If 15 is the lowest number and 200 the highest number, what formula do I need to use to define a multiplication pattern that applies the 28 values needed to complete this multiplication table?
I would like to learn how to create this pattern to use, I tried it
through percentages but I was not successful exactly because it had
the minimum and the maximum, if it was only the minimum or only the
maximum, just multiply it by the percentage.
if you want to paste those values in the same column you need to do:
then the formula will be:
=ARRAYFORMULA(INDIRECT("A"&
MAX(IF(A3:A<>"", ROW(A3:A), )))+SORT(ROW(INDIRECT("A1:A"&
MAX(IF(A3:A<>"", ROW(A3:A), ))-2)), 1, 0)*(A1-INDIRECT("A"&
MAX(IF(A3:A<>"", ROW(A3:A), ))))/(
MAX(IF(A3:A<>"", ROW(A3:A), ))-1))
Try this formula in A2 =$A1+($A$30-$A$1)/29 and then drag down to A29
I am trying to use an array to highlight specific rows in my data.
As shown in the picture, i want to extract the maximum positive profit($) and maximum negative profit($) for each respective serial numbers and highlight it in yellow. For example, Tom123's maximum +ve profit = 100 and -ve profit =-74 , thus both values are highlighted in yellow.
I already know how to highlight cells in vba, but i do not know how to get the rows that i need to highlight. Any help is much appreciated.
Add a column to identify if the column C = max profit or max loss
step 1
In D2, write =IF(OR(C2=MAX(IF((A:A=A2)*(C:C>=0),C:C,"")),C2=MIN(IF((A:A=A2)*(C:C<0),C:C,""))),TRUE,"") then press ctrl+shift+enter.
step 2
drag D2 down.
then you will see:
step 3
highlight A:C if D == true with conditional formatting
Background
Following on from a question I asked a while ago about getting an array of different (but not necessarily unique) random numbers to which the answer was this:
=RANDBETWEEN(ROW(A1:A10)^0,10)
To get an array of 10 random numbers between 1 and 10
The Problem
If I create a named range (called "randArray") with the formula above I hoped I would be able to reference randArray a number of times and get the same set of random numbers. Granted, they would change each time I press F9 or update the worksheet -- but change together.
This is what I get instead, two completely different sets of random numbers
I'm not surprised by this behavior but how can I achieve this without using VBA and without putting the random numbers onto the worksheet?
If you're interested
This example is intended to be MCVE. In my actual case, I am using random numbers to estimate Pi. The user stipulates how many random points to apply and gets an accordingly accurate estimation. The problem arises because I also graph the points and when there are a small number of points it's very clear to see that the estimation and the graph don't represent the same dataset
Update
I have awarded the initial bounty to #Michael for providing an interesting and different solution. I am still looking for a complete solution which allows the user to stipulate how many random points to use, and although there might not be a perfect answer I'm still interested in any other possible solutions and more than happy to put up further bounties.
Thank you to everyone who has contributed so far.
This solution generates 10 seemingly random numbers between 1 and 10 that persist for nearly 9 seconds at a time. This allows repeated calls of the same formula to return the same set of values in a single refresh.
You can modify the time frame if required. Shorter time periods allow for more frequent updates, but also slightly increase the extremely unlikely chance that some calls to the formula occur after the cutover point resulting in a 2nd set of 10 random numbers for subsequent calls.
Firstly, define an array "Primes" with 10 different prime numbers:
={157;163;167;173;179;181;191;193;197;199}
Then, define this formula that will return an array of 10 random numbers:
=MOD(ROUND(MOD(ROUND(NOW(),4)*70000,Primes),0),10)+1
Explanation:
We need to build our own random number generator that we can seed with the same value for an amount of time; long enough for the called formula to keep returning the same value.
Firstly, we create a seed: ROUND(NOW(),4) creates a new seed number every 0.0001 days = 8.64 seconds.
We can generate rough random numbers using the following formula:
Random = Seed * 7 mod Prime
https://cdsmith.wordpress.com/2011/10/10/build-your-own-simple-random-numbers/
Ideally, a sequence of random numbers is generated by taking input from the previous output, but we can't do that in a single function. So instead, this uses 10 different prime numbers, essentially starting 10 different random number generators. Now, this has less reliability at generating random numbers, but testing results further below shows it actually seems to do a pretty good job.
ROUND(NOW(),4)*70000 gets our seed up to an integer and multiplies by 7 at the same time
MOD(ROUND(NOW(),4)*70000,Prime) generates a sequence of 10 random numbers from 0 to the respective prime number
ROUND(MOD(ROUND(NOW(),4)*70000,Prime),0) is required to get us back to an integer because Excel seems to struggle with apply Mod to floating point numbers.
