I have a list of names in cells A1:A100, and I'm looking for a formula to enter into a cell that will select a random name from that range. I have spent much time searching to no avail, all solutions seem to be derived from the RANDBETWEEN function which is not part of Excel 2003, but only newer versions. This seems like a very common thing to want to do so I'm really surprised there is no easy-to-find solution. Help is much appreciated =)
If you can't use RandBetween(), you can still use Rand(), which delivers a decimal number. Round the decimal number to 2 places, then multiply by 100 to get a number between 0 and 99. Add 1 to get a number between 1 and 100
=(ROUND(RAND(),2)*100)+1
Feed that into an Index function like this
=Index(A1:A100,(ROUND(RAND(),2)*100)+1)
Done.
Related
I have a couple of rather large nested if functions in my spreadsheet. It sure would be nice to have an alternative method. Problem is I'm using a wildcard (*) in my lookup because the source text has slight variations (date for example).
For example, if my list of data contains:
VENMO PAYMENT 220828 1022093447487 BRENDA HOSPY
VENMO PAYMENT 220813 1031323447487 BRENDA HOSPY
I want these to show in an adjacent column of cells as just Venmo
Currently my if function in that second column of cells is:
=IF(COUNTIF($F10,"*APPLE.COM/BILL*"),"AP",
IF(COUNTIF($F10,"IIA VOYA*"),"VOYA",
IF(COUNTIF($F10,"VENMO PAYMENT*"),"Venmo",
IF(COUNTIF($F10,etc...
This works fine but quickly gets unruly as more things get added.
I've spent a great deal of time searching for functions and processes that would make this easier, or at least more compact, but I can't find a way with typical functions like vlookup or index/match.
If I've explained this in a comprehensible fashion perhaps you've seen or experienced a similar situation and could offer a suggestion. It would be appreciated!
I'm not opposed to using a programming function.
I've looked at, and for, various Excel functions or combinations with no luck on my own or online.
I have created a structure as below
Formula present in B2 is as below
=IFERROR(INDEX($F$2:$F$9,MIN(IF(COUNTIF(A2,"*"&$E$2:$E$9&"*")>0,ROW($E$2:$E$9),9999999)-1)),"---")
Enter it as an Array Formula using Ctrl+Shift+Enter
It will search all the strings present in column E in A2 when found will return all the row numbers of column E where there is a match, i have then used min to get the first one, and if not found it will return 9999999, and as the data is starting from row 2 i have added -1 to make it equal to the data index. after that i have called the index to search value present at that index in column F. and at the end used the if error function to show --- where no match was found and 999999 was returned.
I can't believe I have never had this issue before (nor can I find anyone else with the same issue) but today I have just discovered that SAS sometimes gets simple calculations wrong!?! I noticed that one of my records wasn't getting picked up in the right group based on a value being <3.6 and thought there must be something strange in my data with decimal places. But on investigation I found it was just because SAS was calculating the value wrong! For some reason that I can't fathom, it seems that SAS calculates 90 - 86.4 as 3.59999999999999!!! Simple program below to show this:
code
output
If I alter the calculation to 10 - 6.4 I get the correct value of 3.6000 but for some reason this one is coming out wrong. Could there be some mad setting that is wrong in my installation? I tried both SAS EG and Base SAS and both have the same issue. I feel like I'm going mad! Any help appreciated.
Thanks.
Floating point arithmetic, in any language, will have this same issue. The same issue is possible to understand in human terms, assuming the human doesn't have a concept of infinite. If you only write down 4 digits for your decimals, for example, then:
1 - (1/3) - (1/3) - (1/3)
That's zero, right?
1 - 0.3333 = 0.6667
0.6667 - 0.3333 = 0.3334
0.3334 - 0.3333 = 0.0001
Nope! Computers do the same thing, but in binary, so they have a different (and larger) set of "problem" numbers. 1/10, for example, is not representable in binary - so adding or subtracting 0.1 is not always a "neat" operation like it is in decimal.
SAS uses 8 byte floating points, and so it gets ~15 digits of accuracy. Assuming you're not working in a field where 15 digits of accuracy is needed, you should simply round.
if round(value,.01) ge 3.6 then ... ;
Most of the time this isn't needed, but strictly speaking you should always compare rounded numbers whenever using floating point numbers (as SAS does). Integers are safe, but if you're working with 0.1 etc., use ROUND or FUZZ for integers.
Sorry Cannot replicate your findings.
data x;
a=90-86.4;
run;
Gives the correct result. Are you using any formats or put function. Share the complete code.
RE: Apache OpenOffice 4.1.7, AOO417m1(Build:9800) - Rev. 46059c9192, 2019-09-03 12:04.
I need to sum non-integer entries across a range of cells, but without including the decimal values (complicated by some cells being text). I started with ROUNDDOWN, then TRUNC, then FLOOR. I'm driving myself nuts trying to find a clean code (or even an arbitrarily extensible ugly code) for what would be the following:
=SUMIF(ISTEXT(R7:CL7);0;TRUNC(R7:CL7))
The above doesn't work, of course, since TRUNC() doesn't apply to ranges, but it conveys what I'm trying to do in a nutshell -- some of the cells contain text, which SUM() ignores (luckily), but they flummox TRUNC, so I needed to handle the text problem.
I started with ISNUMBER, just to get the ball rolling; ISTEXT has fewer characters, but it's not worth fixing that right now.
