I want to access the count and total values from the ProgressBar instance. If they are exposed, I do not know how to get them. When I'm testing a loop, I may only want to iterate 10 times and break. It seems intuitive that either pb.increment would return the value, or that it would be an attr_reader value. Am I missing something?
require 'ruby-progressbar'
pb = ProgressBar.create(
title:'Items',
total:500,
remainder_mark:'.',
format:'%t |%B| %c of %C %p%%',
length:80
)
i = 0
500.times {|x|
pb.increment
break if (i+=1) > 100
}
pb.stop
As you can see, I have to create and maintain a separate counter. It just seems so intuitive to say "break if pb.increment > 10", or at least do "pb.increment; break if pb.count > 10".
Thanks, dvn
Update
You can access progress from the pb object, just as Anthony suggested:
break if pb.progress > 100
According to the docs, you can get all values as a hash with to_h and then use the 'progress' value:
break if pb.to_h['progress'] > 100
Related
I currently have a very large array of permutations, which is currently using a significant amount of RAM. This is the current code I have which SHOULD:
Count all but the occurrences where more than one '1' exists or three '2's exist in a row.
arr = [*1..3].repeated_permutation(30).to_a;
count = 0
arr.each do |x|
if not x.join('').include? '222' and x.count(1) < 2
count += 1
end
end
print count
So basically this results in a 24,360 element array, each of which have 30 elements.
I've tried to run it through Terminal but it literally ate through 14GB of RAM, and didn't move for 15 minutes, so I'm not sure whether the process froze while attempting to access more RAM or if it was still computing.
My question being: is there a faster way of doing this?
Thanks!
I am not sure what problem you try to solve. If your code is just an example for a more complex problem and you really need to check programatically every single permumation, then you might want to experiment with lazy:
[*1..3].repeated_permutation(30).lazy.each do ||
# your condition
end
Or you might want to make the nested iteratior very explicit:
[1,2,3].each do |x1|
[1,2,3].each do |x2|
[1,2,3].each do |x3|
# ...
[1,2,3].each do |x30|
permutation = [x1,x2,x3, ... , x30]
# your condition
end
end
end
end
end
But it feels wrong to me to solve this kind of problem with Ruby enumerables at all. Let's have a look at your strings:
111111111111111111111111111111
111111111111111111111111111112
111111111111111111111111111113
111111111111111111111111111121
111111111111111111111111111122
111111111111111111111111111123
111111111111111111111111111131
...
333333333333333333333333333323
333333333333333333333333333331
333333333333333333333333333332
333333333333333333333333333333
I suggest to just use enumerative combinatorics. Just look at the patterns and analyse (or count) how often your condition can be true. For example there are 28 indexes in your string at which a 222 substring could be place, only 27 for the 2222 substring... If you place a substring how likely is it that there is no 1 in the other parts of the string?
I think your problem is a mathematics problem, not a programming problem.
NB This is an incomplete answer, but I think the idea might give a push to the proper solution.
I can think of a following approach: let’s represent each permutation as a value in ternary number base, padded by zeroes:
1 = 000..00001
2 = 000..00002
3 = 000..00010
4 = 000..00011
5 = 000..00012
...
Now consider we restated the original task, treating zeroes as ones, ones as twos and twos as threes. So far so good.
The whole list of permutations would be represented by:
(1..3**30-1).map { |e| x = e.to_s(3).rjust(30, '0') }
Now we are to apply your conditions:
def do_calc permutation_count
(1..3**permutation_count-1).inject do |memo, e|
x = e.to_s(3).rjust(permutation_count, '0')
!x.include?('111') && x.count('0') < 2 ? memo + 1 : memo
end
Unfortunately, even for permutation_count == 20 it takes more than 5 minutes to calculate, so probably some additional steps are required. I will be thinking of further optimization. Currently I hope this will give you a hint to find the good approach yourself.
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.
UPDATE 2: For posterity, this is how I've settled on doing it (thanks to Jorg's input):
100.step(2, -2) do |x|
# my code
end
(Obviously there are plenty of ways to do this; but it sounds like this is the most "Ruby" way to do it; and that's exactly what I was after.)
UPDATE: OK, so what I was looking for was step:
(2..100).step(2) do |x|
# my code
end
But it turns out that I wasn't 100% forthcoming in my original question. I actually want to iterate over this range backwards. To my surprise, a negative step isn't legal.
