I am searching for a way to use a formatter to put a space between two characters. i thought it would be easy with a string formatter.
here is what i am trying to accomplish:
given: "AB" it will produce "A B"
Here is what i have tried so far:
"AB".format("%#s")
but this keep returning "AB" i want "A B". i thought the number sign could be used for space.
i also tried this:
"26".format("%#d") but its still prints "26"
is there anyway to do this with string.formatter.
It is kind of possible with the string formatter although not directly with a pattern.
jshell> String.format("%1$c %2$c", "AB".chars().boxed().toArray())
$10 ==> "A B"
We need to turn the string into an object array so it can be passed in as varargs and the formatter pattern can extract characters based on index (1$ and 2$) and format them as characters (c).
A much simpler regex solution is the following which scales to any number of characters:
jshell> "ABC^&*123".replaceAll(".", "$0 ").trim()
$3 ==> "A B C ^ & * 1 2 3"
All single characters are replaced with them-self ($0) followed by a space. Then the last extra space is removed with the trim() call.
I could not find way to do this using String#format. But here is a way to accomplish this using regex replacement:
String input = "AB";
String output = input.replaceAll("(?<=[A-Z])(?=[A-Z])", " ");
System.out.println(output);
The regex pattern (?<=[A-Z])(?=[A-Z]) will match every position in between two capital letters, and interpolate a space at that point. The above script prints:
A B
Related
Assume there's a string
"An example striiiiiing with other words"
I need to replace the 'i's with '*'s like 'str******ng'. The number of '*' must be same as 'i'. This replacement should happen only if there are consecutive 'i' greater than or equal to 3. If the number of 'i' is less than 3 then there is a different rule for that. I can hard code it:
import re
text = "An example striiiiing with other words"
out_put = re.sub(re.compile(r'i{3}', re.I), r'*'*3, text)
print(out_put)
# An example str***iing with other words
But number of i could be any number greater than 3. How can we do that using regex?
The i{3} pattern only matches iii anywhere in the string. You need i{3,} to match three or more is. However, to make it all work, you need to pass your match into a callable used as a replacement argument to re.sub, where you can get the match text length and multiply correctly.
Also, it is advisable to declare the regex outside of re.sub, or just use a string pattern since patterns are cached.
Here is the code that fixes the issue:
import re
text = "An example striiiiing with other words"
rx = re.compile(r'i{3,}', re.I)
out_put = rx.sub(lambda x: r'*'*len(x.group()), text)
print(out_put)
# => An example str*****ng with other words
I want to write a function that, given a string, returns a new string in which occurences of a sequence of the same consonant with 2 or more elements are replaced with the same sequence except the first consonant - which should be replaced with the character 'm'.
The explanation was probably very confusing, so here are some examples:
"hello world" should return "hemlo world"
"Hannibal" should return "Hamnibal"
"error" should return "emror"
"although" should return "although" (returns the same string because none of the characters are repeated in a sequence)
"bbb" should return "mbb"
I looked into using regex but wasn't able to achieve what I wanted. Any help is appreciated.
Thank you in advance!
Regex is probably the best tool for the job here. The 'correct' expression is
test = """
hello world
Hannibal
error
although
bbb
"""
output = re.sub(r'(.)\1+', lambda g:f'm{g.group(0)[1:]}', test)
# '''
# hemlo world
# Hamnibal
# emror
# although
# mbb
# '''
The only real complicated part of this is the lambda that we give as an argument. re.sub() can accept one as its 'replacement criteria' - it gets passed a regex object (which we call .group(0) on to get the full match, i.e. all of the repeated letters) and should output a string, with which to replace whatever was matched. Here, we use it to output the character 'm' followed by the second character onwards of the match, in an f-string.
The regex itself is pretty straightforward as well. Any character (.), then the same character (\1) again one or more times (+). If you wanted just alphanumerics (i.e. not to replace duplicate whitespace characters), you could use (\w) instead of (.)
I'm having issues when trying to remove the first space of a string if that string has 2 spaces in it. For example it should be turning "Fully Functional Method" into "FullyFunctional Method", but "Functional Method" should not be changed because it only has 1 space. I can't really think of a way to remove first space if the string contains 2 spaces.
I don't know exactly what you want to do, but you may search into RegExp and String.replace() to replace some stuff in a String.
Here is another link to understand the Characters, metacharacters, and metasequences.
var myPattern1:RegExp = / /g;
var str1:String = "This is a string that contains double spaces.";
trace(str1.replace(myPattern1, " "));
//this replaces all " " by " "...
//outputs : This is a string that contains double spaces.
Or in your case (I suppose) something like this
var myPattern2:RegExp = / /;
var str2:String = "Fully Functional Method";
trace(str2.replace(myPattern2, ""));
//If you omit the g, only the first space will be replaced by ""
//outputs : FullyFunctional Method
There is so much things you can do by using RegExp, that I will not explain this here...
Just check on the Adobe website...
This is a quick and efficient way to work on Strings.
I hope this will help.
Since you check at those links, you will understand that my example is pure rough and should be modified to have a FullyFunctional Method. :D
Do a linear scan through the string. Count the number of spaces and record the index of the first space, if any. If there are two spaces, return a string that is the concatenation of the characters up to but not including the first space, and the characters after the first space.
Keep it simple. It is possible to solve your problem with regex, but keep in mind that the worst case time complexity of finding a particular character in an unsorted set is always going to be O(N), so it won't be faster.
I have a problem with splitting string into two parts on special character.
For example:
12345#data
or
1234567#data
I have 5-7 characters in first part separated with "#" from second part, where are another data (characters,numbers, doesn't matter what)
I need to store two parts on each side of # in two variables:
x = 12345
y = data
without "#" character.
I was looking for some Lua string function like splitOn("#") or substring until character, but I haven't found that.
Use string.match and captures.
Try this:
s = "12345#data"
a,b = s:match("(.+)#(.+)")
print(a,b)
See this documentation:
First of all, although Lua does not have a split function is its standard library, it does have string.gmatch, which can be used instead of a split function in many cases. Unlike a split function, string.gmatch takes a pattern to match the non-delimiter text, instead of the delimiters themselves
It is easily achievable with the help of a negated character class with string.gmatch:
local example = "12345#data"
for i in string.gmatch(example, "[^#]+") do
print(i)
end
See IDEONE demo
The [^#]+ pattern matches one or more characters other than # (so, it "splits" a string with 1 character).
I have a large body of text and I print only lines that contain one of several strings. Each line can contain more than one string.
Example of the rule:
(house|mall|building)
I want to mark the found string for making the result easier to read.
Example of the result I want:
New record: Two New York houses under contract for nearly $5 million each.
New record: Two New York #house#s under contract for nearly $5 million each.
I know I can find the location, trim, add marker, add string etc.
I am asking if there is a way to mark the found string in one command.
Thanks.
http://pubs.opengroup.org/onlinepubs/009695399/utilities/awk.html
gsub(ere, repl[, in])
Behave like sub (see below), except that it shall replace all occurrences of the regular expression ...
sub(ere, repl[, in ])
Substitute the string repl in place of the first instance of the
extended regular expression ERE in string in and return the number of
substitutions. An ampersand ( '&' ) appearing in the string repl shall
be replaced by the string from in that matches the ERE ...
BEGIN {
r = "house|mall|building"
s = "Two New York houses under contract for nearly $5 million each."
gsub(r, "#&#", s)
print s
}