I have a database of "formulas" stored as strings. Let's assume for simplicity, that each formula contains 2 variables denoted by a and b, and that the formulas are all wellformed and it is ensured that it consists only of characters from the set ()ab+-*.
At runtime, formulas are fetched from this database, and from another source, numeric values for a and b are fetched, and the formulas are evaluated. The evaluation can be programmed like this:
# This is how it works right now
formula = fetch_formula(....)
a = fetch_left_arg(....)
b = fetch_right_arg(....)
result = eval(formula)
This design works, but I'm not entirely happy with it. It requires that my program names the free variables exactly the same as they are named in the formula, which is ugly.
If my "formula" would not be a string, but a Proc object or Lambda which accepts two parameters, I could do something like
# No explicitly named variables
result = fetch_proc(...).call(fetch_left_arg(....),fetch_right_arg(....))
but unfortunately, the formulas have to be strings.
I tried to experiment in the following way: What if the method, which fetches the formula from the database, would wrap the string into something, which behaves like a block, and where I could pass parameters to it?
# This does not work of course, but maybe you get the idea:
block_string = "|a,b| #{fetch_formula(....)}"
Of course I can't eval such a block_string, but is there something similar which I could use? I know that instance_eval can pass parameters, but what object should I apply it to? So this is perhaps not an option either....
This is very nasty approach, but for simple formulas you’ve mentioned it should work:
▶ formula = 'a + b'
▶ vars = formula.scan(/[a-z]+/).uniq.join(',') # getting vars names
#⇒ "a,b"
▶ pr = eval("proc { |#{vars}| #{formula} }") # preparing proc
▶ pr.call 3, 5
#⇒ 8
Here we rely on the fact, that parameters are passed to the proc in the same order, as they appear in the formula.
If I get your question correctly, it is something that I have done recently, and is fairly easy. Given a string:
s = "{|x, y| x + y}"
You can create a proc by doing:
eval("Proc.new#{s}")
One way to avoid creating the variables in the local scope could be to use a Binding:
bind = binding
formula = fetch_formula(....)
bind.local_variable_set :a, fetch_left_arg(....)
bind.local_variable_set :b, fetch_right_arg(....)
result = bind.eval(formula)
The variables a and b now only exist in the binding, and do not pollute the rest of your code.
You can create a lambda from string, as shown below:
formula = "a + b"
lambda_template = "->(a,b) { %s }"
formula_lambda = eval(lambda_template % formula)
p formula_lambda.call(1,2)
#=> 3
Related
I have a three lines of code here like shown below
local = headers.zip(*data_rows).transpose
local = local[1..-1].map {|dataRow| local[0].zip(dataRow).to_h}
p local
Now if you watch the above three lines, I have to store the result of the first line in the variable called local since it would be used in two places in the second line as I have shown,So Can't I cascade the second line with first line anyway? I tried using tap like this
local = headers.zip(*data_rows).transpose.tap{|h|h[1..-1].map {|dataRow| h[0].zip(dataRow).to_h}}
tap is returning the self as explained in the document so can't I get the result final result when I use tab? Anyway other way to achieve this result in one single line so that I don't have to use local variable?
If you're on Ruby 2.5.0 or later, you can use yield_self for this.
local = headers.zip(*data_rows).transpose.yield_self { |h| h[1..-1].map { |dataRow| h[0].zip(dataRow).to_h } }
yield_self is similar to tap in that they both yield self to the block. The difference is in what is returned by each of the two methods.
Object#tap yields self to the block and then returns self. Kernel#yield_self yields self to the block and then returns the result of the block.
Here's an answer to a previous question where I gave a couple of further examples of where each of these method can be useful.
It's often helpful to execute working code with data, to better understand what is to be computed. Seeing transpose and zip, which are often interchangeable, used together, was a clue that a simplification might be possible (a = [1,2,3]; b = [4,5,6]; a.zip(b) => [[1, 4], [2, 5], [3, 6]] <= [a,b].transpose).
