I'd like to split this string by ' ' only if it has ':':
"A:Hey B:Are C:You there"
C:You there should be not split. The result should be:
["A:Hey", "B:Are", "C:You there"]
How can I do this?
\s+(?=\S*:)
You can split by this.
See demo.
https://regex101.com/r/hF7zZ1/4
This basically use a lookahead to make sure that the space which is being split upon is followed by non space characters and : .So it will work as you want.
Related
I have seen many methods for removing some characters based on regular string or special match using tr etc. But I didn't find any way to delete a character based on accurate index in a string.
For example: In var="hello world",I want to delete the 5th char 'o' and then var should be like "hell world". The code should apply to all other normal strings. Thanks in advance.
One method is:
n=5
var="hello world"
var=${var:0:n-1}${var:n}
I have a string like this:
"Jim-Bob's email ###hl###address###endhl### is: jb#example.com"
I want to replace all non-word characters (symbols and whitespace), except the ### delimiters.
I'm currently using:
str.gsub(/[^\w#]+/, 'X')
which yields:
"JimXBobXsXemailX###hl###address###endhl###XisXjb#exampleXcom"
In practice, this is good enough, but it offends me for two reasons:
The # in the email address is not replaced.
The use of [^\w] instead of \W feels sloppy.
How do I replace all non-word characters, unless those characters make up the ###hl### or ###endhl### delimiter strings?
str.gsub(/(###.*?###|\w+)|./) { $1 || "X" }
# => "JimXBobXsXemailX###hl###address###endhl###XisXXjbXexampleXcom"
This approach uses the fact that alternations work like case structure: the first matching one consumes the corresponding string, then no further matching is done on it. Thus, ###.*?### will consume a marker (like ###hl###; nothing else will be matched inside it. We also match any sequence of word characters. If any of those are captured, we can just return them as-is ($1). If not, then we match any other character (i.e. not inside a marker, and not a word character) and replace it with "X".
Regarding your second point, I think you are asking too much; there is no simple way to avoid that.
Regarding the first point, a simple way is to temporarily replace "###" with a character that you will never use (let's say you are using a system without "\r", so that that character is not used; we can use that as a temporal replacement).
"Jim-Bob's email ###hl###address###endhl### is: jb#example.com"
.gsub("###", "\r").gsub(/[^\w\r]/, "X").gsub("\r", "###")
# => "JimXBobXsXemailX###hl###address###endhl###XisXXjbXexampleXcom"
A word processor program features a search and replace function. However, partial words (character combinations found within words) are also replaced. To fix this, I plan to remove extra spaces and use the split function to change the string into an array of words by using " " as a delimiter.
However, once I search through the array, replace the appropriate words, and put the array back into a string separated by spaces, the original formatting of the user will be lost. For example, if the original string was "This is a sentence." and the user wanted "a" to be replaced with "the", the output will be "This is the sentence.", with no additional spaces.
So, my question is whether there is any way to search and replace entire words only while still preserving the formatting (extra spaces) of the user in Visual Basic.
What about using a regex?
In a regex the code \b is a word boundary so for example the regex \ba\b will match a only when a is a whole word.
So for example your code would be:
Dim strPattern As String: strPattern = "\ba\b"
Dim regex As New RegExp
regex.Global = True
regex.Pattern = strPattern
result = regex.Replace("This is a sentence.", "the")
If you use the Split function without removing your extra spaces first your array will have empty items in it so you would not lose the extra spaces and can reconstruct your document with the original formatting in tact.
Why is your formatting lost? If you split the text by space, just attach a space after each element when composing it back from an array. But you will also have to take into account words that end not with a space but punctuation.
in "This is a simple sentence, eh?", "eh" will be stored as "eh?" because u split by space. So you will have to program a complex punctuation-friendly formula or simply use regex. Be prepared - regex is... tricky.
I need to split a string into two variables. For example, the following would work fine:
first,second = "red,blue".split(',')
I would like to split user input, which might have an optional space after the comma. How do I write it so a space after the comma is absorbed? I need to correctly handle all these possibilities:
"red,blue" # first="red" second="blue"
"red, blue" # first="red" second="blue"
"red,dark blue" # first="red" second="dark blue"
"red, light blue" # first="red" second="light blue"
Just trim the resulting entries. The way you do this depends on whether you want to support exactly one space after the comma, or whether you want to remove all leading whitespace (and maybe trailing whitespace too). If your goal is to get words, like it looks like in your sample, you should just remove all surrounding whitespace.
first,second = "red, blue".split(',').map(&:strip)
There is no regexp in your code - you split using a string, which makes a difference.
"red,blue".split(/\s*,\s*/) should work as you expect.
list.split(/, */)
This is a regular expression that works with or without a space after the comma.
I need to filter all lines with words starting with a letter followed by zero or more letters or numbers, but no special characters (basically names which could be used for c++ variable).
egrep '^[a-zA-Z][a-zA-Z0-9]*'
This works fine for words such as "a", "ab10", but it also includes words like "b.b". I understand that * at the end of expression is problem. If I replace * with + (one or more) it skips the words which contain one letter only, so it doesn't help.
EDIT:
I should be more precise. I want to find lines with any number of possible words as described above. Here is an example:
int = 5;
cout << "hello";
//some comments
In that case it should print all of the lines above as they all include at least one word which fits the described conditions, and line does not have to began with letter.
Your solution will look roughly like this example. In this case, the regex requires that the "word" be preceded by space or start-of-line and then followed by space or end-of-line. You will need to modify the boundary requirements (the parenthesized stuff) as needed.
'(^| )[a-zA-Z][a-zA-Z0-9]*( |$)'
Assuming the line ends after the word:
'^[a-zA-Z][a-zA-Z0-9]+|^[a-zA-Z]$'
You have to add something to it. It might be that the rest of it can be white spaces or you can just append the end of line.(AFAIR it was $ )
Your problem lies in the ^ and $ anchors that match the start and end of the line respectively. You want the line to match if it does contain a word, getting rid of the anchors does what you want:
egrep '[a-zA-Z][a-zA-Z0-9]+'
Note the + matches words of length 2 and higher, a * in that place would signel chars too.