If cell contains '?' then formula X if not then copy value - if-statement

At the moment I am busy with a spreadsheet to analyse results per url. The problem is that when I want to make a list of unique urls the urls with a parameter behind it (for example '?fbads') will be seen as unique, instead of that I need these results to be blended together with the main url. See example below:
https://www.holidayguru.nl/deal/accommodatie/luxe-strandvakantie-in-ijmuiden-5e25ba62-e001-4072-8eb5-b6c3b0e7e66f/?fbclid=IwA
&
https://www.holidayguru.nl/deal/accommodatie/luxe-strandvakantie-in-ijmuiden-5e25ba62-e001-4072-8eb5-b6c3b0e7e66f/
Should both be: https://www.holidayguru.nl/deal/accommodatie/luxe-strandvakantie-in-ijmuiden-5e25ba62-e001-4072-8eb5-b6c3b0e7e66f/
I already fixed this with a formula but I need one list with all urls. So I'm look for two options. Or in the
=LEFT(A11,FIND("?",A11)-1)
That I use right now I need to find a way how I can say. If you don't find a '?' than just copy cell A11
Or...
I have to work with an if fuction to say, if A11 contains '?' than execute =left fuction otherwise use A11.
I can't manage to get the formula working. Demo sheet is down below :). Thanks!
Example spreadsheet

Delete everything from Sheet1!A:A (including the header) and place the following in Sheet1!A1:
=ArrayFormula({"UNIQUE URLS"; UNIQUE(FILTER(REGEXEXTRACT(URLs!A2:A,"[^\?]+"),URLs!A2:A<>""))})
This will create the header (which you can change as you like within the formula itself) and a unique list of URLs as determined only by the portion before a question mark (if a question mark exists) or to the end of the original URL.
For your reference, the expression [^\?]+ means "a string of the greatest length that can be extracted without containing a literal question mark."
[ ] = "any of the characters contained herein"
[^ ] = "not any of these characters"
\ = literal marker (i.e., whatever is next will be treated as a literal character)
\? = literal question mark (using the literal marker before the ? is necessary, since alone, the ? has a separate special meaning in REGEX-type expressions)
+ = "one or more of the preceding character or group of characters"

Related

Regex to detect string is x.x.x where x is a digit from 1-3 digits

I have values 1000+ rows with variable values entered as below
5.99
5.188.0
v5.33
v.440.0
I am looking in Gsheet another column to perform following operations:
Remove the 'v' character from the values
if there is 2nd '.' missing as so string can become 5.88 --> 5.88.0
Can help please in the regex and replace logic as tried this but new to regex making. Thanks for the help given
=regexmatch(<cellvalue>,"^[0-9]{1}\.[0-9]{1,3}\.[0-9]{1,3}$")
I have done till finding the value as 5.88.0 returns TRUE and 5.99 returns false, need to now append ".0" so 5.99 --> 5.99.0 and remove 'v' if found.
You can use a combination of functions, it may not be pretty, but it does the work
Replace any instance of v with an empty string using substitute, by making the content of the cell upper case, if we don't put UPPER(CELL) we could exclude any upper case V or lower case v(it will depend which function you use)
SUBSTITUTE(text_to_search, search_for, replace_with, [occurrence_number])
=SUBSTITUTE(UPPER(A1),"V","")
Look for cell missing the last block .xxx, you need to update a bit your regex to specified that the last group it's not present
^([0-9]{1}\.[0-9]{1,3} ( \.[0-9]{1,3}){0} )$
Using REGEXMATCH and IF we can then CONCATENATE the last group as .0
REGEXMATCH(text, regular_expression)
CONCATENATE(string1, [string2, ...])
=IF(REGEXMATCH(substitute(upper(A2),"V",""),"^([0-9]{1}\.[0-9]{1,3}(\.[0-9]{1,3}){0})$"),concatenate(A2,".0"), A2)
The last A2 will be replace with something similar than what we have until now, but before that we need to make small change in the regex, we want to look for the groups you specified were the last group it's present, that's your orignal regex, if it meets the regex it will put it in the cell, otherwise it will put INVALID, you can change that to anything you want it to be
^([0-9]{1}.[0-9]{1,3}.[0-9]{1,3})$
This it's the piece we are putting instead of the last A2
IF(REGEXMATCH(substitute(upper(A2),"V",""),"^([0-9]{1}\.[0-9]{1,3}\.[0-9]{1,3})$"),substitute(upper(A2),"V",""),"INVALID")
With this the final code to put in your cell will be:
=IF(REGEXMATCH(substitute(upper(A2),"V",""),"^([0-9]{1}\.[0-9]{1,3}(\.[0-9]{1,3}){0})$"),concatenate(SUBSTITUTE(UPPER(A2),"V",""),".0"),IF(REGEXMATCH(substitute(upper(A2),"V",""),"^([0-9]{1}\.[0-9]{1,3}\.[0-9]{1,3})$"),substitute(upper(A2),"V",""),"INVALID"))

How to remove/replace specials characters from a 'dynamic' regex/string on ruby?

