The string looks like this "[xx],[xx],[xx]"
Where xx is a ploygon like this "(1.0,2.3),(2.0,3)...
Basically, we are looking for a way to get the string between each pair of square brackets into an array.
E.g. String source = "[hello],[1,2],[(1,2),(2,4)]"
would result in an object a such that:
a[0] == 'hello'
a[1] == '1,2'
a[2] == '(1,2),(2,4)'
We have tried various strategies, including using groovy regex:
def p = "[12],[34]"
def points = p =~ /(\[([^\[]*)\])*/
println points[0][2] // yields 12
However,this yields the following 2 dim array:
[[12], [12], 12]
[, null, null]
[[34], [34], 34]
so if we took the 3rd item from every even rows we would be ok, but this does look very correct. We are not talking into account the ',' and we are not sure why we are getting "[12]" twice, when it should be zero times?
Any regex experts out there?
I think that this is what you're looking for:
def p = "[hello],[1,2],[(1,2),(2,4)]"
def points = p.findAll(/\[(.*?)\]/){match, group -> group }
println points[0]
println points[1]
println points[2]
This scripts prints:
hello
1,2
(1,2),(2,4)
The key is the use of the .*? to make the expression non-greedy to found the minimum between chars [] to avoid that the first [ match with the last ] resulting match in hello],[1,2],[(1,2),(2,4) match... then with findAll you returns only the group captured.
Hope it helps,
Related
Examples
"123456" would be ["123", "456"].
"1234567891011" would be ["123", "456", "789", "10", "11"].
I have come up with this logic using regex to solve the challenge but I am being asked if there is a way to make the logic shorter.
def ft(str)
end
The result from the scan gives a lot of whitespaces so after the join operation, I am left with either a double dash or triple dashes so I used this .gsub(/-+/, '-') to fix that. I also noticed sometimes there is a dash at the begin or the end of the string, so I used .gsub(/^-|-$/, '') to fix that too
Any Ideas?
Slice the string in chunks of max 3 digits. (s.scan(/.{1,3}/)
Check if the last chunk has only 1 character. If so, take the last char of the chunk before and prepend it to the last.
Glue the chunks together using join(" ")
Inspired by #steenslag's recommendation. (There are quite a few other ways to achieve the same with varying levels of verbosity and esotericism)
Here is how I would go about it:
def format_str(str)
numbers = str.delete("^0-9").scan(/.{1,3}/)
# there are a number of ways this operation can be performed
numbers.concat(numbers.pop(2).join.scan(/../)) if numbers.last.length == 1
numbers.join('-')
end
Breakdown:
numbers = str.delete("^0-9") - delete any non numeric character from the string
.scan(/.{1,3}/) - scan them into groups of 1 to 3 characters
numbers.concat(numbers.pop(2).join.scan(/../)) if numbers.last.length == 1 - If the last element has a length of 1 then remove the last 2 elements join them and then scan them into groups of 2 and add these groups back to the Array
numbers.join('-') - join the numbers with a hyphen to return a formatted String
Example:
require 'securerandom'
10.times do
s = SecureRandom.hex
puts "Original: #{s} => Formatted: #{format_str(s)}"
end
# Original: fd1bbce41b1c784ce6ad5303d868bbe9 => Formatted: 141-178-465-303-86-89
# Original: af04bd4d4d6beb5a0412a692d5d3d42d => Formatted: 044-465-041-269-253-42
# Original: 9a1833a43cbef51c3f3c21baa66fe996 => Formatted: 918-334-351-332-166-996
# Original: 4104ae13c998cec896997b9919bdafb3 => Formatted: 410-413-998-896-997-991-93
# Original: 0eb49065472240ba32b3c029f897b30d => Formatted: 049-065-472-240-323-029-897-30
# Original: 4c68f9f68e8f6132c0ed5b966d639cf4 => Formatted: 468-968-861-320-596-663-94
# Original: 65987ee04aea8fb533dbe38c0fea7d63 => Formatted: 659-870-485-333-807-63
# Original: aa8aaf1cf59b52c9ad7db6d4b1ae0cbb => Formatted: 815-952-976-410
# Original: 8eb6b457059f91fd06ccbac272db8f4e => Formatted: 864-570-599-106-272-84
# Original: 1c65825ed59dcdc6ec18af969938ea57 => Formatted: 165-825-596-189-699-38-57
That being said to modify your existing code this will work as well:
def format_str(str)
str
.delete("^0-9")
.scan(/(?=\d{5})\d{3}|(?=\d{3}$)\d{3}|\d{2}/)
.join('-')
end
Here are three more ways to do that.
