ruby parallel assignment, step question - ruby

so, i'm trying to learn ruby by doing some project euler questions, and i've run into a couple things i can't explain, and the comma ?operator? is in the middle of both. i haven't been able to find good documentation for this, maybe i'm just not using the google as I should, but good ruby documentation seems a little sparse . . .
1: how do you describe how this is working? the first snippet is the ruby code i don't understand, the second is the code i wrote that does the same thing only after painstakingly tracing the first:
#what is this doing?
cur, nxt = nxt, cur + nxt
#this, apparently, but how to describe the above?
nxt = cur + nxt
cur = nxt - cur
2: in the following example, how do you describe what the line with 'step' is doing? from what i can gather, the step command works like (range).step(step_size), but this seems to be doing (starting_point).step(ending_point, step_size). Am i right with this assumption? where do i find good doc of this?
#/usr/share/doc/ruby1.9.1-examples/examples/sieve.rb
# sieve of Eratosthenes
max = Integer(ARGV.shift || 100)
sieve = []
for i in 2 .. max
sieve[i] = i
end
for i in 2 .. Math.sqrt(max)
next unless sieve[i]
(i*i).step(max, i) do |j|
sieve[j] = nil
end
end
puts sieve.compact.join(", ")

1: It's called parallel assignment. Ruby cares to create temporal variables and not override your variables with incorrect values. So this example:
cur, nxt = nxt, cur + nxt
is the same as:
tmp = cur + nxt
cur = nxt
nxt = tmp
bur more compact, without place to make stupid mistake and so on.
2: There are 2 step methods in ruby core library. First is for Numeric class (every numbers), so you could write:
5.step(100, 2) {}
and it starts at 5 and takes every second number from it, stops when reaches 100.
Second step in Ruby is for Range:
(5..100).step(2) {}
and it takes range (which has start and end) and iterates through it taking every second element. It is different, because you could pass it not necessarily numeric range and it will take every nth element from it.
Take a look at http://ruby-doc.org/core-1.8.7/index.html

This a parallel assignment. In your example, Ruby first evaluates nxt and cur + nxt. It then assigns the results to cur and nxt respectively.
The step method in the code is actually Numeric#step (ranges are constructed with (n..m)). The step method of Numeric iterates using the number it is called on as a starting point. The arguments are the limit and step size respectively. The code above will therefore invoke the block with i * i and then each successive increment of i until max is reached.
A good starting point for Ruby documentation is the ruby-doc.org site.

Related

Filtering a range in Julia

I have some MWE below. What I want is to have a subsection of a range, interact with the rest of the range, but not itself.
For instance if the range is 1:100, I want to have a for loop that will have each index in 4:6, interact with all values of 1:100 BUT NOT 4:6.
I want to do this using ranges/filters to avoid generating temporary arrays.
In my case the total range is the number of atoms in the system. The sub-range, is the atoms in a specific molecule. I need to do calculations where each atom in a molecule interacts with all other atoms, but not the atoms in the same molecule.
Further
I am trying to avoid using if statements because that messes up parallel codes. Doing this with an if statement would be
for i=4:6
for j = 1:100
if j == 4 || j==5 || j==6
continue
end
println(i, " ", j)
end
end
I have actual indexing in my code, I would never hardcode values like the above... But I want to avoid that if statement.
Trials
The following does what I want, but I now realize that using filter is bad when it comes to memory and the amount used scales linearly with b.
a = 4:6
b = 1:100
for i in a
for j in filter((b) -> !(b in a),b)
print(i, " ", j)
end
end
Is there a way to get the double for loop I want where the outer is a sub-range of the inner, but the inner does not include the outer sub-range and most importantly is fast and does not create alot of memory usage like filter does?
If memory usage is really a concern, consider two for loops using the range components:
systemrange = 1:50
moleculerange = 4:12
for i in systemrange[1]:moleculerange[1]-1
println(i)
end
for i in moleculerange[end]+1:systemrange[end]
println(i)
end
You might be able to do each loop in its own thread.
What about creating a custom iterator?
Note that example below needs some adjustments depending on how you define the exception lists (for example for long list with non continues indices you should use binary search).
struct RangeExcept
start::Int
stop::Int
except::UnitRange{Int}
end
function Base.iterate(it::RangeExcept, (el, stop, except)=(it.except.start > 1 ? it.start : it.except.stop+1, it.stop, it.except))
new_el = el+1
if new_el in except
new_el = except.stop+1
end
el > stop && return nothing
return (el, (new_el, stop,except))
end
Now let us test the code:
julia> for i in RangeExcept(1,10,3:7)
println(i)
end
1
2
8
9
10

Ruby elegant alternative to ++ in nested loops?

