I have 5 temperature sensors. I want to calculate the mean temperature of 4 - excluding the most extreme value (high or low).
Firstly: will std:unique(id) create a window of the last temperature readings for each id 1-5?
select
avg(tempEvent.temp) as meantemp
from
Event(id in (1, 2, 3, 4, 5)).std:unique(id) as tempEvent
Secondly: how could I change the select statement (possibly using an expression if necessary) to only calculate the mean of four values excluding the most extreme?
The background is, I want to know the deviations of each temperature from the average, but I don't want the average to include an anomalous id. Otherwise all temperatures will look like they are deviating from the average but really only one is.
Finding the average of the middle four values is simple enough, though not as elegant as your solution. The code below will work for any number of temps.
SELECT
AVG(temp) AS meantemp
FROM (
SELECT
temp,
COUNT(temp) AS c,
RANK () OVER (PARTITION BY temp ORDER BY temp) AS r
FROM
[table]
)
WHERE
r > 1
AND r < (c-1)
;
As for your second question, I'm not sure I understand. Do you want the value from among the four middle values that has the greatest absolute deviation from the mean of those four values?
I have a graph below.
I would like to calculate lapsed rate which is sum of lapsed value divided by sum of inforce value. I use the formula below in calculated field.
abs(sum(if[Status]='lapsed'then[TotalAmount]end))/abs(sum(if[Status]='inforce'then[TotalAmount]end))
However that formula will also pick the value from Q2 (quarter 2) 2016. What I want to do is to tell tableau to check first if any quarter does not contain both inforce value and lapsed value then skip that quarter. In this case I need to calculate lapsed rate which does not include Q2 2016. How do i do this?
I'm using Tableau v.10.
Thanks.
This is just a quick approach and may not be the most efficient but it seems to work.
You need to use a combination of a row level calculation and a level of detail calculation.
The level of detail calculation can be used to flag the quarters which have both a lapsed and inforced status. Once these quarters are flagged you can calculate the lapsed rate at a row level which can then be rolled up using a sum.
Create a calculated field as follows:
Let me know if you have any questions or need any tweaks.
if
avg(
// Calculate the number of Inforce/Lapsed occurences per Quarter
IF
[Status] = 'Inforce'
or
[Status] = 'Lapsed'
then
{ FIXED
DATEPART('quarter', [Date]):
countd([Status])
}
else
0
end)
//
= 2
then
// Calculate the Lapsed Rate as both statuses exist in the quarter
sum((if
[Status] = 'Lapsed'
then [Total Amount]
END))
/
sum([Total Amount])
END
I have the following chart, which uses Quick Table calculation to count the percentage of the values across the group Category for each month.
What I want is to count the percentage of the monthly running total of the values across the group Category for each month. I can create the quick calculation for monthly running total, separately, but I can't seem to combine it with this below because one requires quick calculation across the table and another across the group. Can anyone advise if this is doable?
The workbook attached here, with the problem above in the second tab.
Edit: Let me give example what I meant here:
April: A = 1000, B=500, C= 200, D=300 -> total is 2000
percentage in April in chart is A = 1000/2000 = 50%, B = 25%, C = 10%, D = 15%, total = 100%
May: A=2000, B=2000, C=1800, D=200 ->
cumulative April+May A=3000, B=2500, C=2000, D=500 -> total is 8000
percentage in May in chart is A=3000/8000 = 37.5%, B = 31.25%, C = 25%, D = 6.25%, total = 100%
I got an answer elsewhere and thought I should update here. The solution was to use a secondary table calculation. Basically, after putting SUM(Value) pill in Row:
right click on it to select Add Table Calculation
set as primary calculation the Running Total Sum of Table (Across)
check the box to Perform a secondary calculation on the result
set as secondary calculation the Percent of Total of Table (Down)
See below screenshot of the setting.
Are you looking for ways to calculate the sum of the running total across the category? it is not clear when you say to count the percentage? However if you are looking for sum of percentage running total across your category... you can edit your calculation and set the calculation type as running total, the summarize value ( you can choose, sum , average, min,...max) and the running along option choose category.
I have daily river flow data for 1975-2009 and I am asked to find the 7 consecutive days within each year that have the smallest flows.
Any advice how to start this? I've only been using MATLAB for a couple weeks.
Thanks!
You could convolve the data with ones(1,7) and look for the minimum, which will yield the starting day of your dry period:
[~,startingDay] = min(conv(flow,ones(1,7),'valid'))
(This is basically a moving average filter without the normalization).
Loop through the years to get each year's result.
Start by finding cumulative sum with cumsum. The difference between cumulative sums 7 days apart will give you the total for those 7 days. Then pick the minimum of those.
a = cumsum(flow);
b = a(8:end) - a(1:end-7);
[m,i] = min(b);
Here m holds the smallest total over 7 consecutive days, and i is a vector of indices telling you when they occurred.
