I would like to isolate all operands from a formula (in the form of a string) by taking out the arithmetic operators so take out: "+","-","/","*","**2"
the formula string is something like:
"y=A+B1*options+B2*items**2+B3*factor+B4"
However: I can manage for most arithmetic operators, except for the exponents "**2" part. It has to be a wildcard search or so (not positional), because the whole formula might change in future and also might have another exponent (eg **5 or **54)
What would be the easiest way to strip "**?" out of the formula where ? can be any number?
To match the pattern you want, use the regex string r"\*\*\d+"
Breakdown:
r"" is the how one denotes regex in python (see the re module for more info)
\* matches a single * character - because the * is a special character in regex, we escape it with the \
\d matches a digit
+ matches the previous pattern at least once greedily: this means it will try to find at least one digit, then keep finding digits until it can find no more. So, it will match **2, **44382, and so on
As for stripping the pattern from the equation, you can do re.sub(pattern, "", equation) - replacing all instances of the pattern with nothing
Related
I take this beautiful formula from JvdV answer:
=TRIM(CONCAT(IF(ISNUMBER(SEARCH(MID(A1,ROW(A$1:INDEX(A:A,LEN(A1))),1),"-./ 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ")),MID(A1,ROW(A$1:INDEX(A:A,LEN(A1))),1)," ")))
This formula replace any non-alphanumeric character (&^%]#$) with simple space " ".
I put in formula some exception (-./ ), but this is not all exceptions.
How about wildcards? How to filter wildcards (~*?) with this formula?
I think: Ok, I will use FIND instead of SEARCH and all will be right, just put lowercase and uppercase alphabet in the FIND index, like this: *"-./ 0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"*
Then I think: But, what if I want to keep not only numeric and regular alphabet? What if I want to keep all diacritics, like this: "ÁÀȦÄǍĀÃÅĄȺẤẦẮẰǠǺǞẪẴẢȀȂẨẲẠḀẬẶĂÂḂɃƁḄḆĆĊĈČÇȻḈƇƆḊĎḐĐƊḌḒḎÐƉÉÈĖÊËĚĔĒẼĘȨɆẾỀḖḔỄḜẺȄȆỂẸḘḚỆÉÈÊËḞƑǴĠĜǦĞḠĢǤƓḢĤḦȞḨĦḤḪⱧÍÌİÏǏĬĪĨĮƗḮỈȈȊỊḬÍÌÏÎȷĴǰḰǨĶƘᶄḲḴⱩꝀꝂꝄĹĿĽⱢⱠĻȽŁḶḼḺḸꝈḾṀṂŃǸṄŇÑŅƝṆṊṈÑŊÓÒȮÔÖǑŎŌÕǪŐỐỒƟØṒṐṌȪỖṎǾȬǬỎȌȎƠỔỌỚỜỠỘỞỢÓÒÔÖÕØṔṖⱣƤƦŔṘŘŖɌⱤȐȒṚṞṜŚṠŜŠṤṦṢṨŞṪŤƬṬƮṰṮȾŢŦÚÙÛÜǓŬŪŨŮŲŰɄǗǛṸṺỦȔȖƯỤṲỨỪṶṴỮỬỰÚÙÛÜṼṾẂẀẆŴẄẈẊẌÝỲẎŶŸȲỸɎỶƳỴÝŹŻẐŽƵẒẔ"
Then lowercase and uppercase alphabet is too much for FIND index.
Ok, for SEARCH index is also too much, because function accept max. 255 length, but lets say we have only 200 characters in index (numbers, alphabet and some diacritics)
So, the question is available:
How to filter (replace with space) wildcards (~*?) with this kind of formula?
As I read this question there are a few problems:
How to include over 255 characters in the 2nd parameter of SEARCH();
How to exclude literal wildcard characters in the 2nd parameter of SEARCH();
One way around the length limit is to feed SEARCH() an array of options, in this case an array of two elements of a lenght of <255:
Formula in C1:
=TRIM(CONCAT(IF(MMULT(IFERROR(SEARCH("~"&MID(A1,ROW(A$1:INDEX(A:A,LEN(A1))),1),{"ÁÀȦÄǍĀÃÅĄȺẤẦẮẰǠǺǞẪẴẢȀȂẨẲẠḀẬẶĂÂḂɃƁḄḆĆĊĈČÇȻḈƇƆḊĎḐĐƊḌḒḎÐƉÉÈĖÊËĚĔĒẼĘȨɆẾỀḖḔỄḜẺȄȆỂẸḘḚỆÉÈÊËḞƑǴĠĜǦĞḠĢǤƓḢĤḦȞḨĦḤḪⱧÍÌİÏǏĬĪĨĮƗḮỈȈȊỊḬÍÌÏÎȷĴǰḰǨĶƘᶄḲḴⱩꝀꝂꝄĹĿĽⱢⱠĻȽŁḶḼḺḸꝈḾṀṂŃǸṄŇÑŅƝṆṊṈÑŊÓÒȮÔÖǑŎŌÕǪŐỐỒƟØṒṐṌȪỖṎǾȬǬỎȌȎƠỔỌỚỜỠỘỞỢÓÒÔÖÕØṔṖⱣƤƦŔṘŘŖɌⱤ";"ȐȒṚṞṜŚṠŜŠṤṦṢṨŞṪŤƬṬƮṰṮȾŢŦÚÙÛÜǓŬŪŨŮŲŰɄǗǛṸṺỦȔȖƯỤṲỨỪṶṴỮỬỰÚÙÛÜṼṾẂẀẆŴẄẈẊẌÝỲẎŶŸȲỸɎỶƳỴÝŹŻẐŽƵẒẔ-./*? 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"}),0),{1,1}),MID(A1,ROW(A$1:INDEX(A:A,LEN(A1))),1)," ")))
What we did here is:
Use an horizontal array {abc;xyz} to check against our characters which was an vertical array {a,b,c}. Note the difference between semi-column and comma.