Before anything, I have read all the answers of Why doesn't Ruby support i++ or i—? and understood why. Please note that this is not just another discussion topic about whether to have it or not.
What I'm really after is a more elegant solution for the situation that made me wonder and research about ++/-- in Ruby. I've looked up loops, each, each_with_index and things alike but I couldn't find a better solution for this specific situation.
Less talk, more code:
# Does the first request to Zendesk API, fetching *first page* of results
all_tickets = zd_client.tickets.incremental_export(1384974614)
# Initialises counter variable (please don't kill me for this, still learning! :D )
counter = 1
# Loops result pages
loop do
# Loops each ticket on the paged result
all_tickets.all do |ticket, page_number|
# For debug purposes only, I want to see an incremental by each ticket
p "#{counter} P#{page_number} #{ticket.id} - #{ticket.created_at} | #{ticket.subject}"
counter += 1
end
# Fetches next page, if any
all_tickets.next unless all_tickets.last_page?
# Breaks outer loop if last_page?
break if all_tickets.last_page?
end
For now, I need counter for debug purposes only - it's not a big deal at all - but my curiosity typed this question itself: is there a better (more beautiful, more elegant) solution for this? Having a whole line just for counter += 1 seems pretty dull. Just as an example, having "#{counter++}" when printing the string would be much simpler (for readability sake, at least).
I can't simply use .each's index because it's a nested loop, and it would reset at each page (outer loop).
Any thoughts?
BTW: This question has nothing to do with Zendesk API whatsoever. I've just used it to better illustrate my situation.
To me, counter += 1 is a fine way to express incrementing the counter.
You can start your counter at 0 and then get the effect you wanted by writing:
p "#{counter += 1} ..."
But I generally wouldn't recommend this because people do not expect side effects like changing a variable to happen inside string interpolation.
If you are looking for something more elegant, you should make an Enumerator that returns integers one at a time, each time you call next on the enumerator.
nums = Enumerator.new do |y|
c = 0
y << (c += 1) while true
end
nums.next # => 1
nums.next # => 2
nums.next # => 3
Instead of using Enumerator.new in the code above, you could just write:
nums = 1.upto(Float::INFINITY)
As mentioned by B Seven each_with_index will work, but you can keep the page_number, as long all_tickets is a container of tuples as it must be to be working right now.
all_tickets.each_with_index do |ticket, page_number, i|
#stuff
end
Where i is the index. If you have more than ticket and page_number inside each element of all_tickets you continue putting them, just remember that the index is the extra one and shall stay in the end.
Could be I oversimplified your example but you could calculate a counter from your inner and outer range like this.
all_tickets = *(1..10)
inner_limit = all_tickets.size
outer_limit = 5000
1.upto(outer_limit) do |outer_counter|
all_tickets.each_with_index do |ticket, inner_counter|
p [(outer_counter*inner_limit)+inner_counter, outer_counter, inner_counter, ticket]
end
# some conditional to break out, in your case the last_page? method
break if outer_counter > 3
end
all_tickets.each_with_index(1) do |ticket, i|
I'm not sure where page_number is coming from...
See Ruby Docs.

Lua - why for loop limit is not calculated dynamically?

Ok here's a basic for loop
local a = {"first","second","third","fourth"}
for i=1,#a do
print(i.."th iteration")
a = {"first"}
end
As it is now, the loop executes all 4 iterations.
Shouldn't the for-loop-limit be calculated on the go? If it is calculated dynamically, #a would be 1 at the end of the first iteration and the for loop would break....
Surely that would make more sense?
Or is there any particular reason as to why that is not the case?
The main reason why numerical for loops limits are computed only once is most certainly for performance.
With the current behavior, you can place arbitrary complex expressions in for loops limits without a performance penalty, including function calls. For example:
local prod = 1
for i = computeStartLoop(), computeEndLoop(), computeStep() do
prod = prod * i
end
The above code would be really slow if computeEndLoop and computeStep required to be called at each iteration.
If the standard Lua interpreter and most notably LuaJIT are so fast compared to other scripting languages, it is because a number of Lua features have been designed with performance in mind.
In the rare cases where the single evaluation behavior is undesirable, it is easy to replace the for loop with a generic loop using while end or repeat until.
local prod = 1
local i = computeStartLoop()
while i <= computeEndLoop() do
prod = prod * i
i = i + computeStep()
end
The length is computed once, at the time the for loop is initialized. It is not re-computed each time through the loop - a for loop is for iterating from a starting value to an ending value. If you want the 'loop' to terminate early if the array is re-assigned to, you could write your own looping code:
local a = {"first", "second", "third", "fourth"}
function process_array (fn)
local inner_fn
inner_fn =
function (ii)
if ii <= #a then
fn(ii,a)
inner_fn(1 + ii)
end
end
inner_fn(1, a)
end
process_array(function (ii)
print(ii.."th iteration: "..a[ii])
a = {"first"}
end)
Performance is a good answer but I think it also makes the code easier to understand and less error-prone. Also, that way you can (almost) be sure that a for loop always terminates.
Think about what would happen if you wrote that instead:
local a = {"first","second","third","fourth"}
for i=1,#a do
print(i.."th iteration")
if i > 1 then a = {"first"} end
end
How do you understand for i=1,#a? Is it an equality comparison (stop when i==#a) or an inequality comparison (stop when i>=#a). What would be the result in each case?
You should see the Lua for loop as iteration over a sequence, like the Python idiom using (x)range:
a = ["first", "second", "third", "fourth"]
for i in range(1,len(a)+1):
print(str(i) + "th iteration")
a = ["first"]
If you want to evaluate the condition every time you just use while:
local a = {"first","second","third","fourth"}
local i = 1
while i <= #a do
print(i.."th iteration")
a = {"first"}
i = i + 1
end

