How can I add Data Annotation for Salutation?
A salutation must begin with Dear Sir or Madam, Mr, Mrs, Dr in lower or uppercase?
I tried the following but it not working for me:
[RegularExpression(#"^(Dr|Mrs?|Ms)\. [A-Za-z] ([A - Za - z] (\s|\.|_)?)+[a-zA-Z]*$", ErrorMessage = "Greeting must begin with Mr., Mrs., Ms., or Dr")]
Use something like this: ^(Mr|Mrs|Ms|Dr)\. [\p{L} '-]+$. The assumption is that a surname can contain letters, not necessarily basic Latin (\p{L}), spaces in strict sense, apostrophes and hyphens. I did not add underscores. In the future you will probably need to extend this set.
This regular expression assumes that the salutation is all that is fed to to regular expression (i.e., that only salutation is subject to data annotation). If you check the whole letter, replace the final $ with an \n (newline); and if there can be an address before the salutation, replace the initial ^ with (?:^|\n). These newlines make sure that the salutation occupies a separate string. Do not use multiline option in this case.
Is the letter is user input, allow for extra spaces: ^\s*(Mr|Mrs|Ms|Dr)\.[ ]+[\p{L} '-]+$.
Also, the full stop after the title can be missing, so: ^\s*(Mr|Mrs|Ms|Dr)\.? [ ]+[\p{L} '-]+$.
You may want to add an optional final comma: ^\s*(Mr|Mrs|Ms|Dr)\.?[ ]+[\p{L} '-]+,?\s*$.
Possible titles are also numerous, like Prof. or military ranks.
Related
I'd like to replace the content of a column in a data frame with only a specific word in that column.
The column always looks like this:
Place(fullName='Würzburg, Germany', name='Würzburg', type='city', country='Germany', countryCode='DE')
Place(fullName='Iphofen, Deutschland', name='Iphofen', type='city', country='Germany', countryCode='DE')
I'd like to extract the city name (in this case Würzburg or Iphofen) into a new column, or replace the entire row with the name of the town. There are many different towns so having a gsub-command for every city name will be tough.
Is there a way to maybe just use a gsub and tell Rstudio to replace whatever it finds inside the first two ' '?
Might it be possible to tell it, "give me the word after "name=' until the next '?
I'm very new to using R so I'm kind of out of ideas.
Thanks a lot for any help!
I know of the gsub command, but I don't think it will be the most appropriate in this case.
Yes, with a regular expression you can do exactly that:
string <- "Place(fullName='Würzburg, Germany', name='Würzburg', type='city', country='Germany', countryCode='DE')"
city <- gsub(".*name='(.*?)'.*", "\\1", string)
The regular expression says "match any characters followed by name=', then capture any characters until the next ' and then match any additional characters". Then you replace all of that with just the captured characters ("\\1").
The parentheses mean "capture this part", and the value becomes "\\1". (You can do multiple captures, with subsequent captures being \\2, \\3, etc.
Note the question mark in (.*?). This means "match as little as possible while still satisfying the rest of the regex". If you don't include the question mark, the regular expression will match "greedily" and you will capture the entire rest of the line instead of just the city since that would also satisfy the regular expression.
More about regular expression (specific to R) can be found here
I'm trying to extract UK postcodes from address strings in R, using the regular expression provided by the UK government here.
