In Stata, I have a dataset with two variables: id and var, and say 1000 observations. The variable var is of type float and takes distinct values for all observations. I would like to keep only the three observations where var is either the minimum of var, the maximum of var, or the median of var.
The way I currently do this:
summarize var, detail
local varmax = r(max)
local varmin = r(min)
local varmedian= r(p50)
keep if inlist(float(var),float(`varmax') , float(`varmedian'), float(`varmin'))
The problem that I face is that sometimes the inlist condition will not match one of the value. E.g. I end up with two observations instead of three, for instance the one with min and the one with max, but not the one with median. I suspect this has to do with a precision problem. As you see, I tried to convert all numbers to float, but this is apparently not sufficient.
Any fix to my solution, or alternative solution would be greatly appreciated (if possible without installing additional packages), thanks!
This is not in the first instance a precision problem.
It is an inevitable problem when (1) the number of values is even and (2) the median is the mean of two central values that are different. Then the median itself is not a value in the dataset and will not be found by keep.
Consider a data set 1, 2, 3, 4. The median 2.5 is not in the data. This is very common; indeed it is what is expected with all values distinct and the number of observations even.
Other problems can arise because two or even three of the minimum, median and maximum could be equal to each other. This is not your present problem, but it can bite with other variables (e.g. indicator variables).
Precision problems are possible.
Here is a general solution purported to avoid all these difficulties.
If you collapse to min, median. max and then reshape you can avoid the problem. You will always get three results, even if they are numerically equal and/or not present in the data.
In the trivial example below, the identifier is needed only to appease reshape. In other problems, you might want to collapse using by() and then your identifier comes ready-made. However, you will be less likely to want to reshape in that case.
. clear
. set obs 4
number of observations (_N) was 0, now 4
. gen y = _n
. collapse (min)ymin=y (max)ymax=y (median)ymedian=y
. gen id = _n
. reshape long y, i(id) j(statistic) string
(note: j = max median min)
Data wide -> long
-----------------------------------------------------------------------------
Number of obs. 1 -> 3
Number of variables 4 -> 3
j variable (3 values) -> statistic
xij variables:
ymax ymedian ymin -> y
-----------------------------------------------------------------------------
. list
+---------------------+
| id statis~c y |
|---------------------|
1. | 1 max 4 |
2. | 1 median 2.5 |
3. | 1 min 1 |
+---------------------+
All that said, having (lots of?) datasets with just three observations sounds poor data management strategy. Perhaps this is extracted from some larger question.
UPDATE
Here is another way to keep precisely 3 observations. Apart from the minimum and maximum, we use the rule that we keep the "low median", i.e. the lower of two values averaged for the median, when the number of observations is even, and a single value that is the median otherwise. (In Stephen Stigler's agreeable terminology, we can talk of "comedians" in the first case.)
. sysuse auto, clear
(1978 Automobile Data)
. sort mpg
. drop if missing(mpg)
(0 observations deleted)
. keep if inlist(_n, 1, cond(mod(_N, 2), ceil(_N/2), floor(_N/2)), _N)
(71 observations deleted)
. l mpg
+-----+
| mpg |
|-----|
1. | 12 |
2. | 20 |
3. | 41 |
+-----+
mod(_N, 2) is 1 if _N is odd and 0 if _N is even. The expression in cond() selects ceil(_N/2) if the number of observations is odd and floor(_N/2) if it is even.
Related
In Stata I would like to create a binary variable median_unemp based on the median value of another variable unemp, grouping the calculation of the median value by region and year. That is, median_unempis 1 when the unemployment for that particular observation is greater than the median unemployment for the region and the year of the observation (and is 0 otherwise).
The code below generates my variable considering the entire dataset, but I want the median to be calculated by subgroups (by region and year):
webuse productivity.dta, clear
summarize unemp, detail
gen median_response = r(p50)
gen median_unemp = (unemp>=median_response)
replace median_unemp =. if unemp==.
On closer inspection of the data, I would like to know if unempfor observation 1 of my dataset (that is in region=1 and year=1970) is greater than the value of median_unemp (calculated for region=1 and year=1970) and so on. If it is greater than the median, then median_unemp==1. If it is lower than the median, then median_unemp==0.
webuse productivity.dta, clear
egen median_unemp = median(unemp), by(region year)
gen high_unemp = (unemp >= median_unemp) if unemp < .
In this dataset, there are no missing values for unemp but separating missings is good practice. Each median is the 5th of 9 values, so setting aside ties 4 values will be less than the median and 5 more than or equal to the median.
I am working on a Scala code that convert set of unique strings to unique IDs. I applied HashCode() but I got negative numbers and I need to work only with positive numbers.
I know that I have to use math.abs to get rid of the negative values but I am not sure if this is the correct solution or not.
If I read before that something like this could solve my problem
math.abs(hashCode()) * constant % size
how can I determine the right constant? and does the size means the total number of strings?
previous questions related to that topic solved the question by using math.abs only but if the total number of string is large an overflow could happen and there is a chance to get a negative number as well. by multiplying the result with constant and take the mod of size could help. This is why I need to understand how to determine the constant and the size?