(100..2).step(-2) do |x|
# ArgumentError: step can't be negative
end
So: how do I do this backwards?
I am completely new to Ruby, so be gentle.
Say I want to iterate over the range of even numbers from 2 to 100; how would I do that?
Obviously I could do:
(2..100).each do |x|
if x % 2 == 0
# my code
end
end
But, obviously (again), that would be pretty stupid.
I know I could do something like:
i = 2
while i <= 100
# my code
i += 2
end
I believe I could also write my own custom class that provides its own each method (?). I am almost sure that would be overkill, though.
I'm interested in two things:
Is it possible to do this with some variation of the standard Range syntax (i.e., (x..y).each)?
Either way, what would be the most idiomatic "Ruby way" of accomplishing this (using a Range or otherwise)? Like I said, I'm new to the language; so any guidance you can offer on how to do things in a more typical Ruby style would be much appreciated.
You can't declare a Range with a "step". Ranges don't have steps, they simply have a beginning and an end.
You can certainly iterate over a Range in steps, for example like this:
(2..100).step(2).reverse_each(&method(:p))
But if all you want is to iterate, then what do you need the Range for in the first place? Why not just iterate?
100.step(2, -2, &method(:p))
This has the added benefit that unlike reverse_each it does not need to generate an intermediate array.
This question answers yours: about ruby range?
(2..100).step(2) do |x|
# your code
end
I had similar issue here are the various ways I found to do the same SIMPLE thing I used step in the end because it allowed for NEGATIVE and FRACTIONAL increments and I had no conditions, other than the bounds to look for
case loop_type
when FOR
# doen't appear to have a negative or larger than 1 step size!
for kg in 50..120 do
kg_to_stones_lbs(kg)
end
when STEP
120.step(70,-0.5){ |kg|
kg_to_stones_lbs(kg)
}
when UPTO
50.upto(120) { |kg|
kg_to_stones_lbs(kg)
}
when DOWNTO
120.downto(50){ |kg|
kg_to_stones_lbs(kg)
}
when RANGE
(50..120).reverse_each{ |kg|
kg_to_stones_lbs(kg)
}
when WHILE
kg = 120
while kg >= 50
kg_to_stones_lbs(kg)
kg -= 0.5
end
end
O/P:
92.0kg - 14st 7lbs
91.5kg - 14st 6lbs
91.0kg - 14st 5lbs
90.5kg - 14st 4lbs
90.0kg - 14st 2lbs
89.5kg - 14st 1lbs
89.0kg - 14st 0lbs
88.5kg - 13st 13lbs
88.0kg - 13st 12lbs
I play around with arrays and hashes quite a lot in ruby and end up with some code that looks like this:
sum = two_dimensional_array.select{|i|
i.collect{|j|
j.to_i
}.sum > 5
}.collect{|i|
i.collect{|j|
j ** 2
}.average
}.sum
(Let's all pretend that the above code sample makes sense now...)
The problem is that even though TextMate (my editor of choice) picks up simple {...} or do...end blocks quite easily, it can't figure out (which is understandable since even I can't find a "correct" way to fold the above) where the above blocks start and end to fold them.
How would you fold the above code sample?
PS: considering that it could have 2 levels of folding, I only care about the outer consecutive ones (the blocks with the i)
To be honest, something that convoluted is probably confusing TextMate as much as anyone else who has to maintain it, and that includes you in the future.
Whenever you see something that rolls up into a single value, it's a good case for using Enumerable#inject.
sum = two_dimensional_array.inject(0) do |sum, row|
# Convert row to Fixnum equivalent
row_i = row.collect { |i| i.to_i }
if (row_i.sum > 5)
sum += row_i.collect { |i| i ** 2 }.average
end
sum # Carry through to next inject call
end
What's odd in your example is you're using select to return the full array, allegedly converted using to_i, but in fact Enumerable#select does no such thing, and instead rejects any for which the function returns nil. I'm presuming that's none of your values.
Also depending on how your .average method is implemented, you may want to seed the inject call with 0.0 instead of 0 to use a floating-point value.
Ok, so say you have a really big Range in ruby. I want to find a way to get the max value in the Range.
