pOutputIndexFile is "G:\new1.dat"
fileNumber is "2.dat"
pOutputIndexFile->replace(QRegExp("\\[0-9]{1,1}\\.dat"),fileNumber);
this doesn't change my string to "G:\new2.dat". How can I do this?
I got it fine with
pOutputIndexFile->replace(QRegExp("\\d\\d?.dat"),fileNumber);
It can be done by:
pOutputIndexFile->replace(QRegExp("\\d{1,10}.dat"),fileNumber);
\d{1,10} means match a sequence of digits that contains at least one digit but no more than ten.
So even when pOutputIndexFile is "G:\new964.dat" and fileNumber is "965.dat" the result is:
"G:\new965.dat"
Your regular expression is looking for files with the extension idx, not dat. Since it doesn't find one, nothing is replaced.
You can try with this :
pOutputIndexFile->replace(QRegExp("[0-9]{1}\.dat"),fileNumber);
[0-9] : means a number
{1} : means exactly one char
\. : protect the .
dat : dat
Related
I have a small Problem. I want to extract a special pattern like this:
v-97bcer
or b-chyfvg or ghd6db
I tried this:
identifier_1 <- "([:alnum:]{6})" # for things like this ghd6db
identifier_2 <- "([:lower:]{1})[- ][:alnum:]{6})" # for things like this v-97bcer or b-chyfvg
The problem is that the first "identifier" works well ok, but extracts for example names as well. In GHD6D8 this example the numbers have no fixed place and can occur everywhere. I do just now that the length is 6.
And the second problem is that for example V-97bcer can occur like v97bcer but I need this format v-97bcer. Here too the numbers are randomly.
If somebody could help or give me a good source for better understanding how to do this. I have not much exp in string matching. Thank you
this should work:
x <- c("v-97bcer", "b-chyfvg", "ghd6db", "v97bcer")
grep("^([a-z].)?[a-z0-9]{6}$", x)
Note that in order to fix the length of the string I provide ^ and $ to the string.
This pattern matches v-97bcer and b-chyfvg and ghd6db but not v97bcer.
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.
This may be a very simple question but I have not much experience with regex expressions. This page is a good source of regex expressions but could not figure out how to include them into my following code:
data %>% filter(grepl("^A01H1", icl))
Question
I would like to extract the values in one column of my data frame starting with this A01H1 up to 2 more digits, for example A01H100, A01H140, A01H110. I could not find a solution despite my few attempts:
Attempts
I looked at this question from which I used ^A01H1[0-9].{2} to select up tot two more digits.
I tried with adding any character ^A01H1[0-9][0-9][x-y] to stop after two digits.
Any help would be much appreciated :)
You can use "^A01H1\\d{1,2}$".
The first part ("^A01H1"), you figured out yourself, so what are we doing in the second part ("\\d{1,2}$")?
\d includes all digits and is equivalent to [0-9], since we are working in R you need to escape \ and thus we use \\d
{1,2} indicates we want to have 1 or 2 matches of \\d
$ specifies the end of the string, so nothing should come afterwards and this prevents to match more than 2 digits
It looks as if you want to match a part of a string that starts with A01H1, then contains 1 or 2 digits and then is not followed with any digit.
You may use
^A01H1\d{1,2}(?!\d)
See the regex demo. If there can be no text after two digits at all, replace (?!\d) with $.
Details
^ - start of strinmg
A01H1 - literal string
\d{1,2} - one to two digits
(?!\d) - no digit allowed immediately to the right
$ - end of string
In R, you could use it like
grepl("^A01H1\\d{1,2}(?!\\d)", icl, perl=TRUE)
Or, with the string end anchor,
grepl("^A01H1\\d{1,2}$", icl)
Note the perl=TRUE is only necessary when using PCRE specific syntax like (?!\d), a negative lookahead.
I have a variable named full.path.
And I am checking if the string contained in it is having certain special character or not.
From my code below, I am trying to grep some special character. As the characters are not there, still the output that I get is true.
Could someone explain and help. Thanks in advance.
full.path <- "/home/xyz"
#This returns TRUE :(
grepl("[?.,;:'-_+=()!##$%^&*|~`{}]", full.path)
By plugging this regex into https://regexr.com/ I was able to spot the issue: if you have - in a character class, you will create a range. The range from ' to _ happens to include uppercase letters, so you get spurious matches.
To avoid this behaviour, you can put - first in the character class, which is how you signal you want to actually match - and not a range:
> grepl("[-?.,;:'_+=()!##$%^&*|~`{}]", full.path)
[1] FALSE
Hi I'm trying to right a regular expression that will take a string and ensure it starts with an 'R' and is followed by 4 numeric digits then anything
eg. RXXXX.................
Can anybody help me with this? This is for ASP.NET
You want it to be at the beginning of the line, not anywhere. Also, for efficiency, you dont want the .+ or .* at the end because that will match unnecessary characters. So the following regex is what you really want:
^R\d{4}
This should do it...
^R\d{4}.*$
\d{4} matches 4 digits
.* is simply a way to match any character 0 or more times
the beginning ^ and end $ anchors ensure that nothing precedes or follows
As Vincent suggested, for your specific task it could even be simplified to this...
^R\d{4}
Because as you stated, it doesn't really matter what follows.
/^R\d{4}.*/ and set the case insensitive option unless you only want capital R's
^R\d{4}.*
The caret ^ matches the position before the first character in the string.
\d matches any numeric character (it's the same as [0-9])
{4} indicates that there must be exactly 4 numbers, and
.* matches 0 or more other characters
To use:
string input = "R0012 etc..";
Match match = Regex.Match(input, #"^R\d{4}.*", RexOptions.IgnoreCase);
if (match.Success)
{
// Success!
}
Note the use of RexOptions.IgnoreCase to ignore the case of the letter R (so it'll match strings which start with r. Leave this out if you don't want to undertake a case insensitive match.