The result will be a 2D-array which MMULT() can sum. Meaning if the character was found in any of the two elements of the array it will return that same character. Otherwise, a space.
The special wildcard characters are now also included with an extra tilde to escape them as with actually all characters.
If Excel doesn't recognize all lowercase diacritics as their uppercase counterparts, just add them to one of the two elements. If need be, add a 3rd. But know that you'd need to extend on the 2nd parameter in MMULT() too then.
To visualize the above:
Remember, you are using Excel 2019 which means you need to CSE-enter this formula. Needles to say that all will be much easier in ms365 using its dynamic array functionality.
I´m working with VBA and trying to split a string into three columns, almost all strings are like Company Name 3567782 Agent Name.pdf
With this pattern I want to match all the text before a space and digits (1st group), the digits (2nd group) and all the text after the space and before the .pdf (3rd group).
strPattern = "^(.+)\n(\d{4,10})\n(.+).pdf"
I recall spaces in python are \s but saw in VBA are \n.
Can you help me find the right pattern for what I´m looking for?
As I put in my comment, I use the https://regex101.com site. There are others but I find this one the most helpful to me.
When I put in your regex
^(.+)\n(\d{4,10})\n(.+).pdf
and test string
Company Name 3567782 Agent Name.pdf
the first thing I notice is that the regex does not match the test string (see right side under MATCH INFORMATION).
Here are a couple things that I saw:
\n is newline, not space. In regex, space is " ".
Your last "." in ".pdf" is not registering as a literal period, it's a token that matches any character. To match a literal period, you need \.
If we change those two things it returns three groups that seem to match what you are looking for.
^(.+) (\d{4,10}) (.+)\.pdf
It looks like for the digits, you are looking for between 4 and 10 digits. If that's correct, it looks like your regex is good. You could put in a handful of example strings into the TEST STRING area and make sure that it works in all cases.
I'd use either of these:
(?:(?:([a-zA-Z]+\.?)|(\d+)))
capture a-Z greedy with a possible . to allow for the .pdf or capture digits
this version excludes the space [ ] or \s
or keep the search structured so you can control what goes in and out of each column
^(\w+\s\w+)|(\d+)|(\w+\s\w+\.\w+$)
\b or ^ - word boundary or start of string
(\w+\s\w+) - 1st capture \w+ - any alpha numeric char greedily, followed by 1 x space (use \s* or \s+ for more), followed again by alpha numeric greedily
|(\d+) - alteration - \d+ - capture just digits
`|(\w+\s\w+.\w+$) - similar to 1st group but allows for the '.' of pdf and bounds to the end of string (\G or $).
you could optionally build the '.' into the 1st group like my top answer, but for neatness and better control I prefer the 2nd.
I'm using the FIND function in Excel to check whether certain characters appear in a string of characters in a cell.
However, this function doesn't work cleanly for certain special characters. Specifically F̌,B̌, and some others. When F̌ appears in the string, FIND recognizes it as both F and F̌.
Notable that this is not the case for characters such as Ď and Č. FIND works nicely for these.
How can I get the formula to always differentiate between characters with and without the hat? Is there a way to work in EXACT?
Thank you!
It is because F̌ is actually two characters.
=LEN("F̌") returns 2 not 1. The second character is the hat.
If you do:
=UNICHAR(70)&UNICHAR(780)
It will return the F̌
And as such =FIND("F","F̌") will return 1 as it is the first letter of a two character string.