The Range is exclusive (defined with three dots) meaning that it does not include the end object in it's results. It could be made up of Integer, String, Time, or really any object that responds to #<=> and #succ. (which are the only requirements for the start/end object in Range)
Here's an example of an exclusive range:
past = Time.local(2010, 1, 1, 0, 0, 0)
now = Time.now
range = past...now
range.include?(now) # => false
Now I know I could just do something like this to get the max value:
range.max # => returns 1 second before "now" using Enumerable#max
But this will take a non-trivial amount of time to execute. I also know that I could subtract 1 second from whatever the end object is is. However, the object may be something other than Time, and it may not even support #-. I would prefer to find an efficient general solution, but I am willing to combine special case code with a fallback to a general solution (more on that later).
As mentioned above using Range#last won't work either, because it's an exclusive range and does not include the last value in it's results.
The fastest approach I could think of was this:
max = nil
range.each { |value| max = value }
# max now contains nil if the range is empty, or the max value
This is similar to what Enumerable#max does (which Range inherits), except that it exploits the fact that each value is going to be greater than the previous, so we can skip using #<=> to compare the each value with the previous (the way Range#max does) saving a tiny bit of time.
The other approach I was thinking about was to have special case code for common ruby types like Integer, String, Time, Date, DateTime, and then use the above code as a fallback. It'd be a bit ugly, but probably much more efficient when those object types are encountered because I could use subtraction from Range#last to get the max value without any iterating.
Can anyone think of a more efficient/faster approach than this?
The simplest solution that I can think of, which will work for inclusive as well as exclusive ranges:
range.max
Some other possible solutions:
range.entries.last
range.entries[-1]
These solutions are all O(n), and will be very slow for large ranges. The problem in principle is that range values in Ruby are enumerated using the succ method iteratively on all values, starting at the beginning. The elements do not have to implement a method to return the previous value (i.e. pred).
The fastest method would be to find the predecessor of the last item (an O(1) solution):
range.exclude_end? ? range.last.pred : range.last
This works only for ranges that have elements which implement pred. Later versions of Ruby implement pred for integers. You have to add the method yourself if it does not exist (essentially equivalent to special case code you suggested, but slightly simpler to implement).
Some quick benchmarking shows that this last method is the fastest by many orders of magnitude for large ranges (in this case range = 1...1000000), because it is O(1):
user system total real
r.entries.last 11.760000 0.880000 12.640000 ( 12.963178)
r.entries[-1] 11.650000 0.800000 12.450000 ( 12.627440)
last = nil; r.each { |v| last = v } 20.750000 0.020000 20.770000 ( 20.910416)
r.max 17.590000 0.010000 17.600000 ( 17.633006)
r.exclude_end? ? r.last.pred : r.last 0.000000 0.000000 0.000000 ( 0.000062)
Benchmark code is here.
In the comments it is suggested to use range.last - (range.exclude_end? ? 1 : 0). It does work for dates without additional methods, but will never work for non-numeric ranges. String#- does not exist and makes no sense with integer arguments. String#pred, however, can be implented.
I'm not sure about the speed (and initial tests don't seem incredibly fast), but the following might do what you need:
past = Time.local(2010, 1, 1, 0, 0, 0)
now = Time.now
range = past...now
range.to_a[-1]
Very basic testing (counting in my head) showed that it took about 4 seconds while the method you provided took about 5-6. Hope this helps.
Edit 1: Removed second solution as it was totally wrong.
I can't think there's any way to achieve this that doesn't involve enumerating the range, at least unless as already mentioned, you have other information about how the range will be constructed and therefore can infer the desired value without enumeration. Of all the suggestions, I'd go with #max, since it seems to be most expressive.
require 'benchmark'
N = 20
Benchmark.bm(30) do |r|
past, now = Time.local(2010, 2, 1, 0, 0, 0), Time.now
#range = past...now
r.report("range.max") do
N.times { last_in_range = #range.max }
end
r.report("explicit enumeration") do
N.times { #range.each { |value| last_in_range = value } }
end
r.report("range.entries.last") do
N.times { last_in_range = #range.entries.last }
end
r.report("range.to_a[-1]") do
N.times { last_in_range = #range.to_a[-1] }
end
end
user system total real
range.max 49.406000 1.515000 50.921000 ( 50.985000)
explicit enumeration 52.250000 1.719000 53.969000 ( 54.156000)
range.entries.last 53.422000 4.844000 58.266000 ( 58.390000)
range.to_a[-1] 49.187000 5.234000 54.421000 ( 54.500000)
I notice that the 3rd and 4th option have significantly increased system time. I expect that's related to the explicit creation of an array, which seems like a good reason to avoid them, even if they're not obviously more expensive in elapsed time.