limit random complex number to a given range - matlab

I can get the real part of a random number to stay withing a given range but the complex part of the number doesn't stay within the range I set. see matlab / octave code below.
xmin=-.5
xmax=1
n=3
x=xmin+rand(1,n)*(xmax-xmin)+(rand(1,n)-(xmax-xmin))*1i
x=x(:)
The real part works but the complex part isn't limited to -0.5 to 1
0.2419028288441536 - 0.6579427654754871i
0.2712527227134944 - 1.451964497492678i
0.3245051849394858 - 1.107556052779179i

You have two mistakes:
x=xmin+rand(1,n)*(xmax-xmin)+(xmin + rand(1,n)*(xmax-xmin))*1i
You should add xmin to the sum and change - to * in the second part.

I've added some spaces to your code so the difference more obvious:
x = xmin+rand(1,n)*(xmax-xmin) + ( rand(1,n)-(xmax-xmin) )*1i
^^^ correct ^^^ not correct: missing `xmin+`
(and as OmG noted, also a `-` instead of a `*`)
One good way to reduce the number of bugs is by avoiding code duplication. You could for example write:
rand_sequence = #(m,xmin,xmax) xmin+rand(1,n)*(xmax-xmin);
x = rand_sequence(n,xmin,xmax) + 1i*rand_sequence(n,xmin,xmax)
(This looks like more code, but the more complicated code logic is not duplicated.)
Or like this:
x = xmin + (rand(1,n)+1i*rand(1,n)) * (xmax-xmin);

Related

Matlab: efficienting portion of code, random start

I have the following code that generates a matrix of 15 blocks that will then be used in a Montecarlo approach as multiple starting points. How can I get the same exact result in a smarter way?
assume that J=15*100 are the total simulation and paramNum the number of parameters
[10^-10*ones(paramNum,round(J/15)) 10^-9*ones(paramNum,round(J/15)) 10^-8*ones(paramNum,round(J/15)) 10^-7*ones(paramNum,round(J/15)) 10^-6*ones(paramNum,round(J/15)) 10^-5*ones(paramNum,round(J/15)) rand*10^-5*ones(paramNum,round(J/15)) 10^-4*ones(paramNum,round(J/15)) rand*10^-4*ones(paramNum,round(J/15)) 10^-3*ones(paramNum,round(J/15)) 10^-2*ones(paramNum,round(J/15)) 10^-1*ones(paramNum,round(J/15)) 10^-abs(randn/2)*ones(paramNum,round(J/15))];
you could do
v = 10.^[-10:-5 rand*10^-5 -4:-1 10^-abs(randn/2)];
repmat(repelem(v, 1, round(J/15)), paramNum) .* ...
repmat(ones(paramNum,round(J/15)), numel(v))
Or mimic the repmat/repelem functionality with a for loop. The first is shorter, the later is more understandable.
By the way... it's less than 15 blocks...

partial Distance Based RDA - Centroids vanished from Plot

I am trying to fir a partial db-RDA with field.ID to correct for the repeated measurements character of the samples. However including Condition(field.ID) leads to Disappearance of the centroids of the main factor of interest from the plot (left plot below).
The Design: 12 fields have been sampled for species data in two consecutive years, repeatedly. Additionally every year 3 samples from reference fields have been sampled. These three fields have been changed in the second year, due to unavailability of the former fields.
Additionally some environmental variables have been sampled (Nitrogen, Soil moisture, Temperature). Every field has an identifier (field.ID).
Using field.ID as Condition seem to erroneously remove the F1 factor. However using Sampling campaign (SC) as Condition does not. Is the latter the rigth way to correct for repeated measurments in partial db-RDA??
set.seed(1234)
df.exp <- data.frame(field.ID = factor(c(1:12,13,14,15,1:12,16,17,18)),
SC = factor(rep(c(1,2), each=15)),
F1 = factor(rep(rep(c("A","B","C","D","E"),each=3),2)),
Nitrogen = rnorm(30,mean=0.16, sd=0.07),
Temp = rnorm(30,mean=13.5, sd=3.9),
Moist = rnorm(30,mean=19.4, sd=5.8))
df.rsp <- data.frame(Spec1 = rpois(30, 5),
Spec2 = rpois(30,1),
Spec3 = rpois(30,4.5),
Spec4 = rpois(30,3),
Spec5 = rpois(30,7),
Spec6 = rpois(30,7),
Spec7 = rpois(30,5))
data=cbind(df.exp, df.rsp)
dbRDA <- capscale(df.rsp ~ F1 + Nitrogen + Temp + Moist + Condition(SC), df.exp); ordiplot(dbRDA)
dbRDA <- capscale(df.rsp ~ F1 + Nitrogen + Temp + Moist + Condition(field.ID), df.exp); ordiplot(dbRDA)
You partial out variation due to ID and then you try to explain variable aliased to this ID, but it was already partialled out. The key line in the printed output was this:
Some constraints were aliased because they were collinear (redundant)
And indeed, when you ask for details, you get
> alias(dbRDA, names=TRUE)
[1] "F1B" "F1C" "F1D" "F1E"
The F1? variables were constant within ID which already was partialled out, and nothing was left to explain.

