{"id":13671,"date":"2022-10-04T22:58:27","date_gmt":"2022-10-04T17:28:27","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-using-the-pig-latin-language-in-ruby-closed\/"},"modified":"2022-10-04T22:58:27","modified_gmt":"2022-10-04T17:28:27","slug":"solved-using-the-pig-latin-language-in-ruby-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-using-the-pig-latin-language-in-ruby-closed\/","title":{"rendered":"[Solved] Using the ‘pig latin language’ in Ruby [closed]"},"content":{"rendered":"

[ad_1]
\n<\/p>\n

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<\/div>\n
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Your translate<\/code> method won’t work. The problem is here:<\/p>\n

if word[0] == \"a\" || \"e\" || \"o\" || \"u\" || \"i\"\n<\/code><\/pre>\n
and<\/p>\n
elsif word[0] != \"a\" || \"e\" || \"o\" || \"u\" || \"i\"\n<\/code><\/pre>\n
You can’t compare that way as the right side of either will not do what you think it will.<\/p>\n
Some simple checks will show why there’s something wrong:<\/p>\n
'abc'[0] == \"a\" || \"e\" || \"o\" || \"u\" || \"i\" # => true\n'efg'[0] == \"a\" || \"e\" || \"o\" || \"u\" || \"i\" # => \"e\"\n'opq'[0] == \"a\" || \"e\" || \"o\" || \"u\" || \"i\" # => \"e\"\n'xyz'[0] == \"a\" || \"e\" || \"o\" || \"u\" || \"i\" # => \"e\"\n\n'abc'[0] != \"a\" || \"e\" || \"o\" || \"u\" || \"i\" # => \"e\"\n'efg'[0] != \"a\" || \"e\" || \"o\" || \"u\" || \"i\" # => true\n'opq'[0] != \"a\" || \"e\" || \"o\" || \"u\" || \"i\" # => true\n'xyz'[0] != \"a\" || \"e\" || \"o\" || \"u\" || \"i\" # => true\n<\/code><\/pre>\n
Why are those wrong? Let’s look at what’s happening:<\/p>\n
When the word starts with ‘a’, the test 'a' == 'a'<\/code> is true:<\/p>\n
'abc'[0] == \"a\" # => true\n<\/code><\/pre>\n
If we ||<\/code> (“or”) true and something, we get true back because it was the first “true” value seen:<\/p>\n
true || \"e\" # => true\n<\/code><\/pre>\n
If the first test failed, then ||<\/code> causes the second test to be evaluated, which in your code was \"e\"<\/code>, and wasn’t a test, but Ruby didn’t know that, and thought it was a “true” return value so it became the result of the expression:<\/p>\n
false || \"e\" # => \"e\"\n<\/code><\/pre>\n
Knowing that, a correct way to write this would be:<\/p>\n
'abc'[0] == \"a\" || 'abc'[0] == \"e\" || 'abc'[0] == \"o\" || 'abc'[0] == \"u\" || 'abc'[0] == \"i\" # => true\n'efg'[0] == \"a\" || 'efg'[0] == \"e\" || 'efg'[0] == \"o\" || 'efg'[0] == \"u\" || 'efg'[0] == \"i\" # => true\n'opq'[0] == \"a\" || 'opq'[0] == \"e\" || 'opq'[0] == \"o\" || 'opq'[0] == \"u\" || 'opq'[0] == \"i\" # => true\n'xyz'[0] == \"a\" || 'xyz'[0] == \"e\" || 'xyz'[0] == \"o\" || 'xyz'[0] == \"u\" || 'xyz'[0] == \"i\" # => false\n\n'abc'[0] != \"a\" && 'abc'[0] != \"e\" && 'abc'[0] != \"o\" && 'abc'[0] != \"u\" && 'abc'[0] != \"i\" # => false\n'efg'[0] != \"a\" && 'efg'[0] != \"e\" && 'efg'[0] != \"o\" && 'efg'[0] != \"u\" && 'efg'[0] != \"i\" # => false\n'opq'[0] != \"a\" && 'opq'[0] != \"e\" && 'opq'[0] != \"o\" && 'opq'[0] != \"u\" && 'opq'[0] != \"i\" # => false\n'xyz'[0] != \"a\" && 'xyz'[0] != \"e\" && 'xyz'[0] != \"o\" && 'xyz'[0] != \"u\" && 'xyz'[0] != \"i\" # => true\n<\/code><\/pre>\n
however, that rapidly becomes hard to read and unwieldy, so something more concise is needed:<\/p>\n
%w[a e o u].include? 'abc'[0] # => true\n%w[a e o u].include? 'efg'[0] # => true\n%w[a e o u].include? 'opq'[0] # => true\n%w[a e o u].include? 'xyz'[0] # => false\n\n!%w[a e o u].include? 'abc'[0] # => false\n!%w[a e o u].include? 'efg'[0] # => false\n!%w[a e o u].include? 'opq'[0] # => false\n!%w[a e o u].include? 'xyz'[0] # => true\n<\/code><\/pre>\n
