Is it implied by default in str.scan? Is it off by default in str[regex] ?
Yes, how often the regex is applied depends on the method used, not on the regex's flags.
scan will return an array containing (or iterate over) all matches of the regex. match and String#[] will return the first match. =~ will return the index of the first match. gsub will replace all occurrences of the regex and sub will replace the first occurence.
smotchkkiss:~$ irb
>> 'Foobar does not like food because he is a fool'.gsub(/foo/i, 'zim')
=> "zimbar does not like zimd because he is a ziml"
Related
Suppose I have the following string:
mystring = "start/abc123/end"
How can you splice out the abc123 with something else, while leaving the "/start/" and "/end" elements intact?
I had the following to match for the pattern, but it replaces the entire string. I was hoping to just have it replace the abc123 with 123abc.
mystring.gsub(/start\/(.*)\/end/,"123abc") #=> "123abc"
Edit: The characters between the start & end elements can be any combination of alphanumeric characters, I changed my example to reflect this.
You can do it using this character class : [^\/] (all that is not a slash) and lookarounds
mystring.gsub(/(?<=start\/)[^\/]+(?=\/end)/,"7")
For your example, you could perhaps use:
mystring.gsub(/\/(.*?)\//,"/7/")
This will match the two slashes between the string you're replacing and putting them back in the substitution.
Alternatively, you could capture the pieces of the string you want to keep and interpolate them around your replacement, this turns out to be much more readable than lookaheads/lookbehinds:
irb(main):010:0> mystring.gsub(/(start)\/.*\/(end)/, "\\1/7/\\2")
=> "start/7/end"
\\1 and \\2 here refer to the numbered captures inside of your regular expression.
The problem is that you're replacing the entire matched string, "start/8/end", with "7". You need to include the matched characters you want to persist:
mystring.gsub(/start\/(.*)\/end/, "start/7/end")
Alternatively, just match the digits:
mystring.gsub(/\d+/, "7")
You can do this by grouping the start and end elements in the regular expression and then referring to these groups in in the substitution string:
mystring.gsub(/(?<start>start\/).*(?<end>\/end)/, "\\<start>7\\<end>")
Take this string.
a = "real-ab(+)real-bc(+)real-cd-xy"
a.scan(/[a-z_0-9]+\-[a-z_0-9]+[\-\[a-z_0-9]+\]?/)
=> ["real-ab", "real-bc", "real-cd-xy"]
But how come this next string gets nothing?
a = "real-a(+)real-b(+)real-c"
a.scan(/[a-z_0-9]+\-[a-z_0-9]+[\-\[a-z_0-9]+\]?/)
=> []
How can I have it so both strings output into a 3 count array?
You've confused parentheses (used for grouping) and square brackets (used for character classes). You want
a.scan(/[a-z_0-9]+-[a-z_0-9]+(?:-[a-z_0-9]+)?/)
(?:...) creates a non-capturing group which is what you need here.
Furthermore, unless you want to disallow uppercase letters explicitly, you can write \w as a shorthand for "a letter, digit or underscore":
a.scan(/\w+-\w+(?:-\w+)?/)
a.scan(/[a-z_0-9]+\-[a-z_0-9]+/)
Why not simply?
a.scan(/[a-z_0-9\-]+/)
Apparently I still don't understand exactly how it works ...
Here is my problem: I'm trying to match numbers in strings such as:
910 -6.258000 6.290
That string should gives me an array like this:
[910, -6.2580000, 6.290]
while the string
blabla9999 some more text 1.1
should not be matched.
The regex I'm trying to use is
/([-]?\d+[.]?\d+)/
but it doesn't do exactly that. Could someone help me ?
It would be great if the answer could clarify the use of the parenthesis in the matching.
Here's a pattern that works:
/^[^\d]+?\d+[^\d]+?\d+[\.]?\d+$/
Note that [^\d]+ means at least one non digit character.