To find "F" in A,B,F̌,F use:
=AGGREGATE(15,7,ROW($ZZ1:INDEX($ZZ:$ZZ,LEN(A1)))/((MID(A1,ROW($ZZ1:INDEX($ZZ:$ZZ,LEN(A1))),1)="F")*(MID(A1,ROW($ZZ2:INDEX($ZZ:$ZZ,LEN(A1)+1)),1)<>UNICHAR(780))),1)
To find either then we need to use IF:
=IF(LEN(A2)=2,FIND(A2,A1),AGGREGATE(15,7,ROW($ZZ$1:INDEX($ZZ:$ZZ,LEN(A1)))/((MID(A1,ROW($ZZ$1:INDEX($ZZ:$ZZ,LEN(A1))),1)=A2)*(MID(A1,ROW($ZZ$2:INDEX($ZZ:$ZZ,LEN(A1)+1)),1)<>UNICHAR(780))),1))
Given that your substrings are comma-separated, look for the character followed by a comma (and add a comma to the end of the string to find the last character).
This allows you to separate multicharacter substrings from uni-character substrings where the latter is contained in the former.
You could use something like:
=FIND("F,",A5&",")
That will find an F in A5, but will not find an F if only F̌ is present
Is it possible to display the line count in the VIM status bar with thousands separators, preferably custom thousands separators?
Example:
set statusline=%L
should lead to "1,234,567" instead of "1234567".
I've found a way but it looks a bit crazy:
set statusline=%{substitute(line('$')\,'\\d\\zs\\ze\\%(\\d\\d\\d\\)\\+$'\,'\,'\,'g')}
The first round of backslashes is just for set (I have to escape , and \ itself).
What I'm actually setting the option to is this string:
%{substitute(line('$'),'\d\zs\ze\%(\d\d\d\)\+$',',','g')}
As a format string, this line contains one formatting code, which is %{...}. Everything in ... is evaluated as an expression and the result substituted back in.
The expression I'm evaluating is (spaces added (if I had added them to the real code, I would've had to escape them for set again, forcing yet more backslashes)):
substitute(line('$'), '\d\zs\ze\%(\d\d\d\)\+$', ',', 'g')
This is a call to the substitute function. The arguments are the source string, the regex, the replacement string, and a list of flags.
The string we're starting with is line('$'). This call returns the number of lines in the current buffer (or rather the number of the last line in the buffer). This is what %L normally shows.
The search pattern we're looking for is \d(\d\d\d)+$ (special vim craziness removed), i.e. a digit followed by 1 or more groups of 3 digits, followed by the end of the string. Grouping is spelled \%( \) in vim, and "1 or more" is \+, which gives us \d\%(\d\d\d\)\+$. The last bit of magic is \zs\ze. \zs sets the start of the matched string; \ze sets the end. This works as if everything before \zs were a look-behind pattern and everything after \ze were a look-ahead pattern.
What this amounts to is: We're looking for every position in the source string that is preceded by a digit and followed by exactly N digits (where N is a multiple of 3). This works like starting at the right and going left, skipping 3 digits each time. These are the positions where we need to insert a comma.
That's what the replacement string is: ',' (a comma). Because we're matching a string of length 0, we're effectively inserting into the source string (by replacing '' with ',').
Finally, the g flag says to do this with all matches, not just the first one.
TL;DR:
line('$') gives us the number of lines
substitute(..., '\d\zs\ze\%(\d\d\d\)\+$', ',', 'g') adds commas where we want them
%{ } lets us embed arbitrary expressions into statusline
I have a very large number (a couple hundred digits long), and I'd like to use vim to add commas to the number in the appropriate manner, i.e. after each group of three digits, moving from right to left. How can I do this efficiently?
Taken from here
Substitue command that adds commas in the right spot.
:%s/\(\d\)\(\(\d\d\d\)\+\d\#!\)\#=/\1,/g
This uses a zero width lookahead to match any number that isn't followed by groups of three numbers followed by one number. (or 3n+1 numbers)
So the numbers that match in are marked with ^. These are then replaced with a comma after it the match.
31415926
^ ^
Which replaces to
31,415,926
A friend of mine suggests using the printf program: ciw<C-r>=system("printf \"%'d\" ".shellescape(#"))<CR>.
This is one way of doing it:
s/\d\{-1,}\ze\(\d\{3}\)\+\s/&,/g
Notes:
\{-1,} is saying match at least 1 but in a non-greedy way (Vim doesn't seem to support the usual \+\? syntax; also, for quantifiers, you just need to escape the opening curly brace)
\ze is saying match the pattern behind this but don't store the match in & (equivalent to positive look-ahead)
\(\d\{3}\)\+\> matches groups of 3 digits that ends with word-nonword boundary (word in this sense means alphanumerical + underscore).
Alternatively, you can use \s for space/tab, or \D for non-digit instead of \>, whichever fits your needs better
The way that I used is to create a macro that adds one single comma, and then invoke the macro a whole bunch of times, like qahhi,<ESC>hq#a#a#a#a…