Also is there another way to get unique numbers for unique strings?
We can phrase your problem another way: How to get an unsigned number from a signed number with the same range?
Suppose you are using an Integer. Its value goes from -2147483648 to 2147483647. Now you need to convert this value into the positive range 0 to 2147483647.
Step 1:
ADD a constant to move the range upwards to 0. You can do this by adding 2147483648 to the value. But now the highest possible value is much greater than the MAX.
Step 2:
So use MODULO to move the value back into the required range.
For example, consider the values -2000 and 2000000000.
| STEP | MIN VALUE | EXAMPLE 1 | EXAMPLE 2 | MAX VALUE |
|-------------------|------------|------------|------------|------------|
| original |-2147483648 | -2000 | 2000000000 | 2147483647 |
| add 2147483648 | 0 | 2147481648 | 4147483648 | 4294967295 |
| modulo 2147483648 | 0 | 2147481648 | 2000000001 | 2147483647 |
So the final formula is:
(NUMBER + 2147483648) % 2147481648
Warning:
Hash codes are not designed to give unique values. There are chances of getting the same hash for two different strings. Also, any scaling operations on the hash (like division, modulo) can further reduce uniqueness.
To strip a sign from an Int, you can just use .abs. It does break on Int.MinValue, but you can just special case it:
def stripSign(n: Int) = math.abs(n) max 0
or simply drop the sign bit:
def stripSign2(n: Int) = n & Int.MaxValue
Or just use negative numbers (what's wrong with them anyway?).
To your other question, you cannot convert a bunch of unique strings to ints, and guarantee that there won't be duplications (for the simple reason that there are more strings than distinct Ints, so, if you wanted to assign an unique int to each of them, you'd run out of ints before you run out of strings), so you have to be able to handle collisions, however infrequent.
You can only shoot for lowering the probability of a collision by making your hash longer (with a 32-bit hash code, you have about 50% probability of at least one collision in a population of approximately 75000 strings, with 31 bits (if you do not want negative numbers), it is 55000, but with a 64-bit hash, the "magic number" is about 5 billion, provided that your hash function is good enough, and produces the numbers that are very evenly distributed).
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.
I have looked thoroughly on the internet for an answer to this question, but it seems to be too specific for an answer anywhere else. This is my last stop.
To preface, this is not a homework problem, but it is adapted from an online Coursera course, whose quiz has already passed. I got the correct answer, but it was mostly luck. Also, it is a more of a general programming question than anything related to the course, so I know for a fact that it is within my right to ask it on a public forum.
The last thing is that I'm trying to do this in MatLab; however if you have an answer that is in C++ or Python or any other high level language, that would be wonderful, as I could easily adapt those solutions to MatLab syntax.
Here it is:
I have two vectors, T and M, each with 600,000 elements/entries/integers.
T is entered as milliseconds from 1 to 600,000 in ascending order, and each element in M represents 'on' or 'off' (entered as 1 or 0 respectively) for each corresponding millisecond entry in T. So there are random 1's and 0's in M that correspond to a particular millisecond from 1 to 600,000 in T.
I need to take, starting with the 150th millisecond of T, and in 150 element/millisecond increments from there on (inclusive), the average millisecond value of those groups of 150 but ONLY of those milliseconds whose entries are 1 in M ('on'). For example, I need to look at the first 150 milliseconds in T, see which ones have a value of 1 in M, and then average them. Then I need to do it again with entries 151 to 300 in T, then 301 to 450, etc. etc. These new averages should also be stored in a new vector. The problem is, the number of corresponding 1's in M isn't going to be the same for every group of 150 milliseconds in T. (And yes, we are trying to average the actual value of the milliseconds, so the values we are using to average and the order of the entries in T will be the same).
My attempt:
It turns out there are only 53,583 random 1's in M (out of the 600,000 entries, the rest are 0). I used a 'find' operator to extract those entries from M that are a 1 into a new vector K that has the millisecond value corresponding from T. So K looks like a bunch of random numbers in ascending order, which is just a list of all the milliseconds in T who are 'on' (assigned a 1 in M).
So K looks something like [2 5 11 27 39 40 79 ...... 599,698 599,727 etc.] (all of the millisecond values who are a 1 in M).
So I have the vector K which is all of the values that I need to average in groups of 150, but the problem is that I need to go in groups of 150 based on the vector T (1 to 600,000), which means there won't always be the same number of 1's (or values in K) in every group of 150 milliseconds in T, which in turn means the number I need to divide by to get the average of each group is going to change for each group of 150. I know I need to use a loop to do the average millisecond value for every 150 entries, but how do I get the dividing number (the number of entries for each group of 150 who is assigned a 1 or 'on') to change on each iteration of the loop? Is there a way to bind T and M together so that they only use the requisite values from K whenever there is a 1 in M, and then just use a simple counter to average?
It's not a complicated problem, but it is very hard to explain. Sorry about that! I hope I explained as clearly as I could. Any help would be appreciated, although I'm sure you'll have questions first.
