set mdraws range using stata - set

I am using the Stata command mdraws to generate random numbers (by default using Halton sequences).
I am wondering if there's a way to set the range for the random numbers? For example, is there anything I could do using mdraws if I want my random numbers to be in the range of 0.05 to 0.5?

Elsewhere in the Stata community you are asked to explain where user-written commands you use come from, and that is a very good convention also for Stack Overflow.
mdraws is a program by Capellari and Jenkins from Stata Journal 6(2), 2006: readers will find that typing findit mdraws in Stata produces pointers to download sources.
The direct answer is No, but the problem is easily tackled by rescaling. For any variable x generated in the interval from 0 to 1, map to 0.05 to 0.5 by
gen x = 0.05 + 0.45 * y
If you have several such variables, use foreach or forval in a loop to rescale.
You didn't spell it out, so I will: this presumes that you want densities uniform on your stated interval.

Related

Generate random number in interval in PostScript

I am struggling to find a way to generate a random number within a given interval in PostScript.
Basically PostScript has three functions to help you generate (pseudo-)random numbers. Those are rand, srand and rrand.
The later two are for passing a seed to the number generator to be able to reproduce specific results. At least that´s what I understood they are for. Anyway they don´t seem suitable for my case.
So rand seems to be the only function I can use to generate a random number, but...
rand returns a random integer in the range 0 to 231 − 1 (From the PostScript Language Reference, page 637 (651 in the PDF))
This is far beyond the the interval I´m looking for. I am more interested in values up to small thousands, maybe 10.000 or something like that and small float values, up to 100, all with the lower limit of 0.
I thought I could just narrow my numbers down by simple divisions and extracting the root but that tends to give me unusable small values in quite a lot cases. I am wondering if there are robust ways to either shrink a large number down to what I need or, I´d prefer that, only generate numbers in the desired interval.
Besides: while-loops are not possible in PostScript, otherwise I´d have written a function to generate numbers until they fit in my interval.
Any hints on what to look for breaking numbers down into my interval?
mod is often good enough and it's fast. But you may get a more uniform distribution by using floating-point ops.
rand 16#7fffffff div 100 mul cvi
This is because mod discards the upper bits of the input. And the PRNG is usually trying to randomize over all the bits. By scaling down then up, they all contribute something in the way of rounding effects.
Just use the modulo operator to get it down to the size you want:
GS>rand 100 mod stack
7

Confirmation required: gnuplot does not change intital fit values

after using gnuplot for years and experiencing many user-related issues, I thought I'd finally know how to fit a function to a dataset.
Today I tried to fit a simple
y = m * x² + b
function. However, gnuplot did not change my 'm' value. It does change 'b' to the correct value however.
I have my dataset uploaded and here is my gnuplot script with which I'm trying to fit, maybe someone can reproduce this on his machine and confirm, that it is not a fault of my computer but some kind of faulty code in the script, or it may even be a bug (I highly doubt that).
set xtics 0.000001
set format x '%10.1E'
set xrange [0:2E-07]
#fit
f(x)=a*(x**2)+b
a=380812
b=1
fit [0:2E-07] f(x) 'GDAMitte1.txt' using ($1+7.6E-06):2 via a,b
plot 'GDAMitte1.txt' using ($1+7.6E-06):2, f(x)
I've pasted the dataset here: http://www.heypasteit.com/clip/29LU
I'd be very thankful for an answer to that, even if it's just a confirmation, that it doesn't fit on your machine as well. Thank you.
Btw: The initial value I've set is pretty much the one it has to be after the fit, but it's not as exact of course. Should be good enough though for gnuplot to get where to go to.
This is because two parameters are of greatly different magnitude, check help fit tips.
You should replace the function with one that has a prefactor built in:
f(x) = a *1e5 * (x**2)+b
a=3.8 # instead of 380000
b=1
fit ....
From gnuplot version 5.0 on, gnuplot by default internally prescales all parameters, so this problem with calculating the residuals should no longer occur for any function, provided your initial values are not off too much.

How to implement a part of histogram equalization in matlab without using for loops and influencing speed and performance

Suppose that I have these Three variables in matlab Variables
I want to extract diverse values in NewGrayLevels and sum rows of OldHistogram that are in the same rows as one diverse value is.
For example you see in NewGrayLevels that the six first rows are equal to zero. It means that 0 in the NewGrayLevels has taken its value from (0 1 2 3 4 5) of OldGrayLevels. So the corresponding rows in OldHistogram should be summed.
So 0+2+12+38+113+163=328 would be the frequency of the gray level 0 in the equalized histogram and so on.
Those who are familiar with image processing know that it's part of the histogram equalization algorithm.
Note that I don't want to use built-in function "histeq" available in image processing toolbox and I want to implement it myself.
I know how to write the algorithm with for loops. I'm seeking if there is a faster way without using for loops.
The code using for loops:
for k=0:255
Condition = NewGrayLevels==k;
ConditionMultiplied = Condition.*OldHistogram;
NewHistogram(k+1,1) = sum(ConditionMultiplied);
end
I'm afraid if this code gets slow for high resolution big images.Because the variables that I have uploaded are for a small image downloaded from the internet but my code may be used for sattellite images.
I know you say you don't want to use histeq, but it might be worth your time to look at the MATLAB source file to see how the developers wrote it and copy the parts of their code that you would like to implement. Just do edit('histeq') or edit('histeq.m'), I forget which.
Usually the MATLAB code is vectorized where possible and runs pretty quick. This could save you from having to reinvent the entire wheel, just the parts you want to change.
I can't think a way to implement this without a for loop somewhere, but one optimisation you could make would be using indexing instead of multiplication:
for k=0:255
Condition = NewGrayLevels==k; % These act as logical indices to OldHistogram
NewHistogram(k+1,1) = sum(OldHistogram(Condition)); % Removes a vector multiplication, some additions, and an index-to-double conversion
end
Edit:
On rereading your initial post, I think that the way to do this without a for loop is to use accumarray (I find this a difficult function to understand, so read the documentation and search online and on here for examples to do so):
NewHistogram = accumarray(1+NewGrayLevels,OldHistogram);
This should work so long as your maximum value in NewGrayLevels (+1 because you are starting at zero) is equal to the length of OldHistogram.
Well I understood that there's no need to write the code that #Hugh Nolan suggested. See the explanation here:
%The green lines are because after writing the code, I understood that
%there's no need to calculate the equalized histogram in
%"HistogramEqualization" function and after gaining the equalized image
%matrix you can pass it to the "ExtractHistogram" function
% (which there's no loops in it) to acquire the
%equalized histogram.
%But I didn't delete those lines of code because I had tried a lot to
%understand the algorithm and write them.
For more information and studying the code, please see my next question.