Having read carefully the previous question
Random numbers that add to 100: Matlab
I am struggling to solve a similar but slightly more complex problem.
I would like to create an array of n elements that sums to 1, however I want an added constraint that the minimum increment (or if you like number of significant figures) for each element is fixed.
For example if I want 10 numbers that sum to 1 without any constraint the following works perfectly:
num_stocks=10;
num_simulations=100000;
temp = [zeros(num_simulations,1),sort(rand(num_simulations,num_stocks-1),2),ones(num_simulations,1)];
weights = diff(temp,[],2);
I foolishly thought that by scaling this I could add the constraint as follows
num_stocks=10;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp2 = [zeros(num_simulations,1),sort(round(rand(num_simulations,num_stocks-1)*scaling)/scaling,2),ones(num_simulations,1)];
weights2 = diff(temp2,[],2);
However though this works for small values of n & small values of increment, if for example n=1,000 & the increment is 0.1% then over a large number of trials the first and last numbers have a mean which is consistently below 0.1%.
I am sure there is a logical explanation/solution to this but I have been tearing my hair out to try & find it & wondered anybody would be so kind as to point me in the right direction. To put the problem into context create random stock portfolios (hence the sum to 1).
Thanks in advance
Thank you for the responses so far, just to clarify (as I think my initial question was perhaps badly phrased), it is the weights that have a fixed increment of 0.1% so 0%, 0.1%, 0.2% etc.
I did try using integers initially
num_stocks=1000;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp = [zeros(num_simulations,1),sort(randi([0 scaling],num_simulations,num_stocks-1),2),ones(num_simulations,1)*scaling];
weights = (diff(temp,[],2)/scaling);
test=mean(weights);
but this was worse, the mean for the 1st & last weights is well below 0.1%.....
Edit to reflect excellent answer by Floris & clarify
The original code I was using to solve this problem (before finding this forum) was
function x = monkey_weights_original(simulations,stocks)
stockmatrix=1:stocks;
base_weight=1/stocks;
r=randi(stocks,stocks,simulations);
x=histc(r,stockmatrix)*base_weight;
end
This runs very fast, which was important considering I want to run a total of 10,000,000 simulations, 10,000 simulations on 1,000 stocks takes just over 2 seconds with a single core & I am running the whole code on an 8 core machine using the parallel toolbox.
It also gives exactly the distribution I was looking for in terms of means, and I think that it is just as likely to get a portfolio that is 100% in 1 stock as it is to geta portfolio that is 0.1% in every stock (though I'm happy to be corrected).
My issue issue is that although it works for 1,000 stocks & an increment of 0.1% and I guess it works for 100 stocks & an increment of 1%, as the number of stocks decreases then each pick becomes a very large percentage (in the extreme with 2 stocks you will always get a 50/50 portfolio).
In effect I think this solution is like the binomial solution Floris suggests (but more limited)
However my question has arrisen because I would like to make my approach more flexible & have the possibility of say 3 stocks & an increment of 1% which my current code will not handle correctly, hence how I stumbled accross the original question on stackoverflow
Floris's recursive approach will get to the right answer, but the speed will be a major issue considering the scale of the problem.
An example of the original research is here
http://www.huffingtonpost.com/2013/04/05/monkeys-stocks-study_n_3021285.html
I am currently working on extending it with more flexibility on portfolio weights & numbers of stock in the index, but it appears my programming & probability theory ability are a limiting factor.......
One problem I can see is that your formula allows for numbers to be zero - when the rounding operation results in two consecutive numbers to be the same after sorting. Not sure if you consider that a problem - but I suggest you think about it (it would mean your model portfolio has fewer than N stocks in it since the contribution of one of the stocks would be zero).
The other thing to note is that the probability of getting the extreme values in your distribution is half of what you want them to be: If you have uniformly distributed numbers from 0 to 1000, and you round them, the numbers that round to 0 were in the interval [0 0.5>; the ones that round to 1 came from [0.5 1.5> - twice as big. The last number (rounding to 1000) is again from a smaller interval: [999.5 1000]. Thus you will not get the first and last number as often as you think. If instead of round you use floor I think you will get the answer you expect.
EDIT
I thought about this some more, and came up with a slow but (I think) accurate method for doing this. The basic idea is this:
Think in terms of integers; rather than dividing the interval 0 - 1 in steps of 0.001, divide the interval 0 - 1000 in integer steps
If we try to divide N into m intervals, the mean size of a step should be N / m; but being integer, we would expect the intervals to be binomially distributed
This suggests an algorithm in which we choose the first interval as a binomially distributed variate with mean (N/m) - call the first value v1; then divide the remaining interval N - v1 into m-1 steps; we can do so recursively.
