I'm having a bit of trouble using parenthesis in a vim string. I just need to add a set of parenthesis around 3 digits, but I can't seem to find where I'm suppose to correctly place them. So for example; I would have to place them around a phone number such as: 2015551212.
Right now I have a strings that separates the numbers and puts a hyphen between them. For example; 201 555-1212. So I just need the parenthesis. The final result should look like: (201) 555-1212
The string I have so far is this: s/\(\d\{3}\)\(\d\{3}\)/\1 \2-/g
How might I go about doing this?
Thanks
Just add the parens around the \1 in your replacement.
s/\(\d\{3\}\)\(\d\{3\}\)/(\1) \2-/g
If you want to go in reverse, and change "(800) 555-1212" to "8005551212", you can use something like this:
s/(\(\d\d\d\))\ \(\d\d\d\)-\(\d\d\d\d\)/\1\2\3/g
Instead of the \d\d\d, you could use \d\{3\}, but that is more trouble to type.
Related
I'm trying to split a given string when two quotes appear and they contain at least 3 characters(I also have to split whenever . or , appear). So, something like hello"example"hello,cat should return [hello;example;hello;cat].
I came up with:
re.split("\'(...+)\'|\.|,","hello'example'hello,cat")
This works fine with the quotes, but whenever it split for . or , this happens:
['hello', 'example', 'hello', None, 'cat']
I found out the capture group is the one that causes it (the None in the middle of the list), but it is the only way I know to keep the content.
Please keep in mind that I have to do as few as possible computations because the program shall work with huge files, also I'm not very experienced with Python so sorry if I did something obvious wrong.
Try just:
re.split("\'|\.|,", "hello'example'hello,cat")
It's tricky because the open quote and close quote is the exact same character. I think you'd have to use a negative look behind to exclude any single quote that is preceded by 0, 1 or 2 characters and another single quote. In addition, you'd have to use a positive lookahead. This works in javascript.
re.split("(?<!'(.?|..))'(?=[^']{3,})|\.|\,", "hello'example'hello,cat")
But it doesn't look like python supports variable-length lookbehinds. Also, this won't work if there is a lone single quote (apostrophe).
str = "fa, (captured)[asd] asf, 31"
for word in str:gmatch("\(%a+\)") do
print(word)
end
Hi! I want to capture a word between parentheses.
My Code should print "captured" string.
lua: /home/casey/Desktop/test.lua:3: invalid escape sequence near '\('
And i got this syntax error.
Of course, I can just find position of parentheses and use string.sub function
But I prefer simple code.
Also, brackets gave me a similar error.
The escape character in Lua patterns is %, not \. So use this:
word=str:match("%((%a+)%)")
If you only need one match, there is no need for a gmatch loop.
To capture the string in square brackets, use a similar pattern:
word=str:match("%[(%a+)%]")
If the captured string is not entirely composed of letters, use .- instead of %a+.
lhf's answer likely gives you what you need, but I'd like to mention one more option that I feel is underused and may work for you as well. One issue with using %((%a+)%) is that it doesn't work for nested parentheses: if you apply it to something like "(text(more)text)", you'll get "more" even though you may expect "text(more)text". Note that you can't fix it by asking to match to the first closing parenthesis (%(([^%)]+)%)) as it will give you "text(more".
However, you can use %bxy pattern item, which balances x and y occurrences and will return (text(more)text) in this case (you'd need to use something like (%b()) to capture it). Again, this may be overkill for your case, but useful to keep in mind and may help someone else who comes across this problem.
I have a pattern where there are double-quotes between numbers in a CSV file.
I can search for the pattern by [0-9]\"[0-9], but how do I retain value while removing the double quote. CSV format is like this:
"1234"5678","Text1","Text2"
"987654321","Text3","text4"
"7812891"3","Text5","Text6"
As you may notice there are double quotes between some numbers which I want to remove.
I have tried the following way, which is incorrect:
:%s/[0-9]\"[0-9]/[0-9][0-9]/g
Is it possible to execute a command at every search pattern, maybe go one character forward and delete it. How can "lx" be embedded in search and replace.
You need to capture groups. Try:
:%s/\(\d\)"\(\d\)/\1\2/g
[A digit can also be denoted by \d.]
I know that this question has been answered already, but here's another approach:
:%s/\d\zs"\ze\d
Explanation:
%s Substitute for the whole buffer
\d look up for a digit
\zs set the start of match here
" look up for a double-quote
\ze set the end of match here
\d look up for a digit
That makes the substitute command to match only the double-quote surrounded by digits.
Omitting the replacement string just deletes the match.
You need boundaries to use in regular expression.
Try this:
:%s/\([0-9]\)"\([0-9]\)/\1\2/g
A bit naive solution:
%s/^"/BEGINNING OF LINE QUOTE MARK/g
%s/\",\"/quote comma quote/g
%s/\"$/quota end of line/g
%s/\"//g
%s/quota end of line/"/g
%s/quote comma quote/","/g
%s/BEGINNING OF LINE QUOTE MARK/"/g
A macro can be created quite easy out of it and invoked as many times as needed.
I'm trying to use both tabs and newlines as delimiters to read from a .txt file. What I have at the moment is:
Scanner fileScanner = new Scanner(new FileReader("propertys.txt"));
fileScanner.useDelimiter("[\\t\\n]");
I've tried:
fileScanner.useDelimiter("\\t|\\n");
and
fileScanner.useDelimiter("[\\t|\\n]");
I've got no idea what's going wrong, I've searched around a lot and it looks like one of those should be working. Clearly I'm doing something wrong.
fileScanner.useDelimiter("\t|\n");
should work.
If you have two slashes "\n" the first acts as an escape and it won't work right.
For the regular expression used as a parameter in useDelimiter method, you should use newline as \n instead of \\n and tab as \t instead of \\t. From Java Pattern class: http://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html.
A part from that, I think you should define you regular expression like, for example, this:
fileScanner.useDelimiter("\\s*[\t\n]\\s*");
to limit strings (\\s) between newline or tab characters.
I'm trying to do a string compare for 'zürich' and 'zurich'
Something like this:
int compareResult = String.Compare(zürich, zurich);
So what happens is that it returns -1, which causes a problem as I'm using compareResult for an if-else later.
Can someone point me to the right direction on why does this happen. Do I need to clean this first before comparing "zürich" or is it something else?
you use the method just fine, but the strings are actually different.
so, in order to make this comparison in your way, you need:
decide if this you want every comparison that uses ü and other "special" latin characters to look at them as they were the simple characters.
i.e. in every time you see ü, it will treat it as a "u"
if so, you need to do pre-processing of both the strings, and replace all special chars with regular ones.
there is another thread about it here:
How can I remove accents on a string?
hope it helped.