneTo(2):
0.7618992223709563
0.5979657793050613
Firstly, there is a mistake in how you are indexing vector_list, where you do
for i in size(vector_list)
Let's see what that does:
julia> x = 1:10;
julia> size(x)
(10,)
The size of x is its length in each dimension, for a vector that is just (10,) since it has only one dimension. Let's try iterating:
julia> for i in size(x)
println(i)
end
10
It just prints out the number 10.
You probably meant
for i in 1:length(vector_list)
but it's better to write
for i in eachindex(vector_list)
since it is more general and safer.
As for your actual question, you can use StaticArrays.SOneTo which provides a static version of [1,2,3]:
function example_svector()
vector_list = [#SVector randn(3) for q in 1:1000]
N = length(eltype(vector_list))
c = SOneTo(N)
for i in eachindex(vector_list)
vector_list[i] = vector_list[i] .* c
end
return vector_list
end
The code at the end of this post constructs a function which is bound to the variables of a given dictionary. Furthermore, the function is not bound to the actual name of the dictionary (as I use the Ref() statement).
An example:
julia> D = Dict(:x => 4, :y => 5)
julia> f= #mymacro4(x+2y, D)
julia> f()
14
julia> DD = D
julia> D = nothing
julia> f()
14
julia> DD[:x] = 12
julia> f()
22
Now I want to be able to construct exactly the same function when I only have access to the expression expr = :(x+2y).
How do I do this? I tried several things, but was not able to find a solution.
julia> f = #mymacro4(:(x+2y), D)
julia> f() ### the function evaluation should also yield 14. But it yields:
:(DR.x[:x] + 2 * DR.x[:y])
(I actually want to use it within another macro in which the dictionary is automatically created. I want to store this dictionary and the function within a struct, such that I'm able to call this function at a later point in time and manipulate the objects in the dictionary. If necessary, I may post the complete example and explain the complete problem.)
_freevars2(literal) = literal
function _freevars2(s::Symbol)
try
if typeof(eval(s)) <: Function
return s
else
return Meta.parse("DR.x[:$s]")
end
catch
return Meta.parse("DR.x[:$s]")
end
end
function _freevars2(expr::Expr)
for (it, s) in enumerate(expr.args)
expr.args[it] = _freevars2(s)
end
return expr
end
macro mymacro4(expr, D)
expr2 = _freevars2(expr)
quote
let DR = Ref($(esc(D)))
function mysym()
$expr2
end
end
end
end
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)
How do I evaluate the function in only one of its variables, that is, I hope to obtain another function after evaluating the function. I have the following piece of code.
deff ('[F] = fun (x, y)', 'F = x ^ 2-3 * y ^ 2 + x * y ^ 3');
fun (4, y)
I hope to get 16-3y ^ 2 + 4y ^ 3
If what you want to do is to write x = f(4,y), and later just do x(2) to get -36, that is called partial application:
Intuitively, partial function application says "if you fix the first arguments of the function, you get a function of the remaining arguments".
This is a very useful feature, and very common Functional Programming Languages, such as Haskell, but even JS and Python now are able to do it. It is also possible to do this in MATLAB and GNU/Octave using anonymous functions (see this answer). In Scilab, however, this feature is not available.
Workround
Nonetheless, Scilab itself uses a workarounds to carry a function with its arguments without fully evaluating. You see this being used in ode(), fsolve(), optim(), and others:
Create a list containing the function and the arguments to partial evaluation: list(f,arg1,arg2,...,argn)
Use another function to evaluate such list and the last argument: evalPartList(list(...),last_arg)
The implementation of evalPartList() can be something like this:
function y = evalPartList(fList,last_arg)
//fList: list in which the first element is a function
//last_arg: last argument to be applied to the function
func = fList(1); //extract function from the list
y = func(fList(2:$),last_arg); //each element of the list, from second
//to last, becomes an argument
endfunction
You can test it on Scilab's console:
--> deff ('[F] = fun (x, y)', 'F = x ^ 2-3 * y ^ 2 + x * y ^ 3');
--> x = list(fun,4)
x =
x(1)
[F]= x(1)(x,y)
x(2)
4.
--> evalPartList(x,2)
ans =
36.
This is a very simple implementation for evalPartList(), and you have to be careful not to exceed or be short on the number of arguments.
In the way you're asking, you can't.
What you're looking is called symbolic (or formal) computational mathematics, because you don't pass actual numerical values to functions.
Scilab is numerical software so it can't do such thing. But there is a toolbox scimax (installation guide) that rely on a the free formal software wxmaxima.
