Create a simple string in Scilab containing a newline.
Seems simple enough, but Scilab only seems to interpret escape sequences through printf style functions and msprintf / sprintf splits the string into a vector of strings at the newline!
The only way I can see to achieve this is to actually write a newline out to a file and read it back in again. Surely there is a simpler way to do this!
Ok, found it. The ascii function will do the job, a newline can be added via its ascii decimal -
str = 'hello' + ascii(10) + 'world'
Related
I have a main string as below
"/tmp/xjtscpdownload/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
From the main string i need to extract a substring starting from the uuid part
"/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
I tried
string.match("/tmp/xjtscpdownload/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/", "/[a-fA-F0-9]{8}-[a-fA-F0-9]{4}-[a-fA-F0-9]{4}-[a-fA-F0-9]{4}-[a-fA-F0-9]{12}/(.)/(.)/$"
But noluck.
if you want to obtain
"/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
from
"/tmp/xjtscpdownload/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
or let's say 7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0, output and 9999.317528060546245771146821638997525068657 as this is what your pattern attempt suggests. Otherwise leave out the parenthesis in the following solution.
You can use a pattern like this:
local text = "/tmp/xjtscpdownload/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
print(text:match("/([%x%-]+)/([^/]+)/([^/]+)"))
"/([^/]+)/" captures at least one non-slash-character between two slashs.
On your attempt:
You cannot give counts like {4} in a string pattern.
You have to escape - with % as it is a magic character.
(.) would only capture a single character.
Please read the Lua manual to find out what you did wrong and how to use string patterns properly.
Try also the code
s="/tmp/xjtscpdownload/7eb17cc6-b3c9-4ebd-945b-c0e0656a33f0/output/9999.317528060546245771146821638997525068657/"
print(s:match("/.-/.-(/.+)$"))
It skips the first two "fields" by using a non-greedy match.
I work with knitr() and I wish to transform inline Latex commands like "\label" and "\ref", depending on the output target (Latex or HTML).
In order to do that, I need to (programmatically) generate valid R strings that correctly represent the backslash: for example "\label" should become "\\label". The goal would be to replace all backslashes in a text fragment with double-backslashes.
but it seems that I cannot even read these strings, let alone process them: if I define:
okstr <- function(str) "do something"
then when I call
okstr("\label")
I directly get an error "unrecognized escape sequence"
(of course, as \l is faultly)
So my question is : does anybody know a way to read strings (in R), without using the escaping mechanism ?
Yes, I know I could do it manually, but that's the point: I need to do it programmatically.
There are many questions that are close to this one, and I have spent some time browsing, but I have found none that yields a workable solution for this.
Best regards.
Inside R code, you need to adhere to R’s syntactic conventions. And since \ in strings is used as an escape character, it needs to form a valid escape sequence (and \l isn’t a valid escape sequence in R).
There is simply no way around this.
But if you are reading the string from elsewhere, e.g. using readLines, scan or any of the other file reading functions, you are already getting the correct string, and no handling is necessary.
Alternatively, if you absolutely want to write LaTeX-like commands in literal strings inside R, just use a different character for \; for instance, +. Just make sure that your function correctly handles it everywhere, and that you keep a way of getting a literal + back. Here’s a suggestion:
okstr("+label{1 ++ 2}")
The implementation of okstr then needs to replace single + by \, and double ++ by + (making the above result in \label{1 + 2}). But consider in which order this needs to happen, and how you’d like to treat more complex cases; for instance, what should the following yield: okstr("1 +++label")?
sorry I am a bit rusty here, how do I concatenate these 2 outputs?
display.println(timeinfo->tm_hour);
display.println(timeinfo->tm_min);
If you just want them to appear in the output one after another on the same line then use print instead of println for the first one. Println adds a newline to the end of the output and print doesn't. It's always good to look stuff like that up before using a function.
If you really want them put together into one string then you will have to show where those strings are coming from. If they are String class objects you can just use + to put them together. If they are proper c-style strings then you will need to use strcat.
How are defining them?
If you have initialized as arrays of characters:
Example: char exampleCString[50] = "This is a C string";
Then you can use strcat() function in C:
strcat(str1,str2);
Note: Make sure "str1" buffer is big enough, because the result goes there.
If on the other hand, you have initialized your strings as objects of String class:
Example: String exampleJavaString="This is a Java String example"
Then just use the + operator to add them:
str1=str1+str2:
I have read some questions and answers on this topic in stack overflow but still don't know how to solve this problem:
My purpose is to transform the file directory strings in windows explorer to the form which is recognizable in R, e.g. C:\Users\Public needs to be transformed to C:/Users/Public, basically the single back slash should be substituted with the forward slash. However the R couldn't store the original string "C:\Users\Public" because the \U and \P are deemed to be escape character.
dirTransformer <- function(str){
str.trns <- gsub("\\", "/", str)
return(str.trns)
}
str <- "C:\Users\Public"
dirTransformer(str)
> Error: '\U' used without hex digits in character string starting ""C:\U"
What I am actually writing is a GUI, where the end effect is, the user types or pastes the directory into a entry field, pushes a button and then the program will process it automatically.
Would someone please suggest to me how to solve this problem?
When you need to use a backslash in the string in R, you need to put double backslash. Also, when you use gsub("\\", "/", str), the first argument is parsed as a regex, and it is not valid as it only contains a single literal backslash that must escape something. In fact, you need to make gsub treat it as a plain text with fixed=TRUE.
However, you might want to use normalizePath, see this SO thread.
dirTransformer <- function(str){
str.trns <- gsub("\\\\", "/", str)
return(str.trns)
}
str <- readline()
C:\Users\Public
dirTransformer(str)
I'm not sure how you intend the user to input the path into the GUI, but when using readline() and then typing C:\Users\Public unquoted, R reads that in as:
> str
[1] "C:\\Users\\Public"
We then want to replace "\\" with "/", but to escape the "\\" we need "\\\\" in the gsub.
I can't be sure how the input from the user is going to be read into R in your GUI, but R will most likely escape the \s in the string like it does when using the readline example. the string you're trying to create "C:\Users\Public" wouldn't normally happen.
In the script, sometimes a newline is added in the beginning of the text field (am using a textArea in adobe flex 3), and later on that newline might need to be removed (after other text has been added). I was wondering how to check if there is a newline at the beginning of the text field and then how to remove it. Thanks in advance.
How about
private function lTrimTextArea(ta:TextArea) {
ta.text = ta.text.replace(/^\n*/,'');
}
To remove all line breaks from the start of a string, regardless of whether they are Windows (CRLF) or UNIX (LF only) line breaks, use:
ta.text = ta.text.replace(/^[\r\n]+/,'');
You should use + in the regex instead of * so that the regex only makes a replacement if there is actually a line break at the start of the string. If you use ^\n* as Robusto suggested the regex will find a zero-length match at the start of the string if the string does not start with a line break, and replace that with nothing. Replacing nothing with nothing is a waste of CPU cycles. It may not matter in this situation, but avoiding unintended zero-length matches is a very good habit when working with regular expressions. In other situations they will bite you.
If you particularly want to check first character here is solution:
if(ta.text.charAt(0) == "\n" || ta.text.charAt(0) == "\r")
{
ta.text.slice(1,ta.text.length-1);
}
slice method will slice that first character and gives your text from second character.
If you want to simply disable the ability to create a carriage return / new line, then all you need to do is disable multiline for that TextField...
(exampleTextField as TextField).multiline = false;
This will still trigger KEY_DOWN and KEY_UP events, however will not append the text with the carriage return.