I'm making a simple android game in Lua, and in one of its steps to set the game is set an word (or sentence; basically, a string) input by the player. The "word" may have spaces, but I want to forbid the player to input a string with two or more spaces in a row, like "fly bird".
I tried using string.match(word, " "), string.match(word, "%s%s")
and string.match(word, "%s+%s+") and none of these worked, and somehow, the last one always "detect" double space, no matter if it has or not spaces.
What can I do to detect if there are multiple spaces in a row in a string? (Just detect, not replace, so I can send a warning message to the player.)
If its exactly two spaces you are interested in, simply use find
word:find(' ')
It will return range of first occurrence of two consecutive spaces.
input = input:gsub("%s+", " ")
The above code should take the input and remove all excessive spacing and replace it with just 1 space.
Related
In the image, 'muddle' is the string containing junk words and the strings I want to extract. There is a fixed list of junk words - the good strings could be literally anything.
You can see this formula has correctly extracted "moo" and "coo", which are not in the list of junk words. The formula is below.
=LET(junkStart,FILTER(SEARCH(Table1[junkwords],Table2[muddle]),ISNUMBER(SEARCH(Table1[junkwords],Table2[muddle]))),
junkEnd,FILTER(SEARCH(Table1[junkwords],Table2[muddle])+LEN(Table1[junkwords])-1,ISNUMBER(SEARCH(Table1[junkwords],Table2[muddle])+LEN(Table1[junkwords])-1)),
goodstart,FILTER(junkEnd+1,(junkEnd+1<=LEN(Table2[muddle]))*(ISERROR(XMATCH(junkEnd+1,junkStart)))),
goodend,FILTER(junkStart-1,(junkStart-1>=LEN(1))*(ISERROR(XMATCH(junkStart-1,junkEnd))))+1,
goodchars,goodend-goodstart,
TEXTJOIN("; ",TRUE,MID(Table2[muddle],goodstart,goodchars)))
This works well, but it falls down if a junk word occurs more than once. See below.
The only difference is that 'woo' occurs twice in the second example.
I need a single cell solution. VBA is not an option for me. Using the name manager would be untidy, as would nested formulas.
I've got this far with formulas, which as far as I can tell is the furthest anyone has got with the 'removing multiple words from a cell' problem. I can see the issue - once SEARCH locates the start of a string in a cell, it doesn't go looking for a second occurrence of that string. But I don't know how to find the start of every instance of every string. Can anyone help?
REDUCE is perfect for this:
=REDUCE(Table2[muddle],Table1[junkwords],LAMBDA(m,j,SUBSTITUTE(m,j,"")))
REDUCE starts at the Table2[muddle] value as m then it substitutes the first value of Table1[junkwords] j with "" the outcome becomes the new m which will get a substitute of the second value of j. The result will be the new m, etc.
If you would want to have it comma separated it becomes more complicated, but you can realize by:
=LET(t,SUBSTITUTE(","&REDUCE(Table2[muddle],Table1[junkwords],LAMBDA(x,y,SUBSTITUTE(x,y,",")))&",",",,",","),
MID(t,2,LEN(t)-3))
This does almost the same as the previous solution, but instead of substituting for blanks it substitutes for , and substitutes all duplicate ,, for singles, so if more substitutes followed eachother it results in one comma. Also, if the first and/or last part got substituted by a single ,, then the result would have a leading and/or trailing ,. This is solved by first adding , in the front and back before substituting the double comma's for singles. the result t is then wrapped in MID, where the first and last character (both being a ,) are removed.
Alternate solution:
=LET(t,REDUCE(Table2[muddle],Table1[junkwords],LAMBDA(x,y,SUBSTITUTE(x,y," "))),
SUBSTITUTE(TRIM(t)," ",","))
Or in one go if you don't want to use LET:
=SUBSTITUTE(TRIM(REDUCE(Table2[muddle],Table1[junkwords],LAMBDA(x,y,SUBSTITUTE(x,y," "))))," ",",")
This replaces the junk words with a space. Regardless how many junk words in between words or how many trailing or leading spaces TRIM will fix it to the words separated by one space only. Substituting the spaces for comma gets to your result.
There's no single-formula solution if the junkwords list is not fixed.
Instead, you may choose to use the Substitute() function on each cell of the "Extracted Strings" column to substitute all occurances of each junk word in muddle, i.e. substitute "boo" muddle, then substitute "voo" in the resulted string, replace "noo" in the resulted string...so on. You will get the last cell.
One point to note though, you need to ensure no substring / partial strings problem in the junkwords or you need to define the rules of processing in order for the solution to be "complete". Consider the followings:
junk words = abc, def, cde
muddle = 1234abcdef5678
if you process the string in the above order, you got "12345678"
if you process the junk words in reverse order, you got "123abf5678"
I have a sequence of a letters, in my case part of a gene. I want to change the first and the last letter in this string of text, but keep the internal characters the same.
For example, if I have the the sequence:
ATCGAATCCATGACG
And I want to change the first letter, in this case A to the word START and change the last letter, G, to STOP all while keeping the internal A's and G's the same. Is this possible to do with the Find and Replace function, or will I have to write a script?
