I have a string that only contains one number on either side of "-", like:
"1-3"
I want to get a result like
"01-03"
If the string had two numbers on one side of the dash like:
"1-10"
then I don't want to make any substitutions. I could do a gsub expression like
str.gsub!(/(^|[^\d]])\d[[:space:]]*\-[[:space:]]*\d([^\d]|$)/, '\1')
but I'm not clear how to do it if there are multiple (e.g. two) things to substitute.
You could probably get away with this:
def dashreplace(str)
str.sub(/\b(\d)\-(\d)\b/) do |s|
'%02d-%02d' % [ $1.to_i, $2.to_i ]
end
end
dashreplace('1-2')
# => "01-02"
dashreplace('1-20')
# => "1-20"
dashreplace('99-1,2-3')
# => "99-1,02-03"
Is there really a need to use regex here, at all? Seems like an over-complication to me. Assuming you know the string will be in the format: <digits><hyphen><digits>, you could do:
def pad_digits(string)
left_digits, right_digits = string.split('-')
if left_digits.length > 1 || right_digits.length > 1
string
else
"%02d-%02d" % [left_digits, right_digits]
end
end
pad_digits("1-3") # => "01-03"
pad_digits("1-10") # => "1-10"
This is a variant of #TomLord's answer.
def pad_single_digits(str)
str.size > 3 ? str : "0%d-0%d" % str.split('-')
end
pad_single_digits "1-3" #=> "01-03"
pad_single_digits "1-10" #=> "1-10"
"0%s-0%s" also works.
You can do:
def nums(s)
rtr=s[/^(\d)(\D+)(\d)$/] ? '0%s%s0%s' % [$1,$2,$3] : s
end
Related
I need to get the expected output in ruby by using any method like scan or match.
Input string:
"http://test.com?t&r12=1&r122=1&r1=1&r124=1"
"http://test.com?t&r12=1&r124=1"
Expected:
r12=1,r122=1, r1=1, r124=1
r12=1,r124=1
How can I get the expected output using regex?
Use regex /r\d+=\d+/:
"http://test.com?t&r12=1&r122=1&r1=1&r124=1".scan(/r\d+=\d+/)
# => ["r12=1", "r122=1", "r1=1", "r124=1"]
"http://test.com?t&r12=1&r124=1".scan(/r\d+=\d+/)
# => ["r12=1", "r124=1"]
You can use join to get a string output. Here:
"http://test.com?t&r12=1&r122=1&r1=1&r124=1".scan(/r\d+=\d+/).join(',')
# => "r12=1,r122=1,r1=1,r124=1"
Update
If the URL contains other parameters that may include r in end, the regex can be made stricter:
a = []
"http://test.com?r1=2&r12=1&r122=1&r1=1&r124=1&ar1=2&tr2=3&xy4=5".scan(/(&|\?)(r+\d+=\d+)/) {|x,y| a << y}
a.join(',')
# => "r12=1,r122=1,r1=1,r124=1"
While input strings are urls with queries, I would safeguard myself from the false positives:
input = "http://test.com?t&r12=1&r122=1&r1=1&r124=1"
query_params = input.split('?').last.split('&')
#⇒ ["t", "r12=1", "r122=1", "r1=1", "r124=1"]
r_params = query_params.select { |e| e =~ /\Ar\d+=\d+/ }
#⇒ ["r12=1", "r122=1", "r1=1", "r124=1"]
r_params.join(',')
#⇒ "r12=1,r122=1,r1=1,r124=1"
It’s safer than just scan the original input for any regexp.