Use String#scan with a regular expression
def fmt(str)
str.delete("^0-9").scan(/\d{2,3}(?!\d\z)/)
end
The regular expression reads, "match two or three digits provided they are not followed by a single digit at the end of the string". (?!\d\z) is a negative lookahead (which is not part of the match). As matches are greedy by default, the regex engine will always match three digits if possible.
Solve by recursion
def fmt(str)
recurse(str.delete("^0-9"))
end
def recurse(s)
case s.size
when 2,3
[s]
when 4
[s[0,2], s[2,2]]
else
[s[0,3], *fmt(s[3..])]
end
end
Determine the last two matches from the size of the string
def fmt(str)
s = str.delete("^0-9")
if s.size % 3 == 1
s[0..-5].scan(/\d{3}/) << s[-4,2] << s[-2,2]
else
s.scan(/\d{2,3}/)
end
end
All methods exhibit the following behaviour.
["5551231234", "18883319", "123456", "1234567891011"].each do |str|
puts "#{str}: #{fmt(str)}"
end
5551231234: ["555", "123", "12", "34"]
18883319: ["188", "833", "19"]
123456: ["123", "456"]
1234567891011: ["123", "456", "789", "10", "11"]
An approach:
def foo(s)
s.gsub(/\D/, '').scan(/\d{1,3}/).each_with_object([]) do |x, arr|
if x.size == 3 || arr == []
arr << x
else
y = arr.last
arr[-1] = y[0...-1]
arr << "#{y[-1]}#{x}"
end
end
end
Remove all non-digits characters, then scan for 1 to 3 digit chunks. Iterate over them. If it's the first time through or the chunk is three digits, add it to the array. If it isn't, take the last digit from the previous chunk and prepend it to the current chunk and add that to the array.
Alternatively, without generating a new array.
def foo(s)
s.gsub(/\D/, '').scan(/\d{1,3}/).instance_eval do |y|
y.each_with_index do |x, i|
if x.size == 1
y[i] = "#{y[i-1][-1]}#{x}"
y[i-1] = y[i-1][0...-1]
end
end
end
end
Without changing your code too much and without adjusting your actual regex, I might suggest replacing scan with split in order to avoid all the extra nil values; replacing gsub with tr which is much faster; and then using reject(&:empty?) to loop through and remove any blank array elements before joining with whatever character you want:
string = "12345fg\n\t\t\t 67"
string.tr("^0-9", "")
.split(/(?=\d{5})(\d{3})|(?=\d{3}$)(\d{3})|(\d{1,2})/)
.reject(&:empty?)
.join("-")
#=> 123-45-67
Not suggesting this is the best approach, but wanted to offer a little food for thought:
You can basically reduce the logic for your challenge to test for 1 single condition and to use 2 very simple pattern matches:
Condition to test for: Number of characters is more than 3 and has a modulo(3) of 1. This condition will require the use of both pattern matches.
All other conditions will use a single pattern match so no reason to test for those.
This could probably be made a little less verbose but it’s all spelled out pretty well for clarity:
def format(s)
n = s.delete("^0-9")
regex_1 = /.{1,3}/
regex_2 = /../
if [n.length-3, 0].max.modulo(3) == 1
a = n[0..-5].scan(regex_1)+n[-4..-1].scan(regex_2)
else a=n.scan(regex_1)
end
a.join("-")
end
I have strings in this format:
object[i].base.base_x[i] and I get lists like List(0,1).
I want to use regular expressions in scala to find the match [i] in the given string and replace the first occurance with 0 and the second with 1. Hence getting something like object[0].base.base_x[1].
