Just note that custom Sheet functions need to be deterministic. 25 to 40) =CONCATENATE(ARRAYFORMULA(MID($A$1, CEILING(RANDARRAY(1, RANDBETWEEN(25, 40)), 1 / LEN($A$1)) * LEN($A$1), 1)))Īpps Script, using javascript, may work. for the OP: ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789, then this would work: Fixed length (e.g. Say you have cell A1 containing your key character set, i.e. :-) =CONCATENATE(ARRAYFORMULA(MID("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789", CEILING(RANDARRAY(1, 8), 1 / 62) * 62, 1))) Which is looks like the most statistically sound so far by the limits of pseudo-random generation. Changing the 2 to 8 yields a length=8 string. This has a neat solution, but only provides a limited character set (see SnosRap75's answer). =CONCATENATE(ARRAYFORMULA(CHOOSE(RANDBETWEEN(SEQUENCE(1, 8, 1, 0),3),Ĭharacters 0-9, A-F equally likely (G-Z & a-z never occur) This is the random string type you get by using the previously suggested methods with RANDBETWEEN(1,3) instead of the more complicated ceiling and rand functions. Every character type equally likely: p()=p()=p()=1/3 To get digits to come up with the right probability (10/62) the choose function needs to be fed with integers 1, 2, 3, this is achieved by applying ceiling with rand and an appropriate offset. This was inspired by Chris River's neat variable length solution - but with RANDARRAY instead of SEQUENCE to suit the more complicated circumstances. In probabilities: p('0')=p('1')=.=p('A')=.p('z')=1/62Īll characters used must be equally likely, so a '1' must occur as often as a 'Z' or a 'd', with a probability of 1 in 62 (A.Z+a.z+0.9). The first solution is one that I've created that hasn't mentioned before and might be what most people expect when they talk about a random string with a number of characters Maximum entropy (every string equally likely) In fact, there isn't one definite definition of random string - there are different definitions used in the answers with different character distributions. &CHOOSE(RANDBETWEEN(1,3),CHAR(RANDBETWEEN(48,57)),CHAR(RANDBETWEEN(65,90)),CHAR(RANDBETWEEN(97,122))īefore I get slated, The reason I posted this as a "duplicate" answer is because I came across this question looking for a simple copy-paste solution, and I couldn't post this as a comment due to being over 600(?) characters.
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