Ruby Array weird syntax [duplicate] - ruby

This question already has answers here:
Can anyone explain this array declaration to me?
(3 answers)
Closed 9 years ago.
Why does this work? (at least on Ruby 2.0)
a = [1,2,]
If I add one more comma I get a syntax error.
Thanks

When defining an array, Ruby allows (but does not require) the last element to have a trailing comma:
a = [1, 2,]
This is especially handy when the array definition is on multiple lines:
a = [
1,
2,
]
With each element on its own line, and each element having a trailing comma, editing the list is trivial: it may be added to, deleted from, reordered, etc., without worrying about the trailing commas, and without having to touch any lines other than the ones you are editing. For example, if you add a new element, you don't have to add a comma to the preceding element.
Two commas in a row are not allowed.
Hashes allow the same convenience:
h = {
:a => 1,
:b => 2,
}

Related

How to remove first 3 characters in a string [duplicate]

This question already has answers here:
How Do I Use VBScript to Strip the First n Characters of a String?
(4 answers)
Closed 6 years ago.
I have a program that prints a string to a notepad file, the output being something random:
#'f7ruhigbergbn
I want to however remove the first 3 characters from the pasted result, how can I do this?
myString = "#'f"
result = workings - myString
This does not work, bare in mind the first 3 characters are always going to be #'f
Any thoughts? thanks
You can use:
result = Mid(workings, 4)
You can use Right function for get the X characters of the right side of string. With Len function you can get the length of the string.
Right(myString,Len(myAtring) - 3)
With this, you get a new string whitout the three first characters, now you can assign to the same string:
myString = Right(myString,Len(myAtring) - 3)
Try this:
mystring = "#'f7ruhigbergbn"
result = Mid(mystring, 3, mystring.length)

Ruby: how to add separator character to the last? [duplicate]

This question already has answers here:
How can I 'join' an array adding to the beginning of the resulting string the first character to join?
(6 answers)
Closed 8 years ago.
Example,
> arr = ['a', 'b', 'c']
> arr.join('-')
=> "a-b-c"
Is there any function to attach one more separator to the last?
> arr.func('-')
=> "a-b-c-"
Thank you.
No, there is no single function like that. You can just hack it like this:
arr.push('').join('-')
If you don't want to change the original array. dup it:
arr.dup.push('').join('-')
You don't actually want a join in this case, you want a reduce (commonly referenced by it's alias, inject):
arr.reduce('') { |concat, entry| concat + entry + '-' }
There are, of course, plenty of other ways of making this work, but spelling it out is less clever, and therefore a lot easier to figure out when you come back to it later (or someone else has to work on it).
Another way (just sayin'):
arr.join.gsub(/./) { |c| c + '-' }
Still like
arr.join('-') << '-'
best for its simplicity.
Modify the array.
arr.map{|c| c.concat("-")}.join

Splitting with empty space in Ruby [duplicate]

This question already has an answer here:
How do I avoid trailing empty items being removed when splitting strings?
(1 answer)
Closed 8 years ago.
In both Ruby and JavaScript I can write expression " x ".split(/[ ]+/)
. In JavaScript I get somehow reasonable result ["", "x", ""], but in Ruby (2.0.0) i get ["", "x"], which is for me quite counterintuitive. I have problems to understand how regular expressions works in Ruby. Why don't I get the same result as in JavaScript or just ["x"]?
From string#split documentation, emphasis my own:
split(pattern=$;, [limit])
If pattern is a String, then its contents are used as the delimiter when splitting str. If pattern is a single space, str is split on whitespace, with leading whitespace and runs of contiguous whitespace characters ignored.
If pattern is a Regexp, str is divided where the pattern matches. Whenever the pattern matches a zero-length string, str is split into individual characters. If pattern contains groups, the respective matches will be returned in the array as well.
If pattern is omitted, the value of $; is used. If $; is nil (which is the default), str is split on whitespace as if ` ' were specified.
If the limit parameter is omitted, trailing null fields are suppressed. If limit is a positive number, at most that number of fields will be returned (if limit is 1, the entire string is returned as the only entry in an array). If negative, there is no limit to the number of fields returned, and trailing null fields are not suppressed.
So if you were to use " x ".split(/[ ]+/, -1) you would get your expected result of ["", "x", ""]
*edited to reflect Wayne's comment
I found this in the C code for String#split, almost right at the end:
if (NIL_P(limit) && lim == 0) {
long len;
while ((len = RARRAY_LEN(result)) > 0 &&
(tmp = RARRAY_AREF(result, len-1), RSTRING_LEN(tmp) == 0))
rb_ary_pop(result);
}
So it actually pops empty strings off the end of the result array before returning! It looks like the creators of Ruby didn't want String#split to return a bunch of empty strings.
Notice the check for NIL_P(limit) -- this accords exactly with what the documentation says, as #dax pointed out.