FLOOR was equally disappointing for ranges:
=SUM(FLOOR(R7:T7;1))
I tried variations of =SUM(IF(... and searches for ROUNDDOWN range (and variations on that) and such pseudocode as "IFTEXT" and "SUMTRUNC" (and variations on that). I found info on ROUNDDOWN(SUM(... and so forth, but not "SUM(ROUNDDOWN(..." or any equivalent.
In my delirium, I got silly and even tried:
=SUMIF(ISTEXT(S7:U7);0;AND(TRUNC(S7);TRUNC(T7);TRUNC(U7)))
To be clear: {2.9→2 + 2.9→2 + 2.9→2 = 6} ≠ {2.9+2.9+2.9 = 8.7→8}. I'm looking for a 6, not an 8 (I'd joke about sixes and sevens, but I'm way past pumpkin o'clock and 2.428571 takes up too much space).
My current test-kludge is:
=SUM(IF(ISNUMBER(R7);ROUNDDOWN(R7);0);IF(ISNUMBER(S7);ROUNDDOWN(S7);0);IF(ISNUMBER(T7);ROUNDDOWN(T7);0); ... ;IF(ISNUMBER(AX7);ROUNDDOWN(AX7);0))
It ends at AX7 only because of the char count. I hope to SUM the whole row in a single sweep, but that ain't gonna cut it. I could do it in large chunks in multiple cells, and then add those cells up, but oy gevalt.
Since it's already ugly anyway, I could use the following to save a few characters, but this would only mean being able to extend the range maybe 6 further cells (not much point in that):
=IF(ISTEXT(R7);0;TRUNC(R7))+IF(ISTEXT(S7);0;TRUNC(S7))+IF(ISTEXT(S7);0;TRUNC(S7))
I'm seriously considering simply going down a bunch of rows (to below my data cells) and entering the following, then copying the cell and pasting it to a complementary range, and telling the SUM cells to just sum up their respectively shadowed rows (instead of the data rows that they sit in):
=IF(ISTEXT(R7);0;TRUNC(R7))
Sorry for the rambling; I need sleep. This started as a need, then multiple failed attempts became a grudge match of principle and obstinacy, and now I'm just plugging away at it out of blind habit developed over the past 2-3 days (hopefully I won't forget what the purpose was).
In summary...: ++?????++ Out of Cheese Error +++DIVIDE BY CUCUMBER.
I'm comfortable enough with macros, though it's been ~7 years (and that was in Excel). Thanks in advance, even if the answer is that I'm stuck with one of these! 🙂
EDIT: I don't see a way to attach a .csv here (though I could open the .csv with Notepad, and copy-and-paste the contents if that would help anyone), so here's a set of pics:
I am new to Stack overflow but I am having logic issues with Google sheets and need some advice on how to proceed.
My goal is I have 3 cup sizes, small medium and large. Example: Small holds 50ML, Medium 100ML,Large 200ML,etc
I want to take a large number and evenly distribute it among the cups to show me how many cups I will need with the least amount of cups used. Example : 170ML = 1 Large, 0 Medium, 0 Small. I only need 1 large to hold everything. Also, 240ML would suggest 1 Large, 0 medium, 1 small. Since small can hold the remaining 40 and medium would be too big of a container.
Problem is I don't understand how to break down my original large number into the smaller number as I have to check and compare if it will fit, I also have to be able to add more if there's a remainder and as far as I know the Google sheet functions only run and represent numbers once.
I've already tried breaking it down to my large container first then in my second row with medium cups I take the first result and subtract from my large number to see if anything is left. If there is, all I can do is add 1 or set the number, I can't seem to scale it up if it requires more than 1 cup which isn't what I want.
I've been going crazy trying to find an easy solution to this but it seems to get more complex as if I need IF statements of some kind.
If anyone has any ideas I'd be happy to hear them out.
you could use IF statement for example like this:
=ARRAYFORMULA(IF(A2:A<>"",
IF(A2:A>=B1, QUOTIENT(A2:A, B1),
IF(A2:A> C1+D1, QUOTIENT(B1, A2:A), 0)), ))
For a school project, I have a simple program, which compares 20x20 photos. I put 20 photos, and then i put 21th photo, which is compared to existing 20, and pops up the answer, which photo i did insert (or which one is most similar). The problem is, my teacher wanted me to use nearest neighbour algorithm, so i am counting distance from every photo. I got everything working, but the thing is, if photos are too similar, i got the problem with saying which one is closer to my one. For example i get these distances with 2 different photos (well, they are ALMOST the same):
0 distance: 1353.07982026191
1 distance: 1353.07982026191
It is 15 digits already, and i am using double type. I was reading that long double is the same. Is there any "easy" way to store numbers with more than 15 digits and do math on them?
I count distance using Euclidean distance
I just need to be more precise, or thats limit i probably wont pass here, and i should talk to my teacher i cant compare such similar photos?
I think you need this: gmplib.org
There's a guide how to install this library on this site too.
And here's article about floats: http://gmplib.org/manual/C_002b_002b-Interface-Floats.html#C_002b_002b-Interface-Floats
Maybe you could use an algebraic approach.
Let us assume that you are trying to calcuate if vector x is closer to a or b. What you need to calculate is the sign of
d2(x, a) - d2(x, b)
Which becomes (I'll omit some passages for brevity)
and then
Which only contains differences between values which should be very similar. Summing over such small values should yield a better precision than working on the aggregate.