=MOD(ROUND(MOD(ROUND(NOW(),4)*70000,Prime),0),10)+1 takes just the value from the ones place (random number from 0 to 9) and shifts it to give us a random number from 1 to 10
Testing results:
I generated 500 lots of 10 random numbers (in columns instead of rows) for seed values incrementing by 0.0001 and counted the number of times each digit occurred for each prime number. You can see that each digit occurred nearly 500 times in total and that the distribution of each digit is nearly equal between each prime number. So, this may be adequate for your purposes.
Looking at the numbers generated in immediate succession you can see similarities between adjacent prime numbers, they're not exactly the same but they're pretty close in places, even if they're offset by a few rows. However, if the refresh is occurring at random intervals, you'll still get seemingly random numbers and this should be sufficient for your purposes. Otherwise, you can still apply this approach to a more complex random number generator or try a different mix of prime numbers that are further apart.
Update 1: Trying to find a way of being able to specify the number of random numbers generated without storing a list of primes.
Attempt 1: Using a single prime with an array of seeds:
=MOD(ROUND(MOD(ROUND(NOW()+ROW(OFFSET(INDIRECT("A1"),0,0,SampleSize))/10000,4)*70000,1013),0),10)+1
This does give you an even distribution, but it really is just repeating the exact same sequence of 10 numbers over and over. Any analysis of the sample would be identical to analysing =MOD(ROW(1:SampleSize),10)+1. I think you want more variation than that!
Attempt 2: Working on a 2-dimensional array that still uses 10 primes....
Update 2: Didn't work. It had terrible performance. A new answer has been submitted that takes a similar but different approach.
OK, here's a solution where users can specify the number of values in defined name SAMPLESIZE
=MOD(ROUND(MOD(ROUND(NOW()+ROW(OFFSET(INDIRECT("A1"),0,0,SampleSize)),4)*10000*163,1013),0)+ROUND(MOD(ROUND(NOW()+ROW(OFFSET(INDIRECT("A1"),0,0,SampleSize))*2,4)*10000*211,1013),0)+ROUND(MOD(ROUND(NOW()+ROW(OFFSET(INDIRECT("A1"),0,0,SampleSize))*3,4)*10000*17,1013),0)+ROUND(MOD(ROUND(NOW()+ROW(OFFSET(INDIRECT("A1"),0,0,SampleSize))*5,4)*10000*179,53),0)+ROUND(MOD(ROUND(NOW()+ROW(OFFSET(INDIRECT("A1"),0,0,SampleSize))*7,4)*10000*6101,1013),0),10)+1
It's a long formula, but has good efficiency and can be used in other functions. Attempts at a shorter formula resulted in unusably poor performance and arrays that for some reason couldn't be used in other functions.
This solution combines 5 different prime number generators to increase variety in the generated random numbers. Some arbitrary constants were introduced to try to reduce repeating patterns.
This has correct distribution and fairly good randomness. Repeated testing with a SampleSize of 10,000 resulted in frequencies of individual numbers varying between 960 and 1040 with no overall favoritism. However it seems to have the strange property of never generating the same number twice in a row!
You can achieve this using just standard spreadsheet formulas.
One way is to use the so called Lehmer random number method. It generates a sequence of random numbers in your spreadsheet that stays the same until you change the "seed number", a number you choose yourself and will recreate a different random sequence for each seed number you choose.
The short version:
In cell B1, enter your "seed" number, it can be any number from 1 to 2,147,483,647
In cell B2 enter the formula =MOD(48271*B1,2^31-1) , this will generate the first random number of your sequence.
Now copy this cell down as far as the the random sequence you want to generate.
That's it. For your named range, go ahead and name the range from B2 down as far as your sequence goes. If you want a different set of numbers, just change the seed in B1. If you ever want to recreate the same set of numbers just use the same seed and the same random sequence will appear.
More details in this tutorial:
How to generate random numbers that don't change in Excel and Google Sheets
It's not a great answer but considering the limitation of a volatile function, it is definitely a possible answer to use the IF formula with Volatile function and a Volatile variable placed somewhere in the worksheet.
I used the below formula to achieve the desired result
=IF(rngIsVolatile,randArray,A1:A10)
I set cell B12 as rngIsVolatile. I pasted the screenshots below to see it in working.
When rngIsVolatile is set to True, it picks up new values from randArray:
When rngIsVolatile is set to False, it picks up old values from A1:A10:
I have an array of N measurements that I should present as a graph, but the graph can be only of M pixels wide and is scrolled only by M pixels.
While M is constant, N can be anything from tens to thousands.
Each time when I need to show the graph, I know what N is, however as N/M can be not integer, there is an accumulated error that I want somehow to compensate.
I am working in a plain C, and no math libraries can be used.
EDIT 2:
The data is relatively homogeneous with peaks once in a while, and I do not want to miss these peaks, while interpolating.
EDIT 3:
I am looking for solution that will work good enough for any N, greater than M and lesser than M.