Efficiently calculate mean value from files

From a Monte-Carlo simulation I have a range of files, say: file_1.mat, file_2.mat,...,file_n.mat, where n is large. Each file contains one or several (maximum 3 if it matters) large 1D arrays in time of interest, say var1, var2, var3.
I am now as always interested in finding the mean value of these variables. My question is now, how do I do this in the most efficient way? The keyword here is efficiency. Below you will find the MWE which is done the standard way, but this is quite time consuming as the files are large and there are many.
I am programming in Matlab, however ideas presented in pseudo code is also very well received.
MWE:(The standard way)
meanVar1 = zeros(1,1e6); %I do not remember the exact size, just use 1e6
meanVar2 = zeros(1,1e6);
meanVar3 = zeros(1,1e6);
for i 1=1:n
load(strcat('file_',int2str(i)),'var1','var2','var3')
meanVar1 = meanVar1 + var1;
meanVar2 = meanVar2 + var2;
meanVar3 = meanVar3 + var3;
end
meanVar1 = meanVar1/n;
meanVar2 = meanVar2/n;
meanVar3 = meanVar3/n;

Function equivalent to SUM() for multiplication in SQL Reporting

I'm looking for a function or solution to the following:
For the chart in SQL Reporting i need to multiply values from a Column A. For summation i would use =SUM(COLUMN_A) for the chart. But what can i use for multiplication - i was not able to find a solution so far?
Currently i am calculating the value of the stacked column as following:
=ROUND(SUM(Fields!Value_Is.Value)/SUM(Fields!StartValue.Value),3)
Instead of SUM i need something to multiply the values.
Something like that:
=ROUND(MULTIPLY(Fields!Value_Is.Value)/MULTIPLY(Fields!StartValue.Value),3)
EDIT #1
Okay tried to get this thing running.
The expression for the chart looks like this:
=Exp(Sum(Log(IIf(Fields!Menge_Ist.Value = 0, 10^-306, Fields!Menge_Ist.Value)))) / Exp(Sum(Log(IIf(Fields!Startmenge.Value = 0, 10^-306, Fields!Startmenge.Value))))
If i calculate my 'needs' manually i have to get the following result:
In my SQL Report i get the following result:
To make it easier, these are the raw values:
and you have the possibility to group the chart by CW, CQ or CY
(The values from the first pictures are aggregated Sum values from the raw values by FertStufe)
EDIT #2
Tried your expression, which results in this:
Just to make it clear:
The values in the column
=Value_IS / Start_Value
in the first picture are multiplied against each other
0,9947 x 1,0000 x 0,59401 = 0,58573
Diffusion Calenderweek 44 Sums
Startvalue: 1900,00 Value Is: 1890,00 == yield:0,99474
Waffer unbestrahlt Calenderweek 44 Sums
Startvalue: 620,00 Value Is: 620,00 == yield 1,0000
Pellet Calenderweek 44 Sums
Startvalue: 271,00 Value Is: 160,00 == yield 0,59041
yield Diffusion x yield Wafer x yield Pellet = needed Value in chart = 0,58730
EDIT #3
The raw values look like this:
The chart ist grouped - like in the image - on these fields
CY (Calendar year), CM (Calendar month), CW (Calendar week)
You can download the data as xls here:
https://www.dropbox.com/s/g0yrzo3330adgem/2013-01-17_data.xls
The expression i use (copy / past from the edit window)
=Exp(Sum(Log(Fields!Menge_Ist.Value / Fields!Startmenge.Value)))
I've exported the whole report result to excel, you can get it here:
https://www.dropbox.com/s/uogdh9ac2onuqh6/2013-01-17_report.xls
it's actually a workaround. But I am pretty sure is the only solution for this infamous problem :D
This is how I did:
Exp(∑(Log(X))), so what you should do is:
Exp(Sum(Log(Fields!YourField.Value)))
Who said math was worth nothing? =D
EDIT:
Corrected the formula.
By the way, it's tested.
Addressing Ian's concern:
Exp(Sum(Log(IIf(Fields!YourField.Value = 0, 10^-306, Fields!YourField.Value))))
The idea is change 0 with a very small number. Just an idea.
EDIT:
Based on your updated question this is what you should do:
Exp(Sum(Log(Fields!Value_IS.Value / Fields!Start_Value.Value)))
I just tested the above code and got the result you hoped for.