If you really need to do it with regex correctly, you'll need to use a regex like this:
puts "http://test.com?t&r12=1&r122=1&r1=1&r124=1".scan(/(?:http.*?\?t|(?<!^)\G)\&*(\br\d*=\d*)(?=.*$)/i).join(',')
puts "http://test.com?t&r12=1&r124=1".scan(/(?:http.*?\?t|(?<!^)\G)\&*(\br\d*=\d*)(?=.*$)/i).join(',')
Sample program output:
r12=1,r122=1,r1=1,r124=1
r12=1,r124=1
I need a one line gsub to replace all the non-digits in a string but only if the non-digits are not more than three and if the total length of the digits is 10
I have this which fits the first condition
p "0177/385490".gsub(/((\d+)\D?(\d+)\D?(\d+)\D?+(\d+))/,'\2\3\4\5')
#=>"0177385490"
but when i try this the {10} check doesn't work
p "0177/385490".gsub(/((\d+)\D?(\d+)\D?(\d+)\D?+(\d+)){10}/,'\2\3\4\5')
#=>"0177/385490"
how to do this please ?
EDIT
i managed to to it like this, but how to do this in a oneline gsub ?
strings = [
"0473/385 490",
"0473/385490",
"0473 38 54 90",
"0473/385 4901" #this one is't captured
]
strings.each do |s|
if /((\d+)\D?(\d+)\D?(\d+)\D?+(\d+))/ =~ s
if "#{$2}#{$3}#{$4}#{$5}".length == 10
puts "#{$2}#{$3}#{$4}#{$5}"
end
end
end
EDIT: to show why it really needs to be a onle line gsub here my routine, there will be more replacements added
def cleanup text
replacements = [
{:pattern => /(04\d{2}) (\d{2}) (\d{2}) (\d{2})/, :replace_with => '\1\2\3\4'},
{:pattern => /(0\d)(\/| |-)(\d{3}) (\d{2}) (\d{2})/, :replace_with => '\1\3\4\5'},
{:pattern => /(\d{6} )(\d{3})-(\d{2})/, :replace_with => '\1\2 \3'},
{:pattern => /(\d{2,4})\D?(\d{2,3})\D?(\d{2,3})/, :replace_with => '\1\2\3'}
].each{|replacement|text.gsub!(replacement[:pattern], replacement[:replace_with])}
text
end
I think a one-line gsub wouldn't be overly readable. Here's my approach:
chars, non_chars = s.each_char.partition { |c| c =~ /\d/ }
puts chars.join if chars.size == 10 && non_chars.size <= 3
Clean and easy to read, without any magic variables. Plus it clearly shows the rules you have imposed on the string.
Here's a one-liner with gsub, mostly to illustrate why Michael Kohl's approach is better:
(digits = s.gsub(/\D/, '')).length == 10 && s.length < 14 ? digits : s
You may use something like this:
puts s.gsub(/\D/, '') if (/\A(\d\D?){10}\z/ =~ s) && (/\A(\d+\D){0,3}\d*\z/ =~ s)
You might also want to know about the scan method.
strings.each do |s|
numbers = s.scan(/\d/).join
non_numbers = s.scan(/\D/)
puts numbers if numbers.length == 10 && non_numbers.length < 4
end
But I like the solution by #MichaelKohl better.
And then a silly example:
strings.select{|s| s.scan(/\D/).length < 4}.map{|s| s.scan(/\d/).join}.select{|s| s.length==10}
Thanks everyone but i can't use the answers because i can't insert them in my routine (edited my answer to make that more clear). Found a sollution myself. I give everyone an upvote who had a one line solution as requested, now i still need to find a way to insert my block as a replacementpattern in the cleanup routine
p "0177/3854901".gsub(/(\d+)\D?(\d+)\D?(\d+)\D?+(\d+)/){ |m| "#{$1}#{$2}#{$3}#{$4}".length==10 ? "#{$1}#{$2}#{$3}#{$4}":m}
#=> "0177/3854901" isn't replaced because it has 11 digits
p "0177/385490".gsub(/(\d+)\D?(\d+)\D?(\d+)\D?+(\d+)/){ |m| "#{$1}#{$2}#{$3}#{$4}".length==10 ? "#{$1}#{$2}#{$3}#{$4}":m}
#=> "0177385490"
I'm playing around with Ruby to do some file versioning for me. I have a string 2.0.0.65 . I split it up, increment the build number (65 --> 66) then I want to replace the 65 with the 66. In this replace though, I only want to replace the last match of the string. What's the best way in Ruby to do this?
version_text = IO.read('C:\\Properties')
puts version_text
version = version_text.match(/(\d+\.\d+\.\d+\.\d+)/)[1]
puts version
build_version = version.split('.')[3]
puts build_version
incremented_version = build_version.to_i + 1
puts incremented_version`
...