Here is my function:
address_to_postcode <- function(addresses) {
# 1. Convert addresses to upper case
addresses = toupper(addresses)
# 2. Regular expression for UK postcodes:
pcd_regex = "[Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([A-Za-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) {0,1}[0-9][A-Za-z]{2})"
# 3. Check if a postcode is present in each address or not (return TRUE if present, else FALSE)
present <- grepl(pcd_regex, addresses)
# 4. Extract postcodes matching the regular expression for a valid UK postcode
postcodes <- regmatches(addresses, regexpr(pcd_regex, addresses))
# 5. Return NA where an address does not contain a (valid format) UK postcode
postcodes_out <- list()
postcodes_out[present] <- postcodes
postcodes_out[!present] <- NA
# 6. Return the results in a vector (should be same length as input vector)
return(do.call(c, postcodes_out))
}
According to the guidance document, the logic this regular expression looks for is as follows:
"GIR 0AA" OR One letter followed by either one or two numbers OR One letter followed by a second letter that must be one of
ABCDEFGHJ KLMNOPQRSTUVWXY (i.e..not I) and then followed by either one
or two numbers OR One letter followed by one number and then another
letter OR A two part post code where the first part must be One letter
followed by a second letter that must be one of ABCDEFGH
JKLMNOPQRSTUVWXY (i.e..not I) and then followed by one number and
optionally a further letter after that AND The second part (separated
by a space from the first part) must be One number followed by two
letters. A combination of upper and lower case characters is allowed.
Note: the length is determined by the regular expression and is
between 2 and 8 characters.
My problem is that this logic is not completely preserved when using the regular expression without the ^ and $ anchors (as I have to do in this scenario because the postcode could be anywhere within the address strings); what I'm struggling with is how to preserve the order and number of characters for each segment in a partial (as opposed to complete) string match.
Consider the following example:
> address_to_postcode("1A noplace road, random city, NR1 2PK, UK")
[1] "NR1 2PK"
According to the logic in the guideline, the second letter in the postcode cannot be 'z' (and there are some other exclusions too); however look what happens when I add a 'z':
> address_to_postcode("1A noplace road, random city, NZ1 2PK, UK")
[1] "Z1 2PK"
... whereas in this case I would expect the output to be NA.
Adding the anchors (for a different usage case) doesn't seem to help as the 'z' is still accepted even though it is in the wrong place:
> grepl("^[Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([A-Za-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) {0,1}[0-9][A-Za-z]{2})$", "NZ1 2PK")
[1] TRUE
Two questions:
Have I misunderstood the logic of the regular expression and
If not, how can I correct it (i.e. why aren't the specified letter
and character ranges exclusive to their position within the regular expression)?
Edit
Since posting this answer, I dug deeper into the UK government's regex and found even more problems. I posted another answer here that describes all the issues and provides alternatives to their poorly formatted regex.
Note
Please note that I'm posting the raw regex here. You'll need to escape certain characters (like backslashes \) when porting to r.
Issues
You have many issues here, all of which are caused by whoever created the document you're retrieving your regex from or the coder that created it.
1. The space character
My guess is that when you copied the regular expression from the link you provided it converted the space character into a newline character and you removed it (that's exactly what I did at first). You need to, instead, change it to a space character.
^([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([AZa-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) [0-9][A-Za-z]{2})$
here ^
2. Boundaries
You need to remove the anchors ^ and $ as these indicate start and end of line. Instead, wrap your regex in (?:) and place a \b (word boundary) on either end as the following shows. In fact, the regex in the documentation is incorrect (see Side note for more information) as it will fail to anchor the pattern properly.
See regex in use here
\b(?:([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([AZa-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) [0-9][A-Za-z]{2}))\b
^^^^^ ^^^
3. Character class oversight
There's a missing - in the character class as pointed out by #deadcrab in his answer here.
\b(?:([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([A-Za-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) [0-9][A-Za-z]{2}))\b
^
4. They made the wrong character class optional!
In the documentation it clearly states:
A two part post code where the first part must be:
One letter followed by a second letter that must be one of ABCDEFGHJKLMNOPQRSTUVWXY (i.e..not I) and then followed by one number and optionally a further letter after that
They made the wrong character class optional!
\b(?:([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([A-Za-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) [0-9][A-Za-z]{2}))\b
^^^^^^
it should be this one ^^^^^^^^
5. The whole thing is just awful...
There are so many things wrong with this regex that I just decided to rewrite it. It can very easily be simplified to perform a fraction of the steps it currently takes to match text.
