Ruby Tips, Part 5

80 points by timblair 12 years ago | 13 comments

bstar77 12 years ago |

My favorite Ruby resources: http://www.confreaks.com | http://www.rubytapas.com | https://www.destroyallsoftware.com/screencasts | http://railscasts.com (dated but still useful) | https://www.ruby-toolbox.com

ultimoo 12 years ago |

I think growing a string with << is an expensive operation. Since once the allocated memory is full, it will cost O(n) to copy the entire string to a new location. If we are carrying out m insertions, this could take O(mn).

The way I have been building strings is to store the m strings in an Array and then join() it at the end (which should be only O(n)). The m small objects need to be allocated in both cases, the only difference being that we hold on to those and join them later on.

Am I missing something? I have never measured the performance on either of the these operations so I could be off the mark here.

matsadler 12 years ago | |

For the cases I have benchmarked concatenation with << has been faster than join on an array.

Here's a simple benchmark:

    require "benchmark"

    n = 100_000
    Benchmark.bm(4) do |x|
      x.report("<<") do
        n.times do
          "aaaaa " << "bbbbbb " << "ccccc " << "ddddd " << "eeeee " << "fffff"
        end
      end

      x.report("join") do
        n.times do
          ["aaaaa", "bbbbbb", "ccccc", "ddddd", "eeeee", "fffff"].join(" ")
        end
      end
    end

The results I get for this are:

               user     system      total        real
    <<     0.140000   0.000000   0.140000 (  0.143750)
    join   0.230000   0.000000   0.230000 (  0.228035)

I'd be interested to see if there were any use cases where the relative performance was reversed.

xentronium 12 years ago | | |

    require "benchmark"
    n = 100_000

    A = "foo " * 20
    B = "bar " * 20
    C = "baz " * 20
    D = "foobar " * 20
    STRINGS = [A, B, C, D]

    Benchmark.bm(4) do |x|
      x.report("<<") do
        n.times do
          "" << A << B << C << D
        end
      end

      x.report("join") do
        n.times do
          STRINGS.join
        end
      end
    end


               user     system      total        real
    <<     0.090000   0.010000   0.100000 (  0.102411)
    join   0.070000   0.010000   0.080000 (  0.071142)

danso 12 years ago |

Something in my computer-science-student background feels unsettled when I can use a language for years and still learn new tricks (or often, that I've forgotten that I've learned) about common functionality such as string interpolation...but I do love a lot of Ruby's tricks.

    if path =~ %r{^/assets/mobile/img}
       ...
    end

I knew about `%r` but hadn't realized that it escapes forward-slashes while leaving other regex symbols intact. But with the previous explanation of `%q`, in which arbitrary delimiters can be used, it makes sense...curly braces simply replace `/` as the delimiter, and `/` simply retains its non-specialness in regex...saves me a lot of writing `Regex.escape("/path/to")`.

Another regex notation that I recently learned (elsewhere) and now love to abuse:

     "He is 32-years old"[/\d+/] # => "32"

as opposed to:

     "He is 32-years old".match(/\d+/)[0]

petercooper 12 years ago | |

The extension of that can also be handy and is often overlooked. Contrived example:

    "28 men are 32 years old"[/(\d+) (year|day|month)/, 1] # => "32"

omaranto 12 years ago | |

This explanation of how %r works here is incorrect:

> I knew about `%r` but hadn't realized that it escapes forward-slashes while leaving other regex symbols intact.

The correct explanation can also be found in the next sentence of your comment:

> But with the previous explanation of `%q`, in which arbitrary delimiters can be used, it makes sense...curly braces simply replace `/` as the delimiter, and `/` simply retains its non-specialness in regex.

jebus989 12 years ago | |

Not sure arbitrary regex delimiters can be counted as one of "Ruby's tricks" given they've been implemented in sed for maybe 4 decades?

Xylakant 12 years ago | |

You can also use

   Regexp.new("/foo/bar")

to achieve the desired result but the %r{} shortcut is handier.

rickdale 12 years ago |

These are some good tips. If you are into the Ruby language and want more, I recommend checking out Ruby Tapas http://www.rubytapas.com/ . I have no affiliation with it, just happy customer.

riffraff 12 years ago |

notice that this is wrong

    x ||= y # equivalent to:
    x || x = y

it's actually something slightly different (or at least it was some years ago).

matsadler 12 years ago | |

Here's an example that shows x ||= y behaves like x || x = y, rather than x = x || y

    class Foo
      def foo=(object)
        puts "foo= called with #{object.inspect}"
        @foo = object
      end

      def foo
        puts "foo called"
        @foo
      end
    end

    puts "x || x = y"

    a = Foo.new

    a.foo || a.foo = 1
    a.foo || a.foo = 2

    puts "\nx = x || y"

    b = Foo.new

    b.foo = b.foo || 1
    b.foo = b.foo || 2

    puts "\nx ||= y"

    c = Foo.new

    c.foo ||= 1
    c.foo ||= 2

This outputs:

    x || x = y
    foo called
    foo= called with 1
    foo called

    x = x || y
    foo called
    foo= called with 1
    foo called
    foo= called with 1

    x ||= y
    foo called
    foo= called with 1
    foo called

and you can see that the output for ||= matches the output for x || x = y

riffraff 12 years ago | | |

I didn't mean that it is "x = x || y", I meant "it's completely ad hoc".

For example, if you try "x || x = y" you will get an exception if x is not defined, but "x ||= y" will work fine. The same is true of constants "X || X = 1" will explode while "X ||= 1" will work fine.

require "benchmark" n = 100_000 Benchmark.bm(4) do |x| x.report("<<") do n.times do "aaaaa " << "bbbbbb " << "ccccc " << "ddddd " << "eeeee " << "fffff" end end x.report("join") do n.times do ["aaaaa", "bbbbbb", "ccccc", "ddddd", "eeeee", "fffff"].join(" ") end end end

require "benchmark" n = 100_000 A = "foo " * 20 B = "bar " * 20 C = "baz " * 20 D = "foobar " * 20 STRINGS = [A, B, C, D] Benchmark.bm(4) do |x| x.report("<<") do n.times do "" << A << B << C << D end end x.report("join") do n.times do STRINGS.join end end end user system total real << 0.090000 0.010000 0.100000 ( 0.102411) join 0.070000 0.010000 0.080000 ( 0.071142)

class Foo def foo=(object) puts "foo= called with #{object.inspect}" @foo = object end def foo puts "foo called" @foo end end puts "x || x = y" a = Foo.new a.foo || a.foo = 1 a.foo || a.foo = 2 puts "\nx = x || y" b = Foo.new b.foo = b.foo || 1 b.foo = b.foo || 2 puts "\nx ||= y" c = Foo.new c.foo ||= 1 c.foo ||= 2

x || x = y foo called foo= called with 1 foo called x = x || y foo called foo= called with 1 foo called foo= called with 1 x ||= y foo called foo= called with 1 foo called