If you just want to increment the integer at the very end of a string then try this:
s = '2.0.0.65'
s.sub(/\d+\Z/) {|x| x.to_i + 1} # => '2.0.0.66'
You can do something like this:
parts = "2.0.0.65".split('.')
parts[3] = parts[3].to_i + 1
puts parts.join(".")
output:
2.0.0.66
This gives you more control over just using a string replacement method, as now you can increment other parts of the version string if needed more easily.
Once you have the string with the build number, you only need to use 'succ' method
'2.0.0.65'.succ()
Which gives you the string
'2.0.0.66'
sample = '2.0.0.65'
def incr_version(version)
parts = version.split('.')
parts[-1] = parts[-1].to_i + 1
parts.join('.')
end
incr_version(sample) # => '2.0.0.66'
For fun, if you want to increment the last integer in any string you could do this:
str = "I have 3 cats and 41 rabbits"
str.reverse.sub(/\d+/){ |s| (s.reverse.to_i+1).to_s.reverse }.reverse
#=> "I have 3 cats and 42 rabbits"
This is only valid when you modify your regex to match the reversed version of the text.
More generally, you can do this:
class String
# Replace the last occurrence of a regex in a string.
# As with `sub` you may specify matches in the replacement string,
# or pass a block instead of the replacement string.
# Unlike `sub` the captured sub-expressions will be passed as
# additional parameters to your block.
def rsub!(pattern,replacement=nil)
if n=rindex(pattern)
found=match(pattern,n)
self[n,found[0].length] = if replacement
replacement.gsub(/\\\d+/){ |s| found[s[1..-1].to_i] || s }
else
yield(*found).to_s
end
end
end
def rsub(pattern,replacement=nil,&block)
dup.tap{ |s| s.rsub!(pattern,replacement,&block) }
end
end
str = "I have 3 cats and 41 rabbits"
puts str.rsub(/(?<=\D)(\d+)/,'xx')
#=> I have 3 cats and xx rabbits
puts str.rsub(/(?<=\D)(\d+)/,'-\1-')
#=> I have 3 cats and -41- rabbits
puts str.rsub(/(?<=\D)(\d+)/){ |n| n.to_i+1 }
#=> I have 3 cats and 42 rabbits
Note that (as with rindex) because the regex search starts from the end of the string you may need to make a slightly more complex regex to force your match to be greedy.
I have such code
reg = /(.+)_path/
if reg.match('home_path')
puts reg.match('home_path')[0]
end
This will eval regex twice :(
So...
reg = /(.+)_path/
result = reg.match('home_path')
if result
puts result[0]
end
But it will store variable result in memory till.
I have one functional-programming idea
/(.+)_path/.match('home_path').compact.each do |match|
puts match[0]
end
But seems there should be better solution, isn't it?
There are special global variables (their names start with $) that contain results of the last regexp match:
r = /(.+)_path/
# $1 - the n-th group of the last successful match (may be > 1)
puts $1 if r.match('home_path')
# => home
# $& - the string matched by the last successful match
puts $& if r.match('home_path')
# => home_path
You can find full list of predefined global variables here.
Note, that in the examples above puts won't be executed at all if you pass a string that doesn't match the regexp.
And speaking about general case you can always put assignment into condition itself:
if m = /(.+)_path/.match('home_path')
puts m[0]
end
Though, many people don't like that as it makes code less readable and gives a good opportunity for confusing = and ==.
My personal favorite (w/ 1.9+) is some variation of:
if /(?<prefix>.+)_path/ =~ "home_path"
puts prefix
end
If you really want a one-liner: puts /(?<prefix>.+)_path/ =~ 'home_path' ? prefix : false
See the Ruby Docs for a few limitations of named captures and #=~.