On second thought, here's a more generic solution that doesn't need to deal with regular expressions:
str.gsub(/[^\d.-]+/, " ").split.collect{|d| d.to_f}
Example:
str = "blabla9999 some more text -1.1"
Parsed:
[9999.0, -1.1]
The parenthesis have different meanings.
[] defines a character class, that means one character is matched that is part of this class
() is defining a capturing group, the string that is matched by this part in brackets is put into a variable.
You did not define any anchors so your pattern will match your second string
blabla9999 some more text 1.1
^^^^ here ^^^ and here
Maybe this is more what you wanted
^(\s*-?\d+(?:\.\d+)?\s*)+$
See it here on Regexr
^ anchors the pattern to the start of the string and $ to the end.
it allows Whitespace \s before and after the number and an optional fraction part (?:\.\d+)? This kind of pattern will be matched at least once.
maybe /(-?\d+(.\d+)?)+/
irb(main):010:0> "910 -6.258000 6.290".scan(/(\-?\d+(\.\d+)?)+/).map{|x| x[0]}
=> ["910", "-6.258000", "6.290"]
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map(&:to_f)
# => [910.0, -6.258, 6.29]
If you don't want integers to be converted to floats, try this:
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map do |ns|
ns[/\./] ? ns.to_f : ns.to_i
end
# => [910, -6.258, 6.29]
In Ruby, what regex will strip out all but a desired string if present in the containing string? I know about /[^abc]/ for characters, but what about strings?
Say I have the string "group=4&type_ids[]=2&type_ids[]=7&saved=1" and want to retain the pattern group=\d, if it is present in the string using only a regex?
Currently, I am splitting on & and then doing a select with matching condition =~ /group=\d/ on the resulting enumerable collection. It works fine, but I'd like to know the regex to do this more directly.
Simply:
part = str[/group=\d+/]
If you want only the numbers, then:
group_str = str[/group=(\d+)/,1]
If you want only the numbers as an integer, then:
group_num = str[/group=(\d+)/,1].to_i
Warning: String#[] will return nil if no match occurs, and blindly calling nil.to_i always returns 0.
You can try:
$str =~ s/.*(group=\d+).*/\1/;
Typically I wouldn't really worry too much about a complex regex. Simply break the string down into smaller parts and it becomes easier:
asdf = "group=4&type_ids[]=2&type_ids[]=7&saved=1"
asdf.split('&').select{ |q| q['group'] } # => ["group=4"]
Otherwise, you can use regex a bunch of different ways. Here's two ways I tend to use:
asdf.scan(/group=\d+/) # => ["group=4"]
asdf[/(group=\d+)/, 1] # => "group=4"
Try:
str.match(/group=\d+/)[0]
I've got a string Unnecessary:12357927251data and I need to select all data after colon and numbers. I will do it using Regexp.
string.scan(/:\d+.+$/)
This will give me :12357927251data, but can I select only needed information .+ (data)?
Anything in parentheses in a regexp will be captured as a group, which you can access in $1, $2, etc. or by using [] on a match object:
string.match(/:\d+(.+)$/)[1]
If you use scan with capturing groups, you will get an array of arrays of the groups:
"Unnecessary:123data\nUnnecessary:5791next".scan(/:\d+(.+)$/)
=> [["data"], ["next"]]
Use parenthesis in your regular expression and the result will be broken out into an array. For example:
x='Unnecessary:12357927251data'
x.scan(/(:\d+)(.+)$/)
=> [[":12357927251", "data"]]
x.scan(/:\d+(.+$)/).flatten
=> ["data"]
Assuming that you are trying to get the string 'data' from your string, then you can use:
string.match(/.*:\d*(.*)/)[1]
String#match returns a MatchData object. You can then index into that MatchData object to find the part of the string that you want.
(The first element of MatchData is the original string, the second element is the part of the string captured by the parentheses)
Try this: /(?<=\:)\d+.+$/
It changes the colon to a positive look-behind so that it does not appear in the output. Note that the colon alone is a metacharacter and so must be escaped with a backslash.
Using IRB
irb(main):004:0> "Unnecessary:12357927251data".scan(/:\d+(.+)$/)
=> [["data"]]