I have the following code:
val stringWithoutIndex = "object[i].base.base_x[i]" // basically this string is generated dynamically
val indexReplacePattern = raw"\[i\]".r
val indexValues = List(0,1) // list generated dynamically
if(indexValues.nonEmpty){
indexValues.map(row => {
indexReplacePattern.replaceFirstIn(stringWithoutIndex , "[" + row + "]")
})
else stringWithoutIndex
Since String is immutable, I cannot update stringWithoutIndex resulting into an output like List("object[0].base.base_x[i]", "object[1].base.base_x[i]").
I tried looking into StringBuilder but I am not sure how to update it. Also, is there a better way to do this? Suggestions other than regex are also welcome.
You couldloop through the integers in indexValues using foldLeft and pass the string stringWithoutIndex as the start value.
Then use replaceFirst to replace the first match with the current value of indexValues.
If you want to use a regex, you might use a positive lookahead (?=]) and a positive lookbehind (?<=\[) to assert the i is between opening and square brackets.
(?<=\[)i(?=])
For example:
val strRegex = """(?<=\[)i(?=])"""
val res = indexValues.foldLeft(stringWithoutIndex) { (s, row) =>
s.replaceFirst(strRegex, row.toString)
}
See the regex demo | Scala demo
How about this:
scala> val str = "object[i].base.base_x[i]"
str: String = object[i].base.base_x[i]
scala> str.replace('i', '0').replace("base_x[0]", "base_x[1]")
res0: String = object[0].base.base_x[1]
This sounds like a job for foldLeft. No need for the if (indexValues.nonEmpty) check.
indexValues.foldLeft(stringWithoutIndex) { (s, row) =>
indexReplacePattern.replaceFirstIn(s, "[" + row + "]")
}
I'm attempting to create a regex pattern that will only match strings containing only periods (.) with a length that is a power of 3. Obviously I could manually repeat length checks against powers of 3 up until the length is no longer feasible, but I would prefer a short pattern.
I wrote this python method to help explain what I want to do:
#n = length
def check(n):
if n == 1:
return True
elif n / 3 != n / 3.0:
return False
else:
return check(n / 3)
To clarify, the regex should only match ., ..., ........., ..........................., (length 1, 3, 9, 27) etc.
I've read up on regex recursion, making use of (?R) but I haven't been able to put anything together that works correctly.
Is this possible?
It can be done using regex (handles any character, not just period):
def length_power_three(string):
regex = r"."
while True:
match = re.match(regex, string)
if not match:
return False
if match.group(0) == string:
return True
regex = r"(?:" + regex + r"){3}"
But I understand you really want to know if it can be done with a single regex.
UPDATE
I forgot that you asked for a recursive solution:
def length_power_three(string, regex = r"."):
match = re.match(regex, string)
if not match:
return False
if match.group(0) == string:
return True
return length_power_three(string, r"(?:" + regex + r"){3}")
I have a string that looks like this:
my_str = "This sentence has a [b|bolded] word, and [b|another] one too!"
And I need it to be converted into this:
new_str = "This sentence has a <b>bolded</b> word, and <b>another</b> one too!"
Is it possible to use Python's string.replace or re.sub method to do this intelligently?
Just capture all the characters before | inside [] into a group . And the part after | into another group. Just call the captured groups through back-referencing in the replacement part to get the desired output.
Regex:
\[([^\[\]|]*)\|([^\[\]]*)\]
Replacemnet string:
<\1>\2</\1>
DEMO
>>> import re
>>> s = "This sentence has a [b|bolded] word, and [b|another] one too!"
>>> m = re.sub(r'\[([^\[\]|]*)\|([^\[\]]*)\]', r'<\1>\2</\1>', s)
>>> m
'This sentence has a <b>bolded</b> word, and <b>another</b> one too!'
Explanation...
Try this expression: [[]b[|](\w+)[]] shorter version can also be \[b\|(\w+)\]
Where the expression is searching for anything that starts with [b| captures what is between it and the closing ] using \w+ which means [a-zA-Z0-9_] to include a wider range of characters you can also use .*? instead of \w+ which will turn out in \[b\|(.*?)\]
Online Demo
Sample Demo:
import re
p = re.compile(ur'[[]b[|](\w+)[]]')
test_str = u"This sentence has a [b|bolded] word, and [b|another] one too!"
subst = u"<bold>$1</bold>"
result = re.sub(p, subst, test_str)
Output:
This sentence has a <bold>bolded</bold> word, and <bold>another</bold> one too!