\b(?:[A-Za-z][A-HJ-Ya-hj-y]?[0-9][0-9A-Za-z]? [0-9][A-Za-z]{2}|[Gg][Ii][Rr] 0[Aa]{2})\b
Answer
As mentioned in the comments below my answer, some postcodes are missing the space character. For missing spaces in the postcodes (e.g. NR12PK), simply add a ? after the spaces as shown in the regex below:
\b(?:[A-Za-z][A-HJ-Ya-hj-y]?[0-9][0-9A-Za-z]? ?[0-9][A-Za-z]{2}|[Gg][Ii][Rr] ?0[Aa]{2})\b
^^ ^^
You may also shorten the regex above with the following and use the case-insensitive flag (ignore.case(pattern) or ignore_case = TRUE in r, depending on the method used.):
\b(?:[A-Z][A-HJ-Y]?[0-9][0-9A-Z]? ?[0-9][A-Z]{2}|GIR ?0A{2})\b
Note
Please note that regular expressions only validate the possible format(s) of a string and cannot actually identify whether or not a postcode legitimately exists. For this, you should use an API. There are also some edge-cases where this regex will not properly match valid postcodes. For a list of these postcodes, please see this Wikipedia article.
The regex below additionally matches the following (make it case-insensitive to match lowercase variants as well):
British Overseas Territories
British Forces Post Office
Although they've recently changed it to align with the British postcode system to BF, followed by a number (starting with BF1), they're considered optional alternative postcodes
Special cases outlined in that article (as well as SAN TA1 - a valid postcode for Santa!)
See this regex in use here.
\b(?:(?:[A-Z][A-HJ-Y]?[0-9][0-9A-Z]?|ASCN|STHL|TDCU|BBND|[BFS]IQ{2}|GX11|PCRN|TKCA) ?[0-9][A-Z]{2}|GIR ?0A{2}|SAN ?TA1|AI-?[0-9]{4}|BFPO[ -]?[0-9]{2,3}|MSR[ -]?1(?:1[12]|[23][135])0|VG[ -]?11[1-6]0|[A-Z]{2} ? [0-9]{2}|KY[1-3][ -]?[0-2][0-9]{3})\b
I would also recommend anyone implementing this answer to read this StackOverflow question titled UK Postcode Regex (Comprehensive).
Side note
The documentation you linked to (Bulk Data Transfer: Additional Validation for CAS Upload - Section 3. UK Postcode Regular Expression) actually has an improperly written regular expression.
As mentioned in the Issues section, they should have:
Wrapped the entire expression in (?:) and placed the anchors around the non-capturing group. Their regular expression, as it stands, will fail in for some cases as seen here.
The regular expression is also missing - in one of the character classes
It also made the wrong character class optional.
here is my regular expression
txt="0288, Bishopsgate, London Borough of Tower Hamlets, London, Greater London, England, EC2M 4QP, United Kingdom"
matches=re.findall(r'[A-Z]{1,2}[0-9][A-Z0-9]? [0-9][ABD-HJLNP-UW-Z]{2}', txt)
I have a question.
My text file contains lines such as:
1.1 Description.
This is the description.
1.1.1 Quality Assurance
Random sentence.
1.6.1 Quality Control. Quality Control is the responsibility of the contractor.
I'm trying to find out how to get:
1.1 Description
1.1.1 Quality Assurance
1.6.1 Quality Control
Right now, I have:
txt1 <- readLines("text1.txt")
txt2<-grep("^[0-9.]+", txt1, value = TRUE)
file<-write(txt2, "text3.txt")
which results in:
1.1 Description.
1.1.1 Quality Assurance
1.6.1 Quality Control. Quality Control is the responsibility of the contractor.
You are using grep with value=TRUE, which
returns a character vector containing the selected elements of x
(after coercion, preserving names but no other attributes).
This means, that if your regular expression matches anything in the line, the all line will be returned. You managed to build your regular expression to match numbers in the begining of the line. So all the lines which begin with numbers get selected.
It seems that your goal is not to select the all line, but to select only until there is a line break or a period.
So, you need to adjust the regular expression to be more specific, and you need to extract only the matching portion of the line.