So I had this code working for a few months already, lets say I have a table called Categories, which has a string column called name, so I receive a string and I want to know if any category was mentioned (a mention occur when the string contains the substring: #name_of_a_category), the approach I follow for this was something like below:
categories.select { |category_i| content_received.downcase.match(/##{category_i.downcase}/)}
That worked pretty well until today suddenly started to receive an exception unmatched close parenthesis, I realized that the categories names can contain special chars so I decided to not consider special chars or spaces anymore (don't want to add restrictions to the user and at the same time don't want to deal with those cases so the policy is just to ignore it).
So the question is there a clean way of removing these special chars (maintaining the #) and matching the string (don't want to modify the data just ignore it while looking for mentions)?
You can also use
prep_content_received = content_received.gsub(/[^\w\s]|_/,'')
p categories.select { |c|
prep_content_received.match?(/\b#{c.gsub(/[^\w\s]|_/, '').strip()}\b/i)
}
See the Ruby demo
Details:
The prep_content_received = content_received.gsub(/[^\w\s]|_/,'') creates a copy of content_received with no special chars and _. Using it once reduced overhead if there are a lot of categories
Then, you iterate over the categories list, and each time check if the prep_content_received matches \b (word boundary) + category with all special chars, _ and leading/trailing whitespace stripped from it + \b in a case insensitive way (see the /i flag, no need to .downcase).
So after looking around I found some answers on the platform but nothing with my specific requirements (maybe I missed something, if so please let me know), and this is how I fix it for my case:
content_received = 'pepe is watching a #comedy :)'
categories = ['comedy :)', 'terror']
temp_content = content_received.downcase
categories.select { |category_i| temp_content.gsub(/[^\sa-zA-Z0-9]/, '#' => '#').match?(/##{category_i.downcase.
gsub(/[^\sa-zA-Z0-9]/, '')}/) }
For the sake of the example, I reduced the categories to a simple array of strings, basically the first gsub, remove any character that is not a letter or a number (any special character) and replace each # with an #, the second gsub is a simpler version of the first one.
You can test the snippet above here

Regex Multiple rows [duplicate]