There is a problem with this though; As the array size increases, more loops are required to compare to the [0]<\/code> value, which slows the code unnecessarily. A regular expression, written correctly, can get rid of that looping so the speed stays very constant:<\/p>\n
'abc'[0][\/[aeou]\/] # => \"a\"\n'efg'[0][\/[aeou]\/] # => \"e\"\n'opq'[0][\/[aeou]\/] # => \"o\"\n'xyz'[0][\/[aeou]\/] # => nil\n<\/code><\/pre>\n
Notice though, that instead of getting true\/false, the results are the character matched by the pattern or nil. In Ruby, only nil and false are considered false values, and everything else is true, so we can translate those into true, true, true, false respectively, but by taking advantage of the !<\/code> operator we can make it even more clear:<\/p>\n
!!'abc'[0][\/[aeou]\/] # => true\n!!'efg'[0][\/[aeou]\/] # => true\n!!'opq'[0][\/[aeou]\/] # => true\n!!'xyz'[0][\/[aeou]\/] # => false\n<\/code><\/pre>\n
It might seem that we’d have to use !!!<\/code> to “not” the results like we’d want when using !=<\/code>, but that isn’t necessary. A single !<\/code> will do the same thing:<\/p>\n
!'abc'[0][\/[aeou]\/] # => false\n!'efg'[0][\/[aeou]\/] # => false\n!'opq'[0][\/[aeou]\/] # => false\n!'xyz'[0][\/[aeou]\/] # => true\n<\/code><\/pre>\n
But wait! There’s more! Even that can be improved upon a slight amount by removing the string slice ([0]<\/code>) and using a regex anchor. Compare these two, and their benchmark:<\/p>\n
require 'fruity'\n\nALPHABET = ('a'..'z').to_a.join\n\ncompare do\n  slice_it  { ALPHABET[0][\/[aeou]\/] }\n  regex_it  { ALPHABET[\/^[aeou]\/] }\nend\n# >> Running each test 8192 times. Test will take about 1 second.\n# >> regex_it is faster than slice_it by 39.99999999999999% \u00b1 10.0%\n<\/code><\/pre>\n
So, using something like:<\/p>\n
'abc'[\/^[aeou]\/]  # => \"a\"\n!'abc'[\/^[aeou]\/] # => false\n!!'abc'[\/^[aeou]\/]  # => true\n<\/code><\/pre>\n
will be fast and compact and let you test to see what a string starts with.<\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            <\/span> <\/p><\/div>\n<\/div>\n
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solved Using the ‘pig latin language’ in Ruby [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Your translate method won’t work. The problem is here: if word[0] == “a” || “e” || “o” || “u” || “i” and elsif word[0] != “a” || “e” || “o” || “u” || “i” You can’t compare that way as the right side of either will not do what you think it will. Some … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[419,455],"class_list":["post-13671","post","type-post","status-publish","format-standard","hentry","category-solved","tag-methods","tag-ruby"],"yoast_head":"\n[Solved] Using the 'pig latin language' in Ruby [closed] - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-using-the-pig-latin-language-in-ruby-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Using the 'pig latin language' in Ruby [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Your translate method won’t work. The problem is here: if word[0] == "a" || "e" || "o" || "u" || "i" and elsif word[0] != "a" || "e" || "o" || "u" || "i" You can’t compare that way as the right side of either will not do what you think it will. Some ... 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