Just for reference, in case you don't want two problems:
Quick answer to your particular problem:
my_str = "This sentence has a [b|bolded] word, and [b|another] one too!"
print my_str.replace("[b|", "<b>").replace("]", "</b>")
# output:
# This sentence has a <b>bolded</b> word, and <b>another</b> one too!
This has the flaw that it will replace all ] to </b> regardless whether it is appropriate or not. So you might want to consider the following:
Generalize and wrap it in a function
def replace_stuff(s, char):
begin = s.find("[{}|".format(char))
while begin != -1:
end = s.find("]", begin)
s = s[:begin] + s[begin:end+1].replace("[{}|".format(char),
"<{}>".format(char)).replace("]", "</{}>".format(char)) + s[end+1:]
begin = s.find("[{}|".format(char))
return s
For example
s = "Don't forget to [b|initialize] [code|void toUpper(char const *s)]."
print replace_stuff(s, "code")
# output:
# "Don't forget to [b|initialize] <code>void toUpper(char const *s)</code>."
There is a string in the following format:
It can start with any number of strings enclosed by double braces, possibly with white space between them (whitespace may or may not occur).
It may also contain strings enclosed by double-braces in the middle.
I am looking for a regular expression that can separate the start from the rest.
For example, given the following string:
{{a}}{{b}} {{c}} def{{g}}hij
The two parts are:
{{a}}{{b}} {{c}}
def{{g}}hij
I tried this:
/^({{.*}})(.*)$/
But, it captured also the g in the middle:
{{a}}{{b}} {{c}} def{{g}}
hij
I tried this:
/^({{.*?}})(.*)$/
But, it captured only the first a:
{{a}}
{{b}} {{c}} def{{g}}hij
This keeps matching {{, any non { or } character 1 or more times, }}, possible whitespace zero or more times and stores it in the first group. Rest of the string will be in the 2nd group. If there are no parts surrounded by {{ and }} the first group will be empty. This was in JavaScript.
var str = "{{a}}{{b}} {{c}} def{{g}}hij";
str.match(/^\s*((?:\{\{[^{}]+\}\}\s*)*)(.*)/)
// [whole match, group 1, group 2]
// ["{{a}}{{b}} {{c}} def{{g}}hij", "{{a}}{{b}} {{c}} ", "def{{g}}hij"]
How about using preg_split:
$str = '{{a}}{{b}} {{c}} def{{g}}hij';
$list = preg_split('/(\s[^{].+)/', $str, -1, PREG_SPLIT_DELIM_CAPTURE | PREG_SPLIT_NO_EMPTY);
print_r($list);
output:
Array
(
[0] => {{a}}{{b}} {{c}}
[1] => def{{g}}hij
)
I think I got it:
var string = "{{a}}{{b}} {{c}} def{{g}}hij";
console.log(string.match(/((\{\{\w+\}\})\s*)+/g));
// Output: [ '{{a}}{{b}} {{c}} ', '{{g}}' ]
Explanation:
( starts a group.
( another;
\{\{\w+\}\} looks for {{A-Za-z_0-9}}
) closes second group.
\s* Counts whitespace if it's there.
)+ closes the first group and looks for oits one or more occurrences.
When it gets any not-{{something}} type data, it stops.
P.S. -> Complex RegEx takes CPU speed.
You can use this:
(java)
string[] result = yourstr.split("\\s+(?!{)");
(php)
$result = preg_split('/\s+(?!{)/', '{{a}}{{b}} {{c}} def{{g}}hij');
print_r($result);
I don´t know exactly why are you want to split, but in case that the string contains always a def inside, and you want to separate the string from there in two halves, then, you can try something like:
string text = "{{a}}{{b}} {{c}} def{{g}}hij";
Regex r = new Regex("def");
string[] split = new string[2];
int index = r.Match(text).Index;
split[0] = string.Join("", text.Take(index).Select(x => x.ToString()).ToArray<string>());
split[1] = string.Join("", text.Skip(index).Take(text.Length - index).Select(x => x.ToString()).ToArray<string>());
// Output: [ '{{a}}{{b}} {{c}} ', 'def{{g}}hij' ]