# How clear is this Ruby code? [closed]

Elsewhere, there was a question about finding an elegant solution to a particular problem, and the following solution was presented.

I'm curious whether this solution is elegant from the perspective of easy to understand for a Ruby programmer (which I am no longer).

merged_file = File.open("merge_out.txt", "w")

files = ARGV.map { |filename| File.open( filename, "r") }

lines = files.map { |file| file.gets }

while lines.any?
next_line = lines.compact.min
file_id = lines.index( next_line )
merged_file.print next_line
lines[ file_id ] = files[ file_id ].gets
end


The question is not whether it is efficient, but just "how long does it take a Ruby programmer to understand what this does?"

## closed as off-topic by nhgrif, Nic Hartley, RubberDuck, rolfl♦Jun 10 '15 at 0:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

• The code does something, it's not hypothetical – Caridorc Jul 9 '15 at 10:16
• Please stop changing the summary of this question. I have not debated its closure because it is not code that I own or maintain. However, the topic that I was asking about was nothing to do with "Merging files", it is to do with "how clear is the code". Placing the answer of what the code does in the title detracts from the point of the question. – GreenAsJade Jul 9 '15 at 10:22
• @GreenAsJade Titles are meant to describe what the code does, instead of what you want to ask. Currently, your title is poor. – Ismael Miguel Jul 9 '15 at 10:29
• @GreenAsJade You should read the help section, about "Titling your question" – Ismael Miguel Jul 9 '15 at 10:33
• I have subsequent to posting this question been educated about this. Changing the title now on this closed question doesn't help one bit. – GreenAsJade Jul 9 '15 at 10:36

I had no trouble figuring it out what it does, but I'm not sure I'd call it "easy to understand".

All the declarations are implicit and thus if you're familiar with all the in's and outs of Array and the map method, it's pretty straight forward what the code is attempting to do. Some of the logic takes a bit of thinking about,

I did puzzle a bit over this line

file_id = lines.index( next_line)


I think the most puzzling part is the unstated requirement that the input files are already sorted. Otherwise this line doesn't work. Once I figured out that this line meant that the input files were sorted, the rest fell into place.

The exit condition is also a bit tricky to figure out.

The code really feels like a translation to ruby from a more explicitly functional language, it doesn't feel "ruby-like" at all.

1. Took around a 100 seconds to understand. Would be twice faster if I had a commentary: # this code merges two presorted files.

2. All is pretty ok, but what you forgot here is to close all opened files. And it is the only reason to have a first variable declared, otherwise I would mix both maps together.

3. If you pass filenames via ARGV, there should be also the merged_filename as the first or last parameter. Or at any position and optional if you start to use some named options parsing.

So we have this:

files_to_merge = ARGV.map &File.method(:open)
lines = files_to_merge.map &:gets
merged_file = File.open "merge_out.txt", "w"
while lines.any?
merged_file.print(next_line = lines.compact.min)
file_id = lines.index next_line
lines[file_id] = files_to_merge[file_id].gets
end
merged_file.close
files_to_merge.each &:close


But what if try to solve the problem of mindblowing file_id = .index?
You could put file handler and current line together into an Array or Hash, but I don't feel like it makes code better:

files_and_lines = ARGV.map(&File.method(:open)).map{ |file| {file:file, line:file.gets} }
...
loop do
files_and_lines.select!{ |tuple| tuple[:line] }
break if files_and_lines.empty?
next_tuple = files_and_lines.min_by{ |tuple| tuple[:line] }
merged_file.print next_tuple[:line]
next_tuple[:line] = next_tuple[:file].gets
end
...


Trying to make it shorter didn't work for me: converting tuples into Hash pair file->line can't easily return file for line and line->file doesn't look better either because to edit a key you have to have some temporal array variable:

files_and_lines = Hash[ ARGV.map(&File.method(:open)).map{ |file| [file.gets, file] } ]
...
loop do
break if files_and_lines.keep_if{ |line, file| line }.empty?
array = files_and_lines.to_a.sort_by &:first
merged_file.print array[0][0]
array[0][0] = array[0][1].gets
files_and_lines = Hash[array]
end
...


So just leave Britney alone don't touch it -- .index is ok.

• Use parentheses for non-system function calls, you don't do this practically anywhere in your code. – Devon Parsons Jun 9 '15 at 13:13
• @DevonParsons, why should I do what some kid on github says? – Nakilon Jun 9 '15 at 14:01
• It's not some kid, it's the community driven accepted style guide. – Devon Parsons Jun 9 '15 at 14:05
• Really, because the rule I linked to tells you exactly what to do. You seem upset about this. – Devon Parsons Jun 9 '15 at 14:58
• The problem of modern coders is that they do not realise who is worth to listen to and who is not. And if you think code reviewing is about parentheses, you probably don't know what the good code actually is. – Nakilon Jun 9 '15 at 15:38

It took me about five minutes to figure out, then I found that I got it wrong. At first, I thought that it would take lines from each named file in turn. Then, I thought that it would take the first paragraph from each file, before realizing that an "empty" line still consists of a "\n" and therefore wouldn't get compacted out. In the end, I decided that it's just concatenating all the files. Then I ran the code to discover that it does one round of mergesort.

• This is exactly how I read the code, except that I never ran it and discovered that it ran just one round of merge sort. The mental disconnect for me was seeing those file opens. I expected the code below to deal with the files in their entirety, not just part of them. – Wayne Conrad Jan 8 '14 at 1:41
• What does "one round of mergesort" mean? The code is supposed to deal with the files in their entirety... my understanding is that it does. – GreenAsJade Jan 8 '14 at 3:49
• I mean that if each of the files is already individually sorted, then the code in this question lists all of the lines in sorted order. – 200_success Jan 8 '14 at 5:35
• That some of us were so confused about what it does is an indication that, at least for us, the code is too clever. – Wayne Conrad Jan 9 '14 at 0:35
• On the other hand, I'm not sure how to go about improving it. – 200_success Jan 9 '14 at 0:38

I am a beginner and I don't understand what it does.

• I deleted this post when I saw it in the low-quality review queue. On further reflection, it is more relevant as an answer than I initially thought. Apologies for any confusion. – rolfl Jun 10 '15 at 0:26
• @rolf I had the same thought and would have voted to close/delete if you didn't comment here. – Heslacher Jun 11 '15 at 8:11