Here's my data:
headers=[1,2,3]
data_rows=[[11,12,13],[21,22,23],[31,32,33],[41,42,43]]
and here's what the working code returns:
local = headers.zip(*data_rows).transpose
local[1..-1].map {|dataRow| local[0].zip(dataRow).to_h}
#=> [{1=>11, 2=>12, 3=>13}, {1=>21, 2=>22, 3=>23},
# {1=>31, 2=>32, 3=>33}, {1=>41, 2=>42, 3=>43}]
It would seem that this might be computed more simply:
data_rows.map { |row| headers.zip(row).to_h }
#=> [{1=>11, 2=>12, 3=>13}, {1=>21, 2=>22, 3=>23},
# {1=>31, 2=>32, 3=>33}, {1=>41, 2=>42, 3=>43}]
I've read a number of articles on the difference between assignment and binding, but it hasn't clicked yet (specifically in the context of an imperative language vs one without mutation).
I asked in IRC, and someone mentioned these 2 examples illustrate the difference, but then I had to go and I didn't see the full explanation.
Can someone please explain how/why this works in a detailed way, to help illustrate the difference?
Ruby
x = 1; f = lambda { x }; x = 2; f.call
#=> 2
Elixir
x = 1; f = fn -> x end; x = 2; f.()
#=> 1
I've heard this explanation before and it seems pretty good:
You can think of binding as a label on a suitcase, and assignment as a
suitcase.
In other languages, where you have assignment, it is more like putting a value in a suitcase. You actually change value that is in the suitcase and put in a different value.
If you have a suitcase with a value in it, in Elixir, you put a label on it. You can change the label, but the value in the suitcase is still the same.
So, for example with:
iex(1)> x = 1
iex(2)> f = fn -> x end
iex(3)> x = 2
iex(4)> f.()
1
You have a suitcase with 1 in it and you label it x.
Then you say, "Here, Mr. Function, I want you to tell me what is in this suitcase when I call you."
Then, you take the label off of the suitcase with 1 in it and put it on another suitcase with 2 in it.
Then you say "Hey, Mr. Function, what is in that suitcase?"
He will say "1", because the suitcase hasn't changed. Although, you have taken your label off of it and put it on a different suitcase.
After a while, I came up with the answer that is probably the best explanation of the difference between “binding” and “assignment”; it has nothing in common with what I have written in another answer, hence it’s posted as a separate answer.
In any functional language, where everything is immutable, there is no meaningful difference between terms “binding” and “assignment.” One might call it either way; the common pattern is to use the word “binding,“ explicitly denoting that it’s a value bound to a variable. In Erlang, for instance, one can not rebound a variable. In Elixir this is possible (why, for God’s sake, José, what for?)
Consider the following example in Elixir:
iex> x = 1
iex> 1 = x
The above is perfectly valid Elixir code. It is evident, one cannot assign anything to one. It is neither assignment nor binding. It is matching. That is how = is treated in Elixir (and in Erlang): a = b fails if both are bound to different values; it returns RHO if they match; it binds LHO to RHO if LHO is not bound yet.
In Ruby it differs. There is a significant difference between assignment (copying the content,) and binding (making a reference.)
Elixir vs Ruby might not be the best contrast for this. In Elixir, we can readily "re-assign" the value of a previously assigned named variable. The two anonymous-function examples you provided demonstrate the difference in how the two languages assign local variables in them. In Ruby, the variable, meaning the memory reference, is assigned, which is why when we change it, the anonymous function returns the current value stored in that memory-reference. While in Elixir, the value of the variable at the time the anonymous function is defined (rather than the memory reference) is copied and stored as the local variable.
In Erlang, Elixir's "parent" language, however, variables as a rule are "bound." Once you've declared the value for the variable named X, you are not allowed to alter it for the remainder of the program and any needed alterations would need to be stored in new named variables. (There is a way to reassign a named variable in Erlang but it is not the custom.)
Binding refers to particular concept used in expression-based languages that may seem foreign if you're used to statement-based languages. I'll use an ML-style example to demonstrate:
let x = 3 in
let y = 5 in
x + y
val it : int = 8
The let... in syntax used here demonstrates that the binding let x = 3 is scoped only to the expression following the in. Likewise, the binding let y = 5 is only scoped to the expression x + y, such that, if we consider another example:
let x = 3 in
let f () =
x + 5
let x = 4 in
f()
val it : int = 8
The result is still 8, even though we have the binding let x = 4 above the call to f(). This is because f itself was bound in the scope of the binding let x = 3.