simple method to keep last n elements in a queue for vb6?

I am trying to keep the last n elements from a changing list of x elements (where x >> n)
I found out about the deque method, with a fixed length, in other programming languages. I was wondering if there is something similar for VB6
Create a Class that extends an encapsulated Collection.
Add at the end (anonymous), retrieve & remove from the beginning (index 1). As part of adding check your MaxDepth property setting (or hard code it if you like) and if Collection.Count exceeds it remove the extra item.
Or just hard code it all inline if a Class is a stumper for you.
This is pretty routine.
The only thing I can think of is possibly looping through the last 5 values of the dynamic array using something like:
For UBound(Array) - 5 To UBound(Array)
'Code to store or do the desired with these values
Loop
Sorry I don't have a definite answer, but hopefully that might help.
Here's my simplest solution to this:
For i = n - 1 To 1 Step -1
arrayX(i) = arrayX(i - 1)
Next i
arrayX(0) = latestX
Where:
arrayX = array of values
n = # of array elements
latestX = latest value of interest (assumes entire code block is also
within another loop)

Can't convert nil into string--Ruby Secret Santa

I wrote a Secret Santa program (ala Ruby Quiz...ish), but occasionally when the program runs, I get an error.
Stats: If there's 10 names in the pot, the error comes up about 5% of the time. If there's 100 names in the pot, it's less than 1%. This is on a trial of 1000 times in bash. I've determined that the gift arrays are coming up nil at some point, but I'm not sure why or how to avoid it.
Providing code...
0.upto($lname.length-1).each do |i|
j = rand($giftlname.length) # should be less each time.
while $giftlname[j] == $lname[i] # redo random if it picks same person
if $lname[i] == $lname.last # if random gives same output again, means person is left with himself; needs to switch with someone
$giftfname[j], $fname[i] = $giftfname[i], $fname[j]
$giftlname[j], $lname[i] = $giftlname[i], $lname[j]
$giftemail[j], $email[i] = $giftemail[i], $email[j]
else
j = rand($giftlname.length)
end
end
$santas.push('Santa ' + $fname[i] + ' ' + $lname[i] + ' sends gift to ' + $giftfname[j] + ' ' + $giftlname[j] + ' at ' + '<' + $giftemail[j] + '>.') #Error here, something is sometimes nil
$giftfname.delete_at(j)
$giftlname.delete_at(j)
$giftemail.delete_at(j)
end
Thanks SO!
I think your problem is right here:
$giftfname[j], $fname[i] = $giftfname[i], $fname[j]
Your i values range between zero to the last index in $fname (inclusive) and, presumably, your $giftfname starts off as a clone of $fname (or at least another array with the same length). But, as you spin through the each, you're shrinking $giftfname so $giftfname[i] will be nil and the swap operation above will put nil into $giftfname[j] (which is supposed to be a useful entry of $giftfname). Similar issues apply to $giftlname and $giftemail.
I'd recommend using one array with three element objects (first name, last name, email) instead of your three parallel arrays. There's also a shuffle method on Array that might be of use to you:
Start with an array of people.
Make copy of that array.
Shuffle the copy until it is different at every index from that original array.
Then zip the together to get your final list of giver/receiver pairs.
Figured it out and used the retry statement. the if statement now looks like this (all other variables have been edited to be non-global as well)
if lname[i] == lname.last
santas = Array.new
giftfname = fname.clone
giftlname = lname.clone
giftemail = email.clone
retry
That, aside from a few other edits, created the solution I needed without breaking apart the code too much again. Will definitely try out mu's solution as well, but I'm just glad I have this running error-free for now.

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