BUT
An ugly, stupid but still sort of working solution is to takes advantages of strings :
function F = fun (x, y) // Here we define a function that may return a constant or string depending on the input
fmt = '%10.3E'
if (type(x)==type('')) & (type(y)==type(0)) // x is string is
ys = msprintf(fmt,y)
F = x+'^2 - 3*'+ys+'^2 + '+x+'*'+ys+'^3'
end
if (type(y)==type('')) & (type(x)==type(0)) // y is string so is F
xs = msprintf(fmt,x)
F = xs+'^2 - 3*'+y+'^2 + '+xs+'*'+y+'^3'
end
if (type(y)==type('')) & (type(x)==type('')) // x&y are strings so is F
F = x+'^2 - 3*'+y+'^2 + '+x+'*'+y+'^3'
end
if (type(y)==type(0)) & (type(x)==type(0)) // x&y are constant so is F
F = x^2 - 3*y^2 + x*y^3
end
endfunction
// Then we can use this 'symbolic' function
deff('F2 = fun2(y)',' F2 = '+fun(4,'y'))
F2=fun2(2) // does compute fun(4,2)
disp(F2)
Im looking for a function like Pythons
"foobar, bar, foo".count("foo")
Could not find any functions that seemed able to do this, in a obvious way. Looking for a single function or something that is not completely overkill.
Julia-1.0 update:
For single-character count within a string (in general, any single-item count within an iterable), one can use Julia's count function:
julia> count(i->(i=='f'), "foobar, bar, foo")
2
(The first argument is a predicate that returns a ::Bool).
For the given example, the following one-liner should do:
julia> length(collect(eachmatch(r"foo", "bar foo baz foo")))
2
Julia-1.7 update:
Starting with Julia-1.7 Base.Fix2 can be used, through ==('f') below, as to shorten and sweeten the syntax:
julia> count(==('f'), "foobar, bar, foo")
2
What about regexp ?
julia> length(matchall(r"ba", "foobar, bar, foo"))
2
I think that right now the closest built-in thing to what you're after is the length of a split (minus 1). But it's not difficult to specifically create what you're after.
I could see a searchall being generally useful in Julia's Base, similar to matchall. If you don't care about the actual indices, you could just use a counter instead of growing the idxs array.
function searchall(s, t; overlap::Bool=false)
idxfcn = overlap ? first : last
r = findnext(s, t, firstindex(t))
idxs = typeof(r)[] # Or to only count: n = 0
while r !== nothing
push!(idxs, r) # n += 1
r = findnext(s, t, idxfcn(r) + 1)
end
idxs # return n
end
Adding an answer to this which allows for interpolation:
julia> a = ", , ,";
julia> b = ",";
julia> length(collect(eachmatch(Regex(b), a)))
3
Actually, this solution breaks for some simple cases due to use of Regex. Instead one might find this useful:
"""
count_flags(s::String, flag::String)
counts the number of flags `flag` in string `s`.
"""
function count_flags(s::String, flag::String)
counter = 0
for i in 1:length(s)
if occursin(flag, s)
s = replace(s, flag=> "", count=1)
counter+=1
else
break
end
end
return counter
end
Sorry to post another answer instead of commenting previous one, but i've not managed how to deal with code blocks in comments :)
If you don't like regexps, maybe a tail recursive function like this one (using the search() base function as Matt suggests) :
function mycount(what::String, where::String)
function mycountacc(what::String, where::String, acc::Int)
res = search(where, what)
res == 0:-1 ? acc : mycountacc(what, where[last(res) + 1:end], acc + 1)
end
what == "" ? 0 : mycountacc(what, where, 0)
end
This is simple and fast (and does not overflow the stack):
function mycount2(where::String, what::String)
numfinds = 0
starting = 1
while true
location = search(where, what, starting)
isempty(location) && return numfinds
numfinds += 1
starting = location.stop + 1
end
end
one liner: (Julia 1.3.1):
julia> sum([1 for i = eachmatch(r"foo", "foobar, bar, foo")])
2
Since Julia 1.3, there has been a count method that does exactly this.
count(
pattern::Union{AbstractChar,AbstractString,AbstractPattern},
string::AbstractString;
overlap::Bool = false,
)
Return the number of matches for pattern in string.
This is equivalent to calling length(findall(pattern, string)) but more
efficient.
If overlap=true, the matching sequences are allowed to overlap indices in the
original string, otherwise they must be from disjoint character ranges.
│ Julia 1.3
│
│ This method requires at least Julia 1.3.
julia> count("foo", "foobar, bar, foo")
2
julia> count("ana", "bananarama")
1
julia> count("ana", "bananarama", overlap=true)
2