Thanks.
One good solution is not to iterate over your input samples, but over your output positions. That is, you will always draw exactly M pixels. To calculate the nearest sample value for the ith pixel, use the array offset:
[(i*N+M/2)/M]
Of course only using the nearest sample will give a very aliased result (discarding most of your samples in the case where N is large). If you're sure N will always be larger than M, a good but simple approach is to average sufficient neighboring samples with a weighted average such that each sample gets a total weight of 1 (with endpoints having their weight split between neighboring output pixels). Of course there are more elaborate resampling algorithms you can use that may be more appropriate (especially if your data is something like audio samples which are more meaningful in the frequency domain), but for an embedded device with tight memory and clock cycle requirements, averaging is likely the approach you want.
I'm playing around a bit with image processing and decided to read up on how color quantization worked and after a bit of reading I found the Modified Median Cut Quantization algorithm.
I've been reading the code of the C implementation in Leptonica library and came across something I thought was a bit odd.
Now I want to stress that I am far from an expert in this area, not am I a math-head, so I am predicting that this all comes down to me not understanding all of it and not that the implementation of the algorithm is wrong at all.
The algorithm states that the vbox should be split along the lagest axis and that it should be split using the following logic
The largest axis is divided by locating the bin with the median pixel
(by population), selecting the longer side, and dividing in the center
of that side. We could have simply put the bin with the median pixel
in the shorter side, but in the early stages of subdivision, this
tends to put low density clusters (that are not considered in the
subdivision) in the same vbox as part of a high density cluster that
will outvote it in median vbox color, even with future median-based
subdivisions. The algorithm used here is particularly important in
early subdivisions, and 3is useful for giving visible but low
population color clusters their own vbox. This has little effect on
the subdivision of high density clusters, which ultimately will have
roughly equal population in their vboxes.
For the sake of the argument, let's assume that we have a vbox that we are in the process of splitting and that the red axis is the largest. In the Leptonica algorithm, on line 01297, the code appears to do the following
Iterate over all the possible green and blue variations of the red color
For each iteration it adds to the total number of pixels (population) it's found along the red axis
For each red color it sum up the population of the current red and the previous ones, thus storing an accumulated value, for each red
note: when I say 'red' I mean each point along the axis that is covered by the iteration, the actual color may not be red but contains a certain amount of red
So for the sake of illustration, assume we have 9 "bins" along the red axis and that they have the following populations
4 8 20 16 1 9 12 8 8
After the iteration of all red bins, the partialsum array will contain the following count for the bins mentioned above
4 12 32 48 49 58 70 78 86
And total would have a value of 86
Once that's done it's time to perform the actual median cut and for the red axis this is performed on line 01346
It iterates over bins and check they accumulated sum. And here's the part that throws me of from the description of the algorithm. It looks for the first bin that has a value that is greater than total/2
Wouldn't total/2 mean that it is looking for a bin that has a value that is greater than the average value and not the median ? The median for the above bins would be 49
The use of 43 or 49 could potentially have a huge impact on how the boxes are split, even though the algorithm then proceeds by moving to the center of the larger side of where the matched value was..
Another thing that puzzles me a bit is that the paper specified that the bin with the median value should be located, but does not mention how to proceed if there are an even number of bins.. the median would be the result of (a+b)/2 and it's not guaranteed that any of the bins contains that population count. So this is what makes me thing that there are some approximations going on that are negligible because of how the split actually takes part at the center of the larger side of the selected bin.
Sorry if it got a bit long winded, but I wanted to be as thoroughas I could because it's been driving me nuts for a couple of days now ;)
In the 9-bin example, 49 is the number of pixels in the first 5 bins. 49 is the median number in the set of 9 partial sums, but we want the median pixel in the set of 86 pixels, which is 43 (or 44), and it resides in the 4th bin.
Inspection of the modified median cut algorithm in colorquant2.c of leptonica shows that the actual cut location for the 3d box does not necessarily occur adjacent to the bin containing the median pixel. The reasons for this are explained in the function medianCutApply(). This is one of the "modifications" to Paul Heckbert's original method. The other significant modification is to make the decision of which 3d box to cut next based on a combination of both population and the product (population * volume), thus permitting splitting of large but sparsely populated regions of color space.
I do not know the algo, but I would assume your array contains the population of each red; let's explain this with an example:
Assume you have four gradations of red: A,B,C and D
And you have the following sequence of red values:
AABDCADBBBAAA
To find the median, you would have to sort them according to red value and take the middle:
median
v
AAAAAABBBBCDD
Now let's use their approach:
A:6 => 6
B:4 => 10
C:1 => 11
D:2 => 13
13/2 = 6.5 => B
I think the mismatch happened because you are counting the population; the average color would be:
(6*A+4*B+1*C+2*D)/13