From the docs: If a block is given, invoke the block with MatchData if match succeed.
So:
/(.+)_path/.match('home_path') { |m| puts m[1] } # => home
/(.+)_path/.match('homepath') { |m| puts m[1] } # prints nothing
How about...
if m=/regex here/.match(string) then puts m[0] end
A neat one-line solution, I guess :)
how about this ?
puts $~ if /regex/.match("string")
$~ is a special variable that stores the last regexp match. more info: http://www.regular-expressions.info/ruby.html
Actually, this can be done with no conditionals at all. (The expression evaluates to "" if there is no match.)
puts /(.+)_path/.match('home_xath').to_a[0].to_s
How can I get the first character in a string using Ruby?
Ultimately what I'm doing is taking someone's last name and just creating an initial out of it.
So if the string was "Smith" I just want "S".
You can use Ruby's open classes to make your code much more readable. For instance, this:
class String
def initial
self[0,1]
end
end
will allow you to use the initial method on any string. So if you have the following variables:
last_name = "Smith"
first_name = "John"
Then you can get the initials very cleanly and readably:
puts first_name.initial # prints J
puts last_name.initial # prints S
The other method mentioned here doesn't work on Ruby 1.8 (not that you should be using 1.8 anymore anyway!--but when this answer was posted it was still quite common):
puts 'Smith'[0] # prints 83
Of course, if you're not doing it on a regular basis, then defining the method might be overkill, and you could just do it directly:
puts last_name[0,1]
If you use a recent version of Ruby (1.9.0 or later), the following should work:
'Smith'[0] # => 'S'
If you use either 1.9.0+ or 1.8.7, the following should work:
'Smith'.chars.first # => 'S'
If you use a version older than 1.8.7, this should work:
'Smith'.split(//).first # => 'S'
Note that 'Smith'[0,1] does not work on 1.8, it will not give you the first character, it will only give you the first byte.
"Smith"[0..0]
works in both ruby 1.8 and ruby 1.9.
For completeness sake, since Ruby 1.9 String#chr returns the first character of a string. Its still available in 2.0 and 2.1.
"Smith".chr #=> "S"
http://ruby-doc.org/core-1.9.3/String.html#method-i-chr
In MRI 1.8.7 or greater:
'foobarbaz'.each_char.first
Try this:
>> a = "Smith"
>> a[0]
=> "S"
OR
>> "Smith".chr
#=> "S"
In Rails
name = 'Smith'
name.first
>> s = 'Smith'
=> "Smith"
>> s[0]
=> "S"
Another option that hasn't been mentioned yet:
> "Smith".slice(0)
#=> "S"
Because of an annoying design choice in Ruby before 1.9 — some_string[0] returns the character code of the first character — the most portable way to write this is some_string[0,1], which tells it to get a substring at index 0 that's 1 character long.
Try this:
def word(string, num)
string = 'Smith'
string[0..(num-1)]
end
If you're using Rails You can also use truncate
> 'Smith'.truncate(1, omission: '')
#=> "S"
or for additional formatting:
> 'Smith'.truncate(4)
#=> "S..."
> 'Smith'.truncate(2, omission: '.')
#=> "S."
While this is definitely overkill for the original question, for a pure ruby solution, here is how truncate is implemented in rails
# File activesupport/lib/active_support/core_ext/string/filters.rb, line 66
def truncate(truncate_at, options = {})
return dup unless length > truncate_at
omission = options[:omission] || "..."
length_with_room_for_omission = truncate_at - omission.length
stop = if options[:separator]
rindex(options[:separator], length_with_room_for_omission) || length_with_room_for_omission
else
length_with_room_for_omission
end
"#{self[0, stop]}#{omission}"
end
Other way around would be using the chars for a string:
def abbrev_name
first_name.chars.first.capitalize + '.' + ' ' + last_name
end
Any of these methods will work:
name = 'Smith'
puts name.[0..0] # => S
puts name.[0] # => S
puts name.[0,1] # => S
puts name.[0].chr # => S