Assignment in statement-based languages is different, because the variables being assigned are not scoped to a particular expression, they are effectively 'global' for whatever block of code they're in, so reassigning the value of a variable changes the result of an evaluation that uses the same variable.
The easiest way to understand the difference, would be to compare the AST that is used by the language interpreter/compiler to produce machine-/byte-code.
Let’s start with ruby. Ruby does not provide the AST viewer out of the box, so I will use RubyParser gem for that:
> require 'ruby_parser'
> RubyParser.new.parse("x = 1; f = -> {x}; x = 2; f.()").inspect
#=> "s(:block, s(:lasgn, :x, s(:lit, 1)),
# s(:lasgn, :f, s(:iter, s(:call, nil, :lambda), 0, s(:lvar, :x))),
# s(:lasgn, :x, s(:lit, 2)), s(:call, s(:lvar, :f), :call))"
The thing we are looking for is the latest node in the second line: there is x variable inside the proc. In other words, ruby expects the bound variable there, named x. At the time the proc is evaluated, x has a value of 2. Hence the the proc returns 2.
Let’s now check Elixir.
iex|1 ▶ quote do
...|1 ▶ x = 1
...|1 ▶ f = fn -> x end
...|1 ▶ x = 2
...|1 ▶ f.()
...|1 ▶ end
#⇒ {:__block__, [],
# [
# {:=, [], [{:x, [], Elixir}, 1]},
# {:=, [], [{:f, [], Elixir}, {:fn, [], [{:->, [], [[], {:x, [], Elixir}]}]}]},
# {:=, [], [{:x, [], Elixir}, 2]},
# {{:., [], [{:f, [], Elixir}]}, [], []}
# ]}
Last node in the second line is ours. It still contains x, but during a compilation stage this x will be evaluated to it’s currently assigned value. That said, fn -> not_x end will result in compilation error, while in ruby there could be literally anything inside a proc body, since it’ll be evaluated when called.
In other words, Ruby uses a current caller’s context to evaluate proc, while Elixir uses a closure. It grabs the context it encountered the function definition and uses it to resolve all the local variables.
Is there a method to overwrite variable without copying its name? For example, when I want to change my_var = '3' to an integer, I must do something like this:
my_var = my_var.to_i
Is there way to do this without copying variable's name? I want to do something like this:
my_var = something_const.to_i
For numbers there exists +=, -= etc, but is there universal way to do this for all methods ?
There is no way to covert a string to an integer like that, without repeating the variable name. Methods such as String#upcase! and Array#flatten! work by mutating the object; however, it is not possible to define such a method like String#to_i! because we are converting the object to an instance of a different class.
For example, here is a (failed) attempt to define such a method:
# What I want to be able to do:
# my_var = "123"
# my_var.to_i! # => my_var == 123
class String
def to_i!
replace(Integer(self))
end
end
my_var = "123"
my_var.to_i! # TypeError: no implicit conversion of Fixnum into String
...And even if this code were valid, it would still offer no performance gain since a new object is still being created.
As for your examples of += and -=, these are in fact simply shorthand for:
x += 1
# Is equivalent to:
x = x + 1
So again, there is no performance gain here either; just slightly nicer syntax. A good question to ask is, why doesn't ruby support a ++ operator? If such an operator existed then it would offer performance gain... But I'll let you research for yourself why this is missing from the language.
So to summarise,
is there universal way to do this for all methods?
No. The special operators like +=, -=, |= and &= are all predefined; there is no "generalised" version such as method_name=.
You can also define methods that mutate the object, but only when appropriate. Such methods are usually named with a !, are called "bang-methods", and have a "non-bang" counterpart. On String objects, for example, there is String#capitalize! (and String#capitalize), String#delete! (and String#delete), String#encode! (and String#encode), .... but no String#to_i! for the reasons discussed above.