How to select the last column of numbers from a table created by FoldList in Mathematica

I am new to Mathematica and I am having difficulties with one thing. I have this Table that generates 10 000 times 13 numbers (12 numbers + 1 that is a starting number). I need to create a Histogram from all 10 000 13th numbers. I hope It's quite clear, quite tricky to explain.
This is the table:
F = Table[(Xi = RandomVariate[NormalDistribution[], 12];
Mu = -0.00644131;
Sigma = 0.0562005;
t = 1/12; s = 0.6416;
FoldList[(#1*Exp[(Mu - Sigma^2/2)*t + Sigma*Sqrt[t]*#2]) &, s,
Xi]), {SeedRandom[2]; 10000}]
The result for the following histogram could be a table that will take all the 13th numbers to one table - than It would be quite easy to create an histogram. Maybe with "select"? Or maybe you know other ways to solve this.
You can access different parts of a list using Part or (depending on what parts you need) some of the more specialised commands, such as First, Rest, Most and (the one you need) Last. As noted in comments, Histogram[Last/#F] or Histogram[F[[All,-1]]] will work fine.
Although it wasn't part of your question, I would like to note some things you could do for your specific problem that will speed it up enormously. You are defining Mu, Sigma etc 10,000 times, because they are inside the Table command. You are also recalculating Mu - Sigma^2/2)*t + Sigma*Sqrt[t] 120,000 times, even though it is a constant, because you have it inside the FoldList inside the Table.
On my machine:
F = Table[(Xi = RandomVariate[NormalDistribution[], 12];
Mu = -0.00644131;
Sigma = 0.0562005;
t = 1/12; s = 0.6416;
FoldList[(#1*Exp[(Mu - Sigma^2/2)*t + Sigma*Sqrt[t]*#2]) &, s,
Xi]), {SeedRandom[2]; 10000}]; // Timing
{4.19049, Null}
This alternative is ten times faster:
F = Module[{Xi, beta}, With[{Mu = -0.00644131, Sigma = 0.0562005,
t = 1/12, s = 0.6416},
beta = (Mu - Sigma^2/2)*t + Sigma*Sqrt[t];
Table[(Xi = RandomVariate[NormalDistribution[], 12];
FoldList[(#1*Exp[beta*#2]) &, s, Xi]), {SeedRandom[2];
10000}] ]]; // Timing
{0.403365, Null}
I use With for the local constants and Module for the things that are other redefined within the Table (Xi) or are calculations based on the local constants (beta). This question on the Mathematica StackExchange will help explain when to use Module versus Block versus With. (I encourage you to explore the Mathematica StackExchange further, as this is where most of the Mathematica experts are hanging out now.)
For your specific code, the use of Part isn't really required. Instead of using FoldList, just use Fold. It only retains the final number in the folding, which is identical to the last number in the output of FoldList. So you could try:
FF = Module[{Xi, beta}, With[{Mu = -0.00644131, Sigma = 0.0562005,
t = 1/12, s = 0.6416},
beta = (Mu - Sigma^2/2)*t + Sigma*Sqrt[t];
Table[(Xi = RandomVariate[NormalDistribution[], 12];
Fold[(#1*Exp[beta*#2]) &, s, Xi]), {SeedRandom[2];
10000}] ]];
Histogram[FF]
Calculating FF in this way is even a little faster than the previous version. On my system Timing reports 0.377 seconds - but such a difference from 0.4 seconds is hardly worth worrying about.
Because you are setting the seed with SeedRandom, it is easy to verify that all three code examples produce exactly the same results.
Making my comment an answer:
Histogram[Last /# F]