how I could use eval(rlang::parse_expr(string))’ or alternative with such expresssion string <-"print('A\s*B')"`? I am getting unrecognized escape character. The expression is evaluated inside function, print is an example, I am using grepl in similar manier.
Simply using second escape slash helped "print('A\s*B')"
Related
Hi i am trying to replace a string with special character at the end with new string. For Example, I want to replace
qwerty_CRS_abc\
to
qwerty_CRS_abc
I tried with this:
:%s/qwerty_CRS_abc\/qwerty_CRS_abc/g
but I'm getting this error:
Pattern not found: padring_CRS_CAN\/padring_CRS_CAN\g
Basically, I just want to remove that backslash in whole file. It should be just
qwerty_CRS_abc
Use:
:%s/qwerty_CRS_abc\\/qwerty_CRS_abc/g
Certain characters such as /&!.^*$\? carry a special significance to the search process and must be escaped using the \ character when they are used in a search. Hence the \\ used to escape the backslash in your example.
I know this is a very specific question by I didn't manage to do the replacement myself, I need to replace this string in groovy:
com.pantest in com/pantest.
I tried this:
groupId =com.pantest
def mavenGroupID = groupId.replaceAll('.','/')
And this is what I get in the output:
echo mavenGroupID is //////////
mavenGroupID is //////////
Is the dot (.) is some kind of special character? I tried to escape it using **** but it also didn't work.
As mentioned in the comments, String.replaceAll is taking regex as the input, so it means you need to escape dot at least, but actually, you have also to escape escaping char- backslash \
(more clues at Regular Expression to match a dot)
so, you can do it like follows:
def test = "aaa.bbb.ccc"
//want to replace ., need to use escape char \, but it needs to be escaped as well , so \\
println test.replaceAll('\\.','/')
Output is as requested aaa/bbb/ccc
replaceAll('\\.','/') is the key
I'm struggling to get an ES6 template literal to produce a single backslash it its result.
> `\s`
's'
> `\\s`
'\\s'
> `\\\s`
'\\s'
> `\\\\s`
'\\\\s'
> `\u005Cs`
'\\s'
Tested with Node 8.9.1 and 10.0.0 by inspecting the value at a Node REPL (rather than printing it using console.log)
If I get your question right, how about \\?
I tried using $ node -i and run
console.log(`\\`);
Which successfully output a backslash. Keep in mind that the output might be escaped as well, so the only way to know you are successfully getting a backslash is getting the character code:
const myBackslash = `\\`;
console.log(myBackslash.charCodeAt(0)); // 92
And to make sure you are not actually getting \\ (i.e. a double-backslash), check the length:
console.log(myBackslash.length); // 1
It is a known issue that unknown string escape sequences lose their escaping backslash in JavaScript normal and template string literals:
When a character in a string literal or regular expression literal is
preceded by a backslash, it is interpreted as part of an escape
sequence. For example, the escape sequence \n in a string literal
corresponds to a single newline character, and not the \ and n
characters. However, not all characters change meaning when used in an
escape sequence. In this case, the backslash just makes the character
appear to mean something else, and the backslash actually has no
effect. For example, the escape sequence \k in a string literal just
means k. Such superfluous escape sequences are usually benign, and do
not change the behavior of the program.
In regular string literals, one needs to double the backslash in order to introduce a literal backslash char:
console.log("\s \\s"); // => s \s
console.log('\s \\s'); // => s \s
console.log(`\s \\s`); // => s \s
There is a better idea: use String.raw:
The static String.raw() method is a tag function of template
literals. This is similar to the r prefix in Python, or the #
prefix in C# for string literals. (But it is not identical; see
explanations in this issue.) It's used to get the raw string form
of template strings, that is, substitutions (e.g. ${foo}) are
processed, but escapes (e.g. \n) are not.
So, you may simply use String.raw`\s` to define a \s text:
console.log(String.raw`s \s \\s`); // => s \s \\s
I have a file containing string like this one :
print $hash_xml->{'div'}{'div'}{'div'}[1]...
I want to replace {'div'}{'div'}{'div'}[1] by something else.
So I tried
%s/{'div'}{'div'}{'div'}[1]/by something else/gc
The strings were not found. I though I had to escape the {,},[ and ]
Still string not found.
So I tried to search a single { and it found them.
Then I tried to search {'div'}{'div'}{'div'} and it found it again.
Then {'div'}{'div'}{'div'}[1 was still found.
To find {'div'}{'div'}{'div'}[1]
I had to use %s/{'div'}{'div'}{'div'}[1\]
Why ?
vim 7.3 on Linux
The [] are used in regular expressions to wrap a range of acceptable characters.
When both are supplied unescaped, vim is treating the search string as a regex.
So when you leave it out, or escape the final character, vim cannot interpret a single bracket in a regex context, so does a literal search (basically the best it can do given the search string).
Personally, I would escape the opening and closing square brace to ensure that the meaning is clear.
That's because the [ and ] characters are used to build the search pattern.
See :h pattern and use the help file pattern.txt to try the following experiment:
Searching for the "[9-0]" pattern (without quotes) using /[0-9] will match every digit from 0 to 9 individually (see :h \[)
Now, if you try /\[0-9] or /[0-9\] you will match the whole pattern: a zero, an hyphen and a nine inside square brackets. That's because when you escape one of [ or ] the operator [*] ceases to exist.
Using your search pattern, /{'div'}{'div'}{'div'}[1\] and /{'div'}{'div'}{'div'}\[1] should match the same pattern which is the one you want, while /{'div'}{'div'}{'div'}[1] matches the string {'div'}{'div'}{'div'}1.
In order to avoid being caught by these special characters in regular expressions, you can try using the very magic flag.
E.g.:
:%s/\V{'div'}[1]/replacement/
Notice the \V flag at the beginning of the line.
Because the square brackets mean that vim thinks you're looking for any of the characters inside. This is known as a 'character class'. By escaping either of the square brackets it lets vim know that you're looking for the literal square string ending with '[1]'.
Ideally you should write your expression as:
%s/{'div'}{'div'}{'div'}\[1\]/replacement string/
to ensure that the meaning is completely clear.
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"))