So I have this string:
{"scores":{"1":["John",60],"2":["Jude",60],"3":["Max",60],"4":["Kyle",60],"5":["Smith",60],"6":["Mark",50],"7":["Luke",40],"8":["Anne",30],"9":["Bruce",20],"10":["kazuo",10]}}
There are a number of integers there that have quotes around them, and I want to get rid of them. How do I do that? I already tried out:
print(string.gsub(string, '/"(\d)"/', "%1"));
but it does not work. :(
Lua does not have regular expressions like Perl, instead, it does have patterns. These are similar with a few differences.
There is no need for delimiting slashes / /, and the escaping character is % but not \. Otherwise, your trial is essentially correct:
print(string.gsub(str, '"(%d+)"', "%1"))
Where str is the variable containing the input string. Also note that string.gsub returns 2 values, which are both printed, the second result being the number of substitutions. Use an extra pair of parentheses to keep only the first result.
You can simplify a little the notation using the colon : operator :
print((str:gsub('"(%d+)"', "%1")))
Related
I'm having a bit of an issue trying to do an ascii art challenge in GS, since it requires you finishing a line with the \ symbol.
The problem is that "\"p breaks the program since it thinks you escaped a quote, and "\\"p prints two backslashes. I've tried string concatenation, removing one character at a time, printing substrings, etcetc - Nothing seems to work!
I need this string to be printed out, so how would this be done?
It seems that the behavior with p is buggy. I'll look for a place to report it.
However, "\\" by itself does not print two backslashes; it prints one.
Here's a test link to prove it.
Output:
\
"\\" creates a string with 1 backslash because strings are escaped. This is the same as languages like Ruby.
p escapes strings, so a string with one backslash will be displayed as two. This is also the same as languages like Ruby.
So if you want to print a single backslash, or print things without the quotes, you need to print unescaped strings. The best way to do this is with implicit IO (anything on the stack that is left over is printed unescaped).
The program
"\\"
Should print
\
You could also use print or puts if you don't want to use implicit IO.
Help me to decide one problem in TCL.
By using my macros I want find string, which contains two asterisk (**).
I tried to used following commands:
string match \*\* string_name
But it doesn't work. Can you explain me where I made a mistake and how to do it correctly?
Thanks in advance!
What you are actually passing to the interpreter is string match ** string_name. You need to pass the actual backslashes to the interpreter so that it then will understand two escaped asterisks, and to do that you need to add a couple more backslashes:
string match \\*\\* $s
Or use braces:
string match {\*\*} $s
Note that the above will match only if $s contains 2 asterisks, and nothing else. To allow for anything before and after the asterisks, you can use more asterisks...
string match {*\*\**} $s
There are a few other ways to check if a string has double asterisks, you can for instance use string first (and since this one does not support expressions, you can actually get away without having to escape anything):
string first ** $s
If you get something greater than -1, then ** is present in $s.
Or if you happen to know some regular expressions:
regexp -- {\*\*} $s
Those are the most common I think.
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.
While running an R-plugin in SPSS, I receive a Windows path string as input e.g.
'C:\Users\mhermans\somefile.csv'
I would like to use that path in subsequent R code, but then the slashes need to be replaced with forward slashes, otherwise R interprets it as escapes (eg. "\U used without hex digits" errors).
I have however not been able to find a function that can replace the backslashes with foward slashes or double escape them. All those functions assume those characters are escaped.
So, is there something along the lines of:
>gsub('\\', '/', 'C:\Users\mhermans')
C:/Users/mhermans
You can try to use the 'allowEscapes' argument in scan()
X=scan(what="character",allowEscapes=F)
C:\Users\mhermans\somefile.csv
print(X)
[1] "C:\\Users\\mhermans\\somefile.csv"
As of version 4.0, introduced in April 2020, R provides a syntax for specifying raw strings. The string in the example can be written as:
path <- r"(C:\Users\mhermans\somefile.csv)"
From ?Quotes:
Raw character constants are also available using a syntax similar to the one used in C++: r"(...)" with ... any character sequence, except that it must not contain the closing sequence )". The delimiter pairs [] and {} can also be used, and R can be used in place of r. For additional flexibility, a number of dashes can be placed between the opening quote and the opening delimiter, as long as the same number of dashes appear between the closing delimiter and the closing quote.
First you need to get it assigned to a name:
pathname <- 'C:\\Users\\mhermans\\somefile.csv'
Notice that in order to get it into a name vector you needed to double them all, which gives a hint about how you could use regex. Actually, if you read it in from a text file, then R will do all the doubling for you. Mind you it not really doubling the backslashes. It is being stored as a single backslash, but it's being displayed like that and needs to be input like that from the console. Otherwise the R interpreter tries (and often fails) to turn it into a special character. And to compound the problem, regex uses the backslash as an escape as well. So to detect an escape with grep or sub or gsub you need to quadruple the backslashes
gsub("\\\\", "/", pathname)
# [1] "C:/Users/mhermans/somefile.csv"
You needed to doubly "double" the backslashes. The first of each couple of \'s is to signal to the grep machine that what next comes is a literal.
Consider:
nchar("\\A")
# returns `[1] 2`
If file E:\Data\junk.txt contains the following text (without quotes): C:\Users\mhermans\somefile.csv
You may get a warning with the following statement, but it will work:
texinp <- readLines("E:\\Data\\junk.txt")
If file E:\Data\junk.txt contains the following text (with quotes): "C:\Users\mhermans\somefile.csv"
The above readlines statement might also give you a warning, but will now contain:
"\"C:\Users\mhermans\somefile.csv\""
So, to get what you want, make sure there aren't quotes in the incoming file, and use:
texinp <- suppressWarnings(readLines("E:\\Data\\junk.txt"))