The following code implements this:
% random integers adding up to a definite sum
function r = randomInt(n, limit)
% returns an array of n random integers
% whose sum is limit
% calls itself recursively; slow but accurate
if n>1
v = binomialRandom(limit, 1 / n);
r = [v randomInt(n-1, limit - v)];
else
r = limit;
end
function b = binomialRandom(N, p)
b = sum(rand(1,N)<p); % slow but direct
To get 10000 instances, you run this as follows:
tic
portfolio = zeros(10000, 10);
for ii = 1:10000
portfolio(ii,:) = randomInt(10, 1000);
end
toc
This ran in 3.8 seconds on a modest machine (single thread) - of course the method for obtaining a binomially distributed random variate is the thing slowing it down; there are statistical toolboxes with more efficient functions but I don't have one. If you increase the granularity (for example, by setting limit=10000) it will slow down more since you increase the number of random number samples that are generated; with limit = 10000 the above loop took 13.3 seconds to complete.
As a test, I found mean(portfolio)' and std(portfolio)' as follows (with limit=1000):
100.20 9.446
99.90 9.547
100.09 9.456
100.00 9.548
100.01 9.356
100.00 9.484
99.69 9.639
100.06 9.493
99.94 9.599
100.11 9.453
This looks like a pretty convincing "flat" distribution to me. We would expect the numbers to be binomially distributed with a mean of 100, and standard deviation of sqrt(p*(1-p)*n). In this case, p=0.1 so we expect s = 9.4868. The values I actually got were again quite close.
I realize that this is inefficient for large values of limit, and I made no attempt at efficiency. I find that clarity trumps speed when you develop something new. But for instance you could pre-compute the cumulative binomial distributions for p=1./(1:10), then do a random lookup; but if you are just going to do this once, for 100,000 instances, it will run in under a minute; unless you intend to do it many times, I wouldn't bother. But if anyone wants to improve this code I'd be happy to hear from them.
Eventually I have solved this problem!
I found a paper by 2 academics at John Hopkins University "Sampling Uniformly From The Unit Simplex"
http://www.cs.cmu.edu/~nasmith/papers/smith+tromble.tr04.pdf
In the paper they outline how naive algorthms don't work, in a way very similar to woodchips answer to the Random numbers that add to 100 question. They then go on to show that the method suggested by David Schwartz can also be slightly biased and propose a modified algorithm which appear to work.
If you want x numbers that sum to y
Sample uniformly x-1 random numbers from the range 1 to x+y-1 without replacement
Sort them
Add a zero at the beginning & x+y at the end
difference them & subtract 1 from each value
If you want to scale them as I do, then divide by y
It took me a while to realise why this works when the original approach didn't and it come down to the probability of getting a zero weight (as highlighted by Floris in his answer). To get a zero weight in the original version for all but the 1st or last weights your random numbers had to have 2 values the same but for the 1st & last ones then a random number of zero or the maximum number would result in a zero weight which is more likely.
In the revised algorithm, zero & the maximum number are not in the set of random choices & a zero weight occurs only if you select two consecutive numbers which is equally likely for every position.
I coded it up in Matlab as follows
function weights = unbiased_monkey_weights(num_simulations,num_stocks,min_increment)
scaling=1/min_increment;
sample=NaN(num_simulations,num_stocks-1);
for i=1:num_simulations
allcomb=randperm(scaling+num_stocks-1);
sample(i,:)=allcomb(1:num_stocks-1);
end
temp = [zeros(num_simulations,1),sort(sample,2),ones(num_simulations,1)*(scaling+num_stocks)];
weights = (diff(temp,[],2)-1)/scaling;
end
Obviously the loop is a bit clunky and as I'm using the 2009 version the randperm function only allows you to generate permutations of the whole set, however despite this I can run 10,000 simulations for 1,000 numbers in 5 seconds on my clunky laptop which is fast enough.
The mean weights are now correct & as a quick test I replicated woodchips generating 3 numbers that sum to 1 with the minimum increment being 0.01% & it also look right
Thank you all for your help and I hope this solution is useful to somebody else in the future
The simple answer is to use the schemes that work well with NO minimum increment, then transform the problem. As always, be careful. Some methods do NOT yield uniform sets of numbers.
Thus, suppose I want 11 numbers that sum to 100, with a constraint of a minimum increment of 5. I would first find 11 numbers that sum to 45, with no lower bound on the samples (other than zero.) I could use a tool from the file exchange for this. Simplest is to simply sample 10 numbers in the interval [0,45]. Sort them, then find the differences.
X = diff([0,sort(rand(1,10)),1]*45);
The vector X is a sample of numbers that sums to 45. But the vector Y sums to 100, with a minimum value of 5.
Y = X + 5;
Of course, this is trivially vectorized if you wish to find multiple sets of numbers with the given constraint.