Using regex backreference value as numeric value in regex [closed]

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 9 years ago.
Improve this question
I've got a string that has variable length sections. The length of the section precedes the content of that section. So for example, in the string:
13JOHNSON,STEVE
The first 2 characters define the content length (13), followed by the actual content. I'd like to be able to parse this using named capture groups with a backreference, but I'm not sure it is possible. I was hoping this would work:
(?<length>\d{2})(?<name>.{\k<length>})
But it doesn't. Seems like the backreference isn't interpreted as a number. This works fine though:
(?<length>\d{2})(?<name>.{13})
No, that will not work of course. You need to recompile your regular expression after extracting the first number.
I would recommend you to use two different expressions:
the first one that extracts number, and the second one that extracts texts basing on the number extracted by the first one.
You can't do that.
>> s = '13JOHNSON,STEVE'
=> "13JOHNSON,STEVE"
>> length = s[/^\d{2}/].to_i # s[0,2].to_i
=> 13
>> s[2,length]
=> "JOHNSON,STEVE"
This really seems like you're going after this the hard way. I suspect the sample string is not as simple as you said, based on:
I've got a string that has variable length sections. The length of the section precedes the content of that section.
Instead I'd use something like:
str = "13JOHNSON,STEVE 08Blow,Joe 10Smith,John"
str.scan(/\d{2}(\S+)/).flatten # => ["JOHNSON,STEVE", "Blow,Joe", "Smith,John"]
If the string can be split accurately, then there's this:
str.split.map{ |s| s[2..-1] } # => ["JOHNSON,STEVE", "Blow,Joe", "Smith,John"]
If you only have length bytes followed by strings, with nothing between them something like this works:
offset = 0
str.delete!(' ') # => "13JOHNSON,STEVE08Blow,Joe10Smith,John"
str.scan(/\d+/).map{ |l| s = str[offset + 2, l.to_i]; offset += 2 + l.to_i ; s }
# => ["JOHNSON,STEVE", "Blow,Joe", "Smith,John"]
won't work if the names have digits in them – tihom
str = "13JOHNSON,STEVE 08Blow,Joe 10Smith,John 1012345,7890"
str.scan(/\d{2}(\S+)/).flatten # => ["JOHNSON,STEVE", "Blow,Joe", "Smith,John", "12345,7890"]
str.split.map{ |s| s[2..-1] } # => ["JOHNSON,STEVE", "Blow,Joe", "Smith,John", "12345,7890"]
With a a minor change, and minor addition it'll continue to work correctly with strings not containing delimiters:
str.delete!(' ') # => "13JOHNSON,STEVE08Blow,Joe10Smith,John1012345,7890"
offset = 0
str.scan(/\d{2}/).map{ |l| s = str[offset + 2, l.to_i]; offset += 2 + l.to_i ; s }.compact
# => ["JOHNSON,STEVE", "Blow,Joe", "Smith,John", "12345,7890"]
\d{2} grabs the numerics in groups of two. For the names where the numeric is a leading length value of two characters, which is according to the OPs sample, the correct thing happens. For a solid numeric "name" several false-positives are returned, which would return nil values. compact cleans those out.
What about this?
a = '13JOHNSON,STEVE'
puts a.match /(?<length>\d{2})(?<name>(.*),(.*))/

Ruby gsub / regex with several arguments [duplicate]

This question already has answers here:
Match a string against multiple patterns
(2 answers)
Closed 8 years ago.
I'm new to ruby and I'm trying to solve a problem.
I'm parsing through several text field where I want to remove the header which has different values. It works fine when the header always is the same:
variable = variable.gsub(/(^Header_1:$)/, '')
But when I put in several arguments it doesn't work:
variable = variable.gsub(/(^Header_1$)/ || /(^Header_2$)/ || /(^Header_3$)/ || /(^Header_4$)/ || /^:$/, '')
You can use Regexp.union:
regex = Regexp.union(
/^Header_1/,
/^Header_2/,
/^Header_3/,
/^Header_4/,
/^:$/
)
variable.gsub(regex, '')
Please note that ^something$ will not work on strings containing something more than something :)
Cause ^ is for matching beginning of string and $ is for end of string.
So i intentionally removed $.
Also you do not need brackets when you only need to remove the matched string.
You can also use it like this:
headers = %w[Header_1 Header_2 Header_3]
regex = Regexp.union(*headers.map{|s| /^#{s}/}, /^\:$/, /etc/)
variable.gsub(regex, '')
And of course you can remove headers without explicitly define them.
Most likely there are a white space after headers?
If so, you can do it as simple as:
variable = "Header_1 something else"
puts variable.gsub(/(^Header[^\s]*)?(.*)/, '\2')
#=> something else
variable = "Header_BLAH something else"
puts variable.gsub(/(^Header[^\s]*)?(.*)/, '\2')
#=> something else
Just use a proper regexp:
variable.gsub(/^(Header_1|Header_2|Header_3|Header_4|:)$/, '')
If the header is always the same format of Header_n, where n is some integer value, then you can simplify your regex greatly:
/Header_\d+/
will find every one of these:
%w[Header_1 Header_2 Header_3].grep(/Header_\d+/)
[
[0] "Header_1",
[1] "Header_2",
[2] "Header_3"
]
Tweaking it to handle finding words, not substrings:
/^Header_\d+$/
or:
/\bHeader_\d+\b/
As mentioned, using Regexp.union is a good start, but, used blindly, can result in very slow or inefficient patterns, so think ahead and help out the engine by giving it useful sub-patterns to work with:
values = %w[foo bar]
/Header_(?:\d+|#{ values.join('|') })/
=> /Header_(?:\d+|foo|bar)/
Unfortunately, Ruby doesn't have the equivalent to Perl's Regexp::Assemble module, which can build highly optimized patterns from big lists of words. Search here on Stack Overflow for examples of what it can do. For instance:
use Regexp::Assemble;
my #values = ('Header_1', 'Header_2', 'foo', 'bar', 'Header_3');
my $ra = Regexp::Assemble->new;
foreach (#values) {
$ra->add($_);
}
print $ra->re, "\n";
=> (?-xism:(?:Header_[123]|bar|foo))

Resources