I'm trying to get the list of all digits preceding a hyphen in a given string (let's say in cell A1), using a Google Sheets regex formula :
=REGEXEXTRACT(A1, "\d-")
My problem is that it only returns the first match... how can I get all matches?
Example text:
"A1-Nutrition;A2-ActPhysiq;A2-BioMeta;A2-Patho-jour;A2-StgMrktg2;H2-Bioth2/EtudeCas;H2-Bioth2/Gemmo;H2-Bioth2/Oligo;H2-Bioth2/Opo;H2-Bioth2/Organo;H3-Endocrino;H3-Génétiq"
My formula returns 1-, whereas I want to get 1-2-2-2-2-2-2-2-2-2-3-3- (either as an array or concatenated text).
I know I could use a script or another function (like SPLIT) to achieve the desired result, but what I really want to know is how I could get a re2 regular expression to return such multiple matches in a "REGEX.*" Google Sheets formula.
Something like the "global - Don't return after first match" option on regex101.com
I've also tried removing the undesired text with REGEXREPLACE, with no success either (I couldn't get rid of other digits not preceding a hyphen).
Any help appreciated!
Thanks :)
You can actually do this in a single formula using regexreplace to surround all the values with a capture group instead of replacing the text:
=join("",REGEXEXTRACT(A1,REGEXREPLACE(A1,"(\d-)","($1)")))
basically what it does is surround all instances of the \d- with a "capture group" then using regex extract, it neatly returns all the captures. if you want to join it back into a single string you can just use join to pack it back into a single cell:
You may create your own custom function in the Script Editor:
function ExtractAllRegex(input, pattern,groupId) {
return [Array.from(input.matchAll(new RegExp(pattern,'g')), x=>x[groupId])];
}
Or, if you need to return all matches in a single cell joined with some separator:
function ExtractAllRegex(input, pattern,groupId,separator) {
return Array.from(input.matchAll(new RegExp(pattern,'g')), x=>x[groupId]).join(separator);
}
Then, just call it like =ExtractAllRegex(A1, "\d-", 0, ", ").
Description:
input - current cell value
pattern - regex pattern
groupId - Capturing group ID you want to extract
separator - text used to join the matched results.
Edit
I came up with more general solution:
=regexreplace(A1,"(.)?(\d-)|(.)","$2")
It replaces any text except the second group match (\d-) with just the second group $2.
"(.)?(\d-)|(.)"
1 2 3
Groups are in ()
---------------------------------------
"$2" -- means return the group number 2
Learn regular expressions: https://regexone.com
Try this formula:
=regexreplace(regexreplace(A1,"[^\-0-9]",""),"(\d-)|(.)","$1")
It will handle string like this:
"A1-Nutrition;A2-ActPhysiq;A2-BioM---eta;A2-PH3-Généti***566*9q"
with output:
1-2-2-2-3-
I wasn't able to get the accepted answer to work for my case. I'd like to do it that way, but needed a quick solution and went with the following:
Input:
1111 days, 123 hours 1234 minutes and 121 seconds
Expected output:
1111 123 1234 121
Formula:
=split(REGEXREPLACE(C26,"[a-z,]"," ")," ")
The shortest possible regex:
=regexreplace(A1,".?(\d-)|.", "$1")
Which returns 1-2-2-2-2-2-2-2-2-2-3-3- for "A1-Nutrition;A2-ActPhysiq;A2-BioMeta;A2-Patho-jour;A2-StgMrktg2;H2-Bioth2/EtudeCas;H2-Bioth2/Gemmo;H2-Bioth2/Oligo;H2-Bioth2/Opo;H2-Bioth2/Organo;H3-Endocrino;H3-Génétiq".
Explanation of regex:
.? -- optional character
(\d-) -- capture group 1 with a digit followed by a dash (specify (\d+-) multiple digits)
| -- logical or
. -- any character
the replacement "$1" uses just the capture group 1, and discards anything else
Learn more about regex: https://twiki.org/cgi-bin/view/Codev/TWikiPresentation2018x10x14Regex
This seems to work and I have tried to verify it.
The logic is
(1) Replace letter followed by hyphen with nothing
(2) Replace any digit not followed by a hyphen with nothing
(3) Replace everything which is not a digit or hyphen with nothing
=regexreplace(A1,"[a-zA-Z]-|[0-9][^-]|[a-zA-Z;/é]","")
Result
1-2-2-2-2-2-2-2-2-2-3-3-
Analysis
I had to step through these procedurally to convince myself that this was correct. According to this reference when there are alternatives separated by the pipe symbol, regex should match them in order left-to-right. The above formula doesn't work properly unless rule 1 comes first (otherwise it reduces all characters except a digit or hyphen to null before rule (1) can come into play and you get an extra hyphen from "Patho-jour").
Here are some examples of how I think it must deal with the text
The solution to capture groups with RegexReplace and then do the RegexExctract works here too, but there is a catch.
=join("",REGEXEXTRACT(A1,REGEXREPLACE(A1,"(\d-)","($1)")))
If the cell that you are trying to get the values has Special Characters like parentheses "(" or question mark "?" the solution provided won´t work.
In my case, I was trying to list all “variables text” contained in the cell. Those “variables text “ was wrote inside like that: “{example_name}”. But the full content of the cell had special characters making the regex formula do break. When I removed theses specials characters, then I could list all captured groups like the solution did.
There are two general ('Excel' / 'native' / non-Apps Script) solutions to return an array of regex matches in the style of REGEXEXTRACT:
