I have an array:
a = ["us.production => 1", "us.stats => 1", "us.stats.total_active => 1", "us.stats.inactive => 0"]
How can I modify it into a hash object? e.g.:
h = {"us.production" => 1, "us.stats" => 1, "us.stats.total_active" => 1, "us.stats.inactive" => 0}
Thank you,
If pattern you are having is proper and constant, you can try following,
h = a.map { |x| x.split(' => ') }.to_h
# => {"us.production"=>"1", "us.stats"=>"1", "us.stats.total_active"=>"1", "us.stats.inactive"=>"0"}
Instead it is better to use split(/\s*=>\s*/) instead of split(' => ')
You can split every string with String#split and then convert an array of pairs into a hash with Array#to_h:
a = ["us.production => 1", "us.stats => 1", "us.stats.total_active => 1", "us.stats.inactive => 0"]
pairs = a.map{|s| s.split(/\s*=>\s*/)}
# => [["us.production", "1"], ["us.stats", "1"], ["us.stats.total_active", "1"], ["us.stats.inactive", "0"]]
pairs.to_h
# => {"us.production"=>"1", "us.stats"=>"1", "us.stats.total_active"=>"1", "us.stats.inactive"=>"0"}
/\s*=>\s*/ is a regular expression that matches first any number of whitespaces with \s*, then => then again any nuber of whitespaces. As it's a String#split delimiter, this part of string won't be present in a string pair.
The other answers to date are incorrect because they leave the values as strings, whereas the spec is that they be integers. That can be easily corrected. One way is to change s.split(/\s*=>\s*/) in #mrzasa's answer to k,v = s.split(/\s*=>\s*/); [k,v.to_i]. Another way is to tack .transform_values(&:to_i) to the ends of the expressions given in those answers. I expect the authors of those answers either didn't notice that integers were required or intended to leave it as an exercise for the OP to do the (rather uninteresting) conversion.
To make a single pass through the array and avoid the creation of a temporary array and local variables (other than block variables), I suggest using Enumerable#each_with_object (rather than map and to_h), and use regular expressions to extract both keys and values (rather than using String#split):
a = ["us.production => 1", "us.stats=>1", "us.stats.total_active => 1"]
a.each_with_object({}) { |s,h| h[s[/.*[^ ](?= *=>)/]] = s[/\d+\z/].to_i }
#=> {"us.production"=>1, "us.stats"=>1, "us.stats.total_active"=>1}
The first regular expression reads, "match zero or more characters (.*) followed by a character that is not a space ([^ ]), provided that is followed by zero or more spaces (*) followed by the string "=>". (?= *=>) is a positive lookahead.
The second regular expression reads, "match one or more digits (\d+) at the end of the string (the anchor \z). If that string could represent a negative integer, change that regex to /-?\d+\z/ (? makes the minus sign optional).
Related
For example, I have this string of only numbers:
0009102
If I convert it to integer Ruby automatically gives me this value:
9102
That's correct. But my program gives me different types of numbers:
2229102 desired output => 9102
9999102 desired output => 102
If you look at them I have treated 2 and 9 as zeros since they are automatically deleted, well, it is easy to delete that with an while but I must avoid it.
In other words, how do you make 'n' on the left be considered a zero for Ruby?
"2229102".sub(/\A(\d)\1*/, "") #=> "9102"`.
The regular expression reads, "match the first digit in the string (\A is the beginning-of-string anchor) in capture group 1 ((\d)), followed by zero or more characters (*) that equal the contents of capture group 1 (\1). String#gsub converts that match to an empty string.
Try with Enumerable#chunk_while:
s = '222910222'
s.each_char.chunk_while(&:==).drop(1).join
#=> "910222"
Where s.each_char.chunk_while(&:==).to_a #=> [["2", "2", "2"], ["9"], ["1"], ["0"], ["2", "2", "2"]]
Similar to the solution of iGian you could also use drop_while.
s = '222910222'
s.each_char.each_cons(2).drop_while { |a, b| a == b }.map(&:last).join
#=> "910222"
# or
s.each_char.drop_while.with_index(-1) { |c, i| i < 0 || c == s[i] }.join
#=> "910222"
You can also try this way:
s = '9999102938'
s.chars.then{ |chars| chars[chars.index(chars.uniq[1])..-1] }.join
=> "102938"
In ruby, I want to substitute some letters in a string, is there a better way of doing this?
string = "my random string"
string.gsub(/a/, "#").gsub(/i/, "1").gsub(/o/, "0")`
And if I want to substitute both "a" and "A" with a "#", I know I can do .gsub(/a/i, "#"), but what if I want to substitute every "a" with an "e" and every "A" with an "E"? Is there a way of abstracting it instead of specifying both like .gsub(/a/, "e").gsub(/A/, "E")?