A regular expression that matches what you want can be:
"^([0-9]\\.?)+ .+?(\\.|$)"
It selects numbers with dots, followed by a space, followed by anything, and stops matching things when a . comes or the line ends. I recommend the following website to better understand what the regex does: https://regexr.com/
The next step is extracting from the given lines only the matching portion, and not the all line where the regex has a match. For this we'll use the function regexpr, which tells us where the matches are, and the function regmatches, which helps us extract those matches:
txt1 <- readLines("text.txt")
regmatches(txt1, regexpr("^([0-9]\\.?)+ .+?(\\.|$)", txt1))
I am trying to figure out how I can put together a find and replace command with wildcards or figure out a way to find and replace the following example:
I would like to find terms that contain double quotes in front of them with a single quote at the end:
Example:
find "joe' and replace with 'joe'
Basically, I'm trying to find all terms with terms having "in front and at the end.'
Check the [x] Regular expression checkbox in textpad's replace dialog and enter the following values:
Find what:
"([^'"]*)'
Replace with:
'\1'
Explanation:
In a regular expression, square brackets are used to indicate character classes. A character class beginning with a caret will match anything not in the class.
Thus [^'"] will match any character except ' and ". The following * indicates that any number of these characters can follow. The ( and ) mark a group. And the group we're looking for starts with " and ends with '. Finally in the replace string we can refer to any group via \n where n is the nth group. In our case it is the first and only group and that is why we used \1.
Updated::
Password strength:
Contain characters from three of the following four categories:
English uppercase characters (A through Z)
English lowercase characters (a through z)
Base 10 digits (0 through 9)
Non-alphabetic characters (for example, !, $, #, %
IS it possible to compare two fields value(entered) with regex...if yes then please add onr another condition to above list.
compare password with username entered they must be different
EDIT: This answer was written before the question was edited. It originally included the requirement to not include the user's account name, and be at least 8 characters long.
Given that you need to use the user's account name as part of it anyway, is there any reason you particularly want to do this as a regular expression? You may want to use regular expressions to express the patterns for the four categories (although there are other ways of doing it too) but I would write the rules out separately. For example:
// Categories is a list of regexes in this case. You could easily change
// it to anything else.
int categories = Categories.Count(regex => regex.IsMatch(password));
bool valid = password.IndexOf(name, StringComparison.OrdinalIgnoreCase) == -1
&& password.Length >= 8
&& categories >= 3;
If you need to do it in one expression it should be something like this:
^(?:(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])|(?=.*[a-z])(?=.*[A-Z])(?=.*[!%,.;:])|(?=.*[a-z])(?=.*[0-9])(?=.*[!%,.;:])|(?=.*[A-Z])(?=.*[0-9])(?=.*[!%,.;:])).{8,}$
See it here on Regexr
Positive lookaheads (the (?=.*[a-z])) are used to check if the string contains the character group you want.
The problem here is, you want 3 out of 4, that means you have to make an alternation with all the allowed combinations.
The last part .{8,} is then matching the string and checking for at least 8 characters.
^ and $ are anchors, that anchor the pattern to the start and the end of the string.
[!%,.;:] is a character class, here you can add all the characters you want to include. Maybe its simpler to use a Unicode script like \p{P} for all punctuation characters. For more details see here on regular-expresssions.info
Update
compare password with username entered they must be different
normally you should be able to build up your regular expression using string concatenation. I have no idea how it is in your case where you put the regex ...
Something like this (pseudo)
String Username = "FooBar";
regex = "^(?:(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])|(?=.*[a-z])(?=.*[A-Z])(?=.*[!%,.;:])|(?=.*[a-z])(?=.*[0-9])(?=.*[!%,.;:])|(?=.*[A-Z])(?=.*[0-9])(?=.*[!%,.;:]))(?i)(?!.*" + Username + ").+$";
I used here also an inline modifier (?i) to match it case independent. The (?!.* is the start of negative lookahead, meaning the string should not contain ...