It is easy to do when I have a handful of sequences, I do it by hand. When I get into the hundreds, it can be very difficult.
Thank you.
The function LEN(text) returns the number of characters within a string of letters. MID(text, start, num_chars) returns the middle section of a string. CONCATENATE(text1, text2, ...) pieces together different strings. We can use these in combination to get what you want:
=CONCATENATE("START", MID(A1,2,LEN(A1)-2), "STOP")
You could use replace, and focus on the left and right side independently, then combine, or you can use left/right to add string of text to the available string minus a character, like:
="START"&LEFT(RIGHT(A1,LEN(A1)-1),LEN(A1)-2)&"STOP"
I used left/right, but mid would also work
just another option:
=REPLACE(REPLACE(A1,LEN(A1),1,"STOP"),1,1,"START")
I've been trying a few different ways to try and search and replace on excell to remove the last couple of characters.
For instance in one column I have product name S
I want to remove the " S" only.
I have tried some if formulas a swell and not had much luck. I'm assuming there is a simple regex that can be used for the search and replace e.g. " S/" that would just replace if its the last characters and has nothing after it.
Try using the SUBSTITUTE function and replace the letters you want to remove with a unique character/ word / space not appearing anywhere else in the booklet, depending on which part of the string you're trying to remove and what format you're trying to keep
then find and replace ( CTRL +F) that word with the black (space) character
see how to use SUBSTITUTE function here:
https://exceljet.net/excel-functions/excel-substitute-function
Since you are only interested in the end of the string, I don't think you need regex or anything too sophisticated.
If I understand correctly, you want to get the original string (product name S) up until but not including something that appears at the end (S). This means that in your example, you want the 12 leftmost digits: the digits of the original string (14) minus the digits of the pattern (2) - this would give you product name. If the original string does not end with the pattern, you want the original string.
Therefore, I suggest the following:
=IF(RIGHT("original string",LEN("pattern"))="pattern",
LEFT("original string",LEN("original string")-LEN("pattern")),
"original string")
Check these examples:
I have a string, and I need to calculate the number of spaces that I remove when I do trimStart.
For example, I have the following string \t\t \tabcs
so I have two tabs and two spaces and another tab that will be removed using trim start (the rest is non space related chars).
I need to know how many spaces will be removed. since I don't know how much is \t, I can't just count it as a single char.
(My purpose is to calculate the column shift of a string due to the trimming action. Obviously comparing the lengths before and after the trim will not return me the desired result.
Do you have any ideas?
Thanks!
I am translating a game, and the game's text box only supports 50 characters max per line. Is there a way to use a formula to split the entire sentence every 50 characters or whole word (49, 48, 47, etc)?
I am currently working with this formula.
=JOIN(CHAR(10),SPLIT(REGEXREPLACE(A1, "(.{50})", "/$1"),"/"))
The problem with this code, is that it splits at exactly 50 characters (one time), and will split in the middle of the word.
So again, my goal is to have it not split on the 50th character IF the 50th character is in the middle of the word, and for the rule to apply for the rest of the lines too because it only applies on the first line.
Please take a look at this test google sheet to get an example of what I am talking about.
If it's impossible to do it on Google Sheets, I don't mind moving to Excel provided I get a functioning code.
For the record, I did ask in Google's product forums 2 days ago, and still haven't received an answer.
=REGEXREPLACE(A1, "(.{1,50})\b", "$1" & CHAR(10))
{50} matches exactly 50 times, but what you need is 50 or less.
\b is word boundary that matches between alphanumeric and non-alphanumeric character.
= REGEXEXTRACT(A1,"(?ism)^"&REPT("([\w\d'\(\),. ]{0,49}\s)", ROUNDUP(LEN(A1)/50,0))&"([\w\d'\(\),. ]{0,49})$")
Tested with various expressions and works as intended. Note that only these characters [a-zA-Z0-9_'(),.] are allowed, Which means - and other characters not mentioned will not work. If you need them, add them inside the REPT expression and finishing regexp formula. Otherwise, This will work perfectly.
You are pretty close. I'm not an expert in Sheets, so not sure if this is the best way, but your Regex is wrong for what you want.
Also, you need to be certain that you don't use a split character that might appear in the phrase itself. However, using CHAR(10) for the replace character allows you to insert LF without going through the JOIN SPLIT sequence.
replace any line feeds, carriage returns and spaces with a single space
Match strings that start with a non-Space character followed by up to 49 more characters which are followed by a space or the end of the string.
replace the capture group with the capturing group followed by the CHAR(10) (and delete the space following).
There will be extra CHAR(10) at the end which you can strip off.
EDIT Regex changed slightly due to a difference in behavior between Google's RE and what I am used to (probably has to do with how a non-backtracking regex works). The problem showed up on your example:
=regexreplace(REGEXREPLACE(REGEXREPLACE(A1 & " ","[\r\n\s]+"," "),"(\S.{0,49})\s","$1" & char(10)),"\n+\z","")