Method 1)
insert a delimiter around matches, remove junk, and call SPLIT
Regexes work by iterating over the string from left to right, and 'consuming'. If we are careful to consume junk values, we can throw them away.
(This gets around the problem faced by the currently accepted solution, which is that as Carlos Eduardo Oliveira mentions, it will obviously fail if the corpus text contains special regex characters.)
First we pick a delimiter, which must not already exist in the text. The proper way to do this is to parse the text to temporarily replace our delimiter with a "temporary delimiter", like if we were going to use commas "," we'd first replace all existing commas with something like "<<QUOTED-COMMA>>" then un-replace them later. BUT, for simplicity's sake, we'll just grab a random character such as  from the private-use unicode blocks and use it as our special delimiter (note that it is 2 bytes... google spreadsheets might not count bytes in graphemes in a consistent way, but we'll be careful later).
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN(""))
)(
REGEXREPLACE(
"xyzSixSpaces:[ ]123ThreeSpaces:[ ]aaaa 12345",".*?( |$)",
"$1"
)
),
""
)
We just use a lambda to define temp="match1match2match3", then use that to remove the last delimiter into "match1match2match3", then SPLIT it.
Taking COLUMNS of the result will prove that the correct result is returned, i.e. {" ", " ", " "}.
This is a particularly good function to turn into a Named Function, and call it something like REGEXGLOBALEXTRACT(text,regex) or REGEXALLEXTRACT(text,regex), e.g.:
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN(""))
)(
REGEXREPLACE(
text,
".*?("&regex&"|$)",
"$1"
)
),
""
)
Method 2)
use recursion
With LAMBDA (i.e. lets you define a function like any other programming language), you can use some tricks from the well-studied lambda calculus and function programming: you have access to recursion. Defining a recursive function is confusing because there's no easy way for it to refer to itself, so you have to use a trick/convention:
trick for recursive functions: to actually define a function f which needs to refer to itself, instead define a function that takes a parameter of itself and returns the function you actually want; pass in this 'convention' to the Y-combinator to turn it into an actual recursive function
The plumbing which takes such a function work is called the Y-combinator. Here is a good article to understand it if you have some programming background.
For example to get the result of 5! (5 factorial, i.e. implement our own FACT(5)), we could define:
Named Function Y(f)=LAMBDA(f, (LAMBDA(x,x(x)))( LAMBDA(x, f(LAMBDA(y, x(x)(y)))) ) ) (this is the Y-combinator and is magic; you don't have to understand it to use it)
Named Function MY_FACTORIAL(n)=
Y(LAMBDA(self,
LAMBDA(n,
IF(n=0, 1, n*self(n-1))
)
))
result of MY_FACTORIAL(5): 120
The Y-combinator makes writing recursive functions look relatively easy, like an introduction to programming class. I'm using Named Functions for clarity, but you could just dump it all together at the expense of sanity...
=LAMBDA(Y,
Y(LAMBDA(self, LAMBDA(n, IF(n=0,1,n*self(n-1))) ))(5)
)(
LAMBDA(f, (LAMBDA(x,x(x)))( LAMBDA(x, f(LAMBDA(y, x(x)(y)))) ) )
)
How does this apply to the problem at hand? Well a recursive solution is as follows:
in pseudocode below, I use 'function' instead of LAMBDA, but it's the same thing:
// code to get around the fact that you can't have 0-length arrays
function emptyList() {
return {"ignore this value"}
}
function listToArray(myList) {
return OFFSET(myList,0,1)
}
function allMatches(text, regex) {
allMatchesHelper(emptyList(), text, regex)
}
function allMatchesHelper(resultsToReturn, text, regex) {
currentMatch = REGEXEXTRACT(...)
if (currentMatch succeeds) {
textWithoutMatch = SUBSTITUTE(text, currentMatch, "", 1)
return allMatches(
{resultsToReturn,currentMatch},
textWithoutMatch,
regex
)
} else {
return listToArray(resultsToReturn)
}
}
Unfortunately, the recursive approach is quadratic order of growth (because it's appending the results over and over to itself, while recreating the giant search string with smaller and smaller bites taken out of it, so 1+2+3+4+5+... = big^2, which can add up to a lot of time), so may be slow if you have many many matches. It's better to stay inside the regex engine for speed, since it's probably highly optimized.
You could of course avoid using Named Functions by doing temporary bindings with LAMBDA(varName, expr)(varValue) if you want to use varName in an expression. (You can define this pattern as a Named Function =cont(varValue) to invert the order of the parameters to keep code cleaner, or not.)
Whenever I use varName = varValue, write that instead.
to see if a match succeeds, use ISNA(...)
It would look something like:
Named Function allMatches(resultsToReturn, text, regex):
UNTESTED:
LAMBDA(helper,
OFFSET(
helper({"ignore"}, text, regex),
0,1)
)(
Y(LAMBDA(helperItself,
LAMBDA(results, partialText,
LAMBDA(currentMatch,
IF(ISNA(currentMatch),
results,
LAMBDA(textWithoutMatch,
helperItself({results,currentMatch}, textWithoutMatch)
)(
SUBSTITUTE(partialText, currentMatch, "", 1)
)
)
)(
REGEXEXTRACT(partialText, regex)
)
)
))
)