Thank you very much!
I think this should work OK.
sz = length(T);
n = sz / 150;
K = T.*M';
t = 1;
aver = zeros(n-1,1); % Your result vector
for i = 1:150:sz-150
aver(t) = mean(K(i:(i+150)-1));
t = t + 1;
end
-Rob
I have production (q) values from 4 different methods stored in the 4 matrices. Each of the 4 matrices contains q values from a different method as:
Matrix_1 = 1 row x 20 column
Matrix_2 = 100 rows x 20 columns
Matrix_3 = 100 rows x 20 columns
Matrix_4 = 100 rows x 20 columns
The number of columns indicate the number of years. 1 row would contain the production values corresponding to the 20 years. Other 99 rows for matrix 2, 3 and 4 are just the different realizations (or simulation runs). So basically the other 99 rows for matrix 2,3 and 4 are repeat cases (but not with exact values because of random numbers).
Consider Matrix_1 as the reference truth (or base case ). Now I want to compare the other 3 matrices with Matrix_1 to see which one among those three matrices (each with 100 repeats) compares best, or closely imitates, with Matrix_1.
How can this be done in Matlab?
I know, manually, that we use confidence interval (CI) by plotting the mean of Matrix_1, and drawing each distribution of mean of Matrix_2, mean of Matrix_3 and mean of Matrix_4. The largest CI among matrix 2, 3 and 4 which contains the reference truth (or mean of Matrix_1) will be the answer.
mean of Matrix_1 = (1 row x 1 column)
mean of Matrix_2 = (100 rows x 1 column)
mean of Matrix_3 = (100 rows x 1 column)
mean of Matrix_4 = (100 rows x 1 column)
I hope the question is clear and relevant to SO. Otherwise please feel free to edit/suggest anything in question. Thanks!
EDIT: My three methods I talked about are a1, a2 and a3 respectively. Here's my result:
ci_a1 =
1.0e+008 *
4.084733001497999
4.097677503988565
ci_a2 =
1.0e+008 *
5.424396063219890
5.586301025525149
ci_a3 =
1.0e+008 *
2.429145282593182
2.838897116739112
p_a1 =
8.094614835195452e-130
p_a2 =
2.824626709966993e-072
p_a3 =
3.054667629953656e-012
h_a1 = 1; h_a2 = 1; h_a3 = 1
None of my CI, from the three methods, includes the mean ( = 3.454992884900722e+008) inside it. So do we still consider p-value to choose the best result?
If I understand correctly the calculation in MATLAB is pretty strait-forward.
Steps 1-2 (mean calculation):
k1_mean = mean(k1);
k2_mean = mean(k2);
k3_mean = mean(k3);
k4_mean = mean(k4);
Step 3, use HIST to plot distribution histograms:
hist([k2_mean; k3_mean; k4_mean]')
Step 4. You can do t-test comparing your vectors 2, 3 and 4 against normal distribution with mean k1_mean and unknown variance. See TTEST for details.
[h,p,ci] = ttest(k2_mean,k1_mean);
EDIT : I misinterpreted your question. See the answer of Yuk and following comments. My answer is what you need if you want to compare distributions of two vectors instead of a vector against a single value. Apparently, the latter is the case here.
Regarding your t-tests, you should keep in mind that they test against a "true" mean. Given the number of values for each matrix and the confidence intervals it's not too difficult to guess the standard deviation on your results. This is a measure of the "spread" of your results. Now the error on your mean is calculated as the standard deviation of your results divided by the number of observations. And the confidence interval is calculated by multiplying that standard error with appx. 2.
This confidence interval contains the true mean in 95% of the cases. So if the true mean is exactly at the border of that interval, the p-value is 0.05 the further away the mean, the lower the p-value. This can be interpreted as the chance that the values you have in matrix 2, 3 or 4 come from a population with a mean as in matrix 1. If you see your p-values, these chances can be said to be non-existent.
So you see that when the number of values get high, the confidence interval becomes smaller and the t-test becomes very sensitive. What this tells you, is nothing more that the three matrices differ significantly from the mean. If you have to choose one, I'd take a look at the distributions anyway. Otherwise the one with the closest mean seems a good guess. If you want to get deeper into this, you could also ask on stats.stackexchange.com
Your question and your method aren't really clear :
Is the distribution equal in all columns? This is important, as two distributions can have the same mean, but differ significantly :
is there a reason why you don't use the Central Limit Theorem? This seems to me like a very complex way of obtaining a result that can easily be found using the fact that the distribution of a mean approaches a normal distribution where sd(mean) = sd(observations)/number of observations. Saves you quite some work -if the distributions are alike! -
Now if the question is really the comparison of distributions, you should consider looking at a qqplot for a general idea, and at a 2-sample kolmogorov-smirnov test for formal testing. But please read in on this test, as you have to understand what it does in order to interprete the results correctly.
On a sidenote : if you do this test on multiple cases, make sure you understand the problem of multiple comparisons and use the appropriate correction, eg. Bonferroni or Dunn-Sidak.