How to do high precision float point arithmetics in mathematica

In Mma, for example, I want to calculate
1.0492843824838929890231*0.2323432432432432^3
But it does not show the full precision. I tried N or various other functions but none seemed to work. How to achieve this? Many thanks.
When you specify numbers using decimal point, it takes them to have MachinePrecision, roughly 16 digits, hence the results typically have less than 16 meaningful digits. You can do infinite precision by using rational/algebraic numbers. If you want finite precision that's better than default, specify your numbers like this
123.23`100
This makes Mathematica interpret the number as having 100 digits of precision. So you can do
ans=1.0492843824838929890231`100*0.2323432432432432`100^3
Check precision of the final answer using Precision
Precision[ans]
Check tutorial/ArbitraryPrecisionNumbers for more details
You may do:
r[x_]:=Rationalize[x,0];
n = r#1.0492843824838929890231 (r#0.2323432432432432)^3
Out:
228598965838025665886943284771018147212124/17369643723462006556253010609136949809542531
And now, for example
N[n,100]
0.01316083216659453615093767083090600540780118249299143245357391544869\
928014026433963352910151464006549
Sometimes you just want to see more of the machine precision result. These are a few methods.
(1) Put the cursor at the end of the output line, and press Enter (not on the numeric keypad) to copy the output to a new input line, showing all digits.
(2) Use InputForm as in InputForm[1.0/7]
(3) Change the setting of PrintPrecision using the Options Inspector.

Generating random number in a given range in Fortran 77

I am a beginner trying to do some engineering experiments using fortran 77. I am using Force 2.0 compiler and editor. I have the following queries:
How can I generate a random number between a specified range, e.g. if I need to generate a single random number between 3.0 and 10.0, how can I do that?
How can I use the data from a text file to be called in calculations in my program. e.g I have temperature, pressure and humidity values (hourly values for a day, so total 24 values in each text file).
Do I also need to define in the program how many values are there in the text file?
Knuth has released into the public domain sources in both C and FORTRAN for the pseudo-random number generator described in section 3.6 of The Art of Computer Programming.
2nd question:
If your file, for example, looks like:
hour temperature pressure humidity
00 15 101325 60
01 15 101325 60
... 24 of them, for each hour one
this simple program will read it:
implicit none
integer hour, temp, hum
real p
character(80) junkline
open(unit=1, file='name_of_file.dat', status='old')
rewind(1)
read(1,*)junkline
do 10 i=1,24
read(1,*)hour,temp,p,hum
C do something here ...
10 end
close(1)
end
(the indent is a little screwed up, but I don't know how to set it right in this weird environment)
My advice: read up on data types (INTEGER, REAL, CHARACTER), arrays (DIMENSION), input/output (READ, WRITE, OPEN, CLOSE, REWIND), and loops (DO, FOR), and you'll be doing useful stuff in no time.
I never did anything with random numbers, so I cannot help you there, but I think there are some intrinsic functions in fortran for that. I'll check it out, and report tomorrow. As for the 3rd question, I'm not sure what you ment (you don't know how many lines of data you'll be having in a file ? or ?)
You'll want to check your compiler manual for the specific random number generator function, but chances are it generates random numbers between 0 and 1. This is easy to handle - you just scale the interval to be the proper width, then shift it to match the proper starting point: i.e. to map r in [0, 1] to s in [a, b], use s = r*(b-a) + a, where r is the value you got from your random number generator and s is a random value in the range you want.
Idigas's answer covers your second question well - read in data using formatted input, then use them as you would any other variable.
For your third question, you will need to define how many lines there are in the text file only if you want to do something with all of them - if you're looking at reading the line, processing it, then moving on, you can get by without knowing the number of lines ahead of time. However, if you are looking to store all the values in the file (e.g. having arrays of temperature, humidity, and pressure so you can compute vapor pressure statistics), you'll need to set up storage somehow. Typically in FORTRAN 77, this is done by pre-allocating an array of a size larger than you think you'll need, but this can quickly become problematic. Is there any chance of switching to Fortran 90? The updated version has much better facilities for dealing with standardized dynamic memory allocation, not to mention many other advantages. I would strongly recommend using F90 if at all possible - you will make your life much easier.
Another option, depending on the type of processing you're doing, would be to investigate algorithms that use only single passes through data, so you won't need to store everything to compute things like means and standard deviations, for example.
This subroutine generate a random number in fortran 77 between 0 and ifin
where i is the seed; some great number such as 746397923
subroutine rnd001(xi,i,ifin)
integer*4 i,ifin
real*8 xi
i=i*54891
xi=i*2.328306e-10+0.5D00
xi=xi*ifin
return
end
You may modifies in order to take a certain range.

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