Spotfire: count the number of a certain character in a string

I am trying to add a new calculated column that counts the number of semi colons in a string and adds one to it. So the column i have contains a bunch of aliases and I need to know how many for each row.
For example,
A; B; C; D
So basically this means there are 4 aliases (3 semi colons + 1)
Need to do this for over 2 million rows. Help please!
Basic idea is to subtract length of your string without ; characters from it's original length:
len([columnName])-len(Substitute([columnName],";",""))+1
Here it is with a regular expression:
Len(RXReplace([Column 1], "(?!;).", "", "gis"))+1
RXReplace takes as arguments:
The string you are wanting to work on (in this case it is on Column 1)
The regular expression you want to use (here it is (?!;). )
What you want to replace matches with (blank in this situation so
that everything that matches the regex is removed)
Finally a parameter saying how you want it to work (we are passing
in gis which means replace all matches not just the first, ignore case, replace newlines)
We wrap this in a Len which gives us the amount of semicolons since that is all that is left and finally we add 1 to it to get the final result.
You can read more about the regular expression here: https://msdn.microsoft.com/en-us/library/az24scfc(v=vs.110).aspx but in a nutshell it says match everything that isn't a semi colon.
You can read more about RXReplace and Len here: https://docs.tibco.com/pub/spotfire/6.0.0-november-2013/userguide-webhelp/ncfe/ncfe_text_functions.htm

Railo, remove some double quotes from SerializeJSON result

Let's say I have:
<cfscript>
arrButtons = [
{
"name" = "Add",
"bclass" = "add",
"onpress" = "addItem"
},
{
"name" = "Edit",
"bclass" = "edit",
"onpress" = "editItem"
},
{
"name" = "Delete",
"bclass" = "delete",
"onpress" = "deleteItem"
}
];
jsButtons = SerializeJSON(arrButtons);
// result :
// [{"onpress":"addItem","name":"Add","bclass":"add"},{"onpress":"editItem","name":"Edit","bclass":"edit"},{"onpress":"deleteItem","name":"Delete","bclass":"delete"}]
</cfscript>
For every onpress item, I need to remove the double quotes from its value to match the JS library requirement (onpress value must a callback function).
How do I remove the double quotes using a regular expression?
The final result must be:
[{"onpress":addItem,"name":"Add","bclass":"add"},{"onpress":editItem,"name":"Edit","bclass":"edit"},{"onpress":deleteItem,"name":"Delete","bclass":"delete"}]
No double quotes surrounding addItem, editItem, and deleteItem.
Edit 2012-07-13
Why I need this? I created a CFML function that the result is a collection of JS that will be used in many files. jsButton object will be used as one part of the options available in the JS library. One of that function's arguments is an array of struct (the default is arrButtons), and the supplied arguments value can merge with the default value.
Since we can't (in CFML) write onpress value without double quotes, so I have to add double quotes to that value, and convert the (CFML) array of struct to JSON (which is just a string) and remove the double quotes before place it in the JS library option.
with Railo, we can declare the struct as a linked struct to make sure we have same ordered key for loop or conversion (from above example onpress always the latest key in the struct). with this linked struct and same key order, we can remove the double quotes with simple Replace function, but of course we can't guarantee every programmer who use the CFML function doesn't forget to use linked struct and key order same as example above
I'm not sure this is actually necessary - depending on how/where you're dealing with the JS callbacks, it might be possible to use the string function names to reference the function without needing to remove the quotes (i.e. object[button.onpress]).
However, since you asked, here is a regex solution:
jsButtons = jsButtons.replaceAll('(?<="onpress":)"([^"]+)"','$1');
The regex there is made up of two parts:
(?<="onpress":) -- lookbehind to ensure we are dealing with the text "onpress":
"([^"]+)" -- match the quotes and capture their contents.
The $1 on the replacement side is to replace the matched text (i.e. the entire quoted value) with the first capture group (i.e. the contents of the quotes).
If case-sensitivity of "onpress" might be an issue, you can prefix the regex with (?i) to ignore case.
If there will be multiple different events (not just "onpress") you can update the relevant part of the expression above to be (?<="on(?:press|hover|squeek)":) etc.
Note: All the above relies on the format output from serializeJson not changing - if it's possible that there might be comments, whitespace, single quotes, or anything else in future then a longer expression would be needed to cater for those - which is part of why you should investigate if you even need regex to solve this problem in the first place.
What you're wanting to output is not JSON, so using SerializeJSON is a kludge.
Is there any reason you are putting it into a ColdFusion Array first, instead of writing the Javascript directly?
JSON is purely meant to be a data description language. Per
http://www.json.org, it is a "lightweight data-interchange format." -
not a programming language.
Per http://en.wikipedia.org/wiki/JSON, the "basic types" supported
are:
Number (integer, real, or floating point)
String (double-quoted Unicode with backslash escaping)
Boolean (true and false)
Array (an ordered sequence of values, comma-separated and enclosed in square brackets)
Object (collection of key:value pairs, comma-separated and enclosed in curly braces)
null
--Source
I guess in this case you can simply use serialize(). That should do the trick...
Gert