You can use a Hash. eg:
h = {'a' => '#', 'b' => 'B', 'A' => 'E'}
"aAbBcC".gsub(/[abA]/, h)
# => "#EBBcC"
Not really an answer to your question, but more an other way to proceed: use the translation:
'aaAA'.tr('aA', 'eE')
# => 'eeEE'
For the same transformation, you can also use the ascii table:
'aaAA'.gsub(/a/i) {|c| (c.ord + 4).chr}
# => 'eeEE'
other example (the last character is used by default):
'aAaabbXXX'.tr('baA', 'B#')
# => '####BBXXX'
Here are two variants of #Santosh's answer:
str ="aAbBcC"
h = {'a' => '#', 'b' => 'B', 'A' => 'E'}
#1
str.gsub(/[#{h.keys.join}]/, h) #=> "#EBBcC"
#2
h.default_proc = ->(_,k) { k }
str.gsub(/./, h) #=> "#EBBcC"
These offer better maintainability should h could change in future
You can also pass gsub a block
"my random string".gsub(/[aoi]/) do |match|
case match; when "a"; "#"; when "o"; "0"; when "i"; "I" end
end
# => "my r#nd0m strIng"
The use of a hash is of course much more elegant in this case, but if you have complex rules of substitution it can come in handy to dedicate a class to it.
"my random string".gsub(/[aoi]/) {|match| Substitute.new(match).result}
# => "my raws0m strAINng"
Based on my hash, I want to match it if it's in the string:
def conv
str = "I only have one, two or maybe sixty"
hash = {:one => 1, :two => 2, :six => 6, :sixty => 60 }
str.match( Regexp.union( hash.keys.to_s ) )
end
puts conv # => <blank>
The above does not work but this only matches "one":
str.match( Regexp.union( hash[0].to_s ) )
Edited:
Any idea how to match "one", "two" and sixty in the string exactly?
If my string has "sixt" it return "6" and that should not happen based on #Cary's answer.
You need to convert each element of hash.keys to a string, rather than converting the array hash.keys to a string, and you should use String#scan rather than String#match. You may also need to play around with the regex until it returns everyhing you want and nothing you don't want.
Let's first look at your example:
str = "I only have one, two or maybe sixty"
hash = {:one => 1, :two => 2, :six => 6, :sixty => 60}
We might consider constructing the regex with word breaks (\b) before and after each word we wish to match:
r0 = Regexp.union(hash.keys.map { |k| /\b#{k.to_s}\b/ })
#=> /(?-mix:\bone\b)|(?-mix:\btwo\b)|(?-mix:\bsix\b)|(?-mix:\bsixty\b)/
str.scan(r0)
#=> ["one", "two", "sixty"]
Without the word breaks, scan would return ["one", "two", "six"], as "sixty" in str would match "six". (Word breaks are zero-width. One before a string requires that the string be preceded by a non-word character or be at the beginning of the string. One after a string requires that the string be followed by a non-word character or be at the end of the string.)
Depending on your requirements, word breaks may not be sufficient or suitable. Suppose, for example (with hash above):
str = "I only have one, two, twenty-one or maybe sixty"
and we do not wish to match "twenty-one". However,
str.scan(r0)
#=> ["one", "two", "one", "sixty"]
One option would be to use a regex that demands that matches be preceded by whitespace or be at the beginning of the string, and be followed by whitespace or be at the end of the string:
r1 = Regexp.union(hash.keys.map { |k| /(?<=^|\s)#{k.to_s}(?=\s|$)/ })
str.scan(r1)
#=> ["sixty"]
(?<=^|\s) is a positive lookbehind; (?=\s|$) is a positive lookahead.
Well, that avoided the match of "twenty-one" (good), but we no longer matched "one" or "two" (bad) because of the comma following each of those words in the string.
Perhaps the solution here is to first remove punctuation, which allows us to then apply either of the above regexes:
str.tr('.,?!:;-','')
#=> "I only have one two twentyone or maybe sixty"
str.tr('.,?!:;-','').scan(r0)
#=> ["one", "two", "sixty"]
str.tr('.,?!:;-','').scan(r1)
#=> ["one", "two", "sixty"]
You may also want to change / at the end of the regex to /i to make the match insensitive to case.1
1 Historical note for readers who want to know why 'a' is called lower case and 'A' is called upper case.
I want to use regexes to check if a given string is composed of certain substrings.
For example, given the regular expression
> regex = /(?:(foo)|(bar)|(baz))*/
I can determine whether a given string matches the pattern:
> regex === "bazbar"
=> true
> regex === "qux"
=> false
But I want to know how to break the string into substrings. I can almost do this with
> regex.match("barbazfoo").captures
=> ["foo", "bar", "baz"]
But here they appear in the order in which I specified them within the regex. I want to return
["bar", "baz", "foo"]
In the order in which they appeared in the string.
You can use String#scan with a modified regular expression:
regex = /foo|bar|baz/
"barbazfoo".scan(regex)
# => ["bar", "baz", "foo"]
UPDATE according to OP's comment.
If some of the strings I'm using are substrings of the others, you need to order the so that all the substrings go last.
"barfoo".scan(/ba|bar|foo/) # without ordering
# => ["ba", "foo"]
words = ['ba', 'bar', 'foo']
pattern = words.map { |word| Regexp.escape(word) }.sort_by { |x| -x.size }.join('|')
"barfoo".scan(Regexp.new(pattern))
# => ["bar", "foo"]
I'd like to take a variable that I have, and turn it into an array separated by the character of my choosing. In the example below, that separator is %
dump = "1%2%3%apple%car%yellow"
into
Array= [1,2,3,apple,car,yellow]
Use String#split:
"1%2%3%apple%car%yellow".split('%')
# => ["1", "2", "3", "apple", "car", "yellow"]
(Note that every element of the returned array is a string, even the ones containing digits.)
From the docs:
split (pattern=$;, [limit]) → anArray
Divides
str into substrings based on a delimiter, returning an array of these
substrings.
You can pass a string like above ('%'), or a regular expression.