I'm trying to set a value to a variable inside a function in Enum.each, but at the end of loop, variable is empty and I don't know exactly why this behaviour.
Code:
base = "master"
candidates = ["stream", "pigeons", "maters"]
return = []
Enum.each(candidates, fn candidate ->
cond do
String.length(base) == String.length(candidate) ->
return = return ++ [candidate]
true ->
true
end
end)
IO.inspect return
At this example, return is expected to be ["stream", "maters"], but instead, it is only an empty list: []
My question is why this happens.
When dealing with languages like Elixir, it is better to think in terms of "values" and "names" instead of "variables".
The reason you cannot do what you want is that Elixir has "lexical scoping".
When you assign to a "variable", you create a new value in the inner scope. You never change the "value" of a "name" defined in the outer scope.
(you probably can get what you want with Enum.filter/2, but I'm guessing this is just an illustrative example)
EDIT:
As of today, Elixir will allow you to write something like this:
if condition_that_evals_to_false do
x = 1
else
x = 2
end
IO.inspect x # => 2
```
But this will be deprecated in Elixir 1.3
Any reason why you don't just filter?
Anyways it seems like you're trying to mutate the value of return which is not possible with Elixir.
base = "master"
candidates = ["stream", "pigeon", "maters"]
result = Enum.filter(candidates, fn(candidate) ->
length(candidate) == length(base)
end
IO.inspect result
Edit: I'd also like to add that based on your logic, all of the candidates would be returned
Not sure, since I've never worked with the language, but a couple things spring to mind:
String.length(base) == String.length(candidate) can be equivalent to true, which is already a pattern in your set.
It could also be a scope issue with the return variable. It could be that the local return is hiding the global return. You could check this by outputting return every iteration. Each iteration the return should contain a single entry.
This is a bug. From Elixir's documentation:
Note: due to a bug in the 0.12.x series, cond‘s conditions actually
leak bindings to the surrounding scope. This should be fixed in
0.13.1.
You should use filtering like #{Christopher Yammine} suggested.
I'm working with ruby and I'm just learning lambdas. I have an array of objects and I want to select certain objects from the array based of different conditions like so:
result = array.select{|ar| ar.number > 4}
I want to put the arguments of the select into a lambda expression i.e. I want to put |ar| ar.number > 4 into a lambda expression. I've tried a few things including:
result = array.select{lambda{|g| g.number > 4}}
But this doesn't work.
I've also tried this:
l = lambda {g.number > 4}
result = array.select{|g| l}
and that also doesn't work. Also I need to pass my lambda express to a function so I don't think the first way I did it would have worked. How would I do this?
Enumerable#select takes a block, not a lambda. So, you need to pass it a block.
Thankfully, there is an operator which will convert a lambda or proc (and in fact anything which responds to to_proc) to a block: the unary prefix & operator, which is only valid in argument lists (where it converts a proc to a block) and in parameter lists (where it converts a block to a proc).
So, if you have a lambda l, you can pass it as a block to method foo like so:
foo(&l)
Your second problem is that your lambda doesn't take any arguments, but the block passed to select takes one argument, so you need to create your lambda with one argument:
l = lambda {|el| el.number > 4 }
# I prefer this syntax, though:
l = -> el { el.number > 4 }
Putting all that together, we have:
result = array.select(&l)
To pass a lambda (or a proc) to a method as a block, you have to use the special &syntax:
result = array.select &mylambda
Also, your lambda (or proc) must specify the name for parameters it will receive and use. Thus:
l = lambda { |g| g.number > 4}
result = array.select &l
A clean way to do this is to use Ruby's "stabby" lambda syntax:
l = ->(ar) { ar.number > 4 }
result = array.select(&l)
The more "traditional" way to define a lambda is as such:
l = lambda { |ar| ar.number > 4 }
result = array.select(&l)
Note that you don't strictly need to use a lambda; you could also use a proc*:
p = proc { |ar| ar.number > 4 }
result = array.select(&p)
* The main difference between proc and lambda is that lambda will throw an error when passed too few or too many arguments, while proc will ignore it. Another subtle difference is that the return statement inside a proc will cause execution to return both from the closure and the enclosing method (if any).