I have a data consist of time and flux (4117 rows x 2 columns) I want to calculate and plot the number distribution of brightness variation between all pairs of two consecutive data points same as the picture distribution of brightness variation
This is the code I used in idl
nx=4117
t=fltarr(nx)
f=fltarr(nx)
df=fltarr(nx-1)
dt=fltarr(nx-1)
n=4116
dff=fltarr(n)
dc=fltarr(n-1)
data=read_table('data.dat')
;print,data(0,*) ;this is t (time)
;print,data(1,*) ;this is f (flux)
; Plot the light curve
window,0
plot,data(0,*)/data(0,0),data(1,*)/data(1,0),yrange=[0.93,1.1],ystyle=1
; calculate the flux difference (dff)
for i=0,nx-2 do begin
df(i)=data(1,i+1)/data(1,0) - data(1,i)/data(1,0)
dt(i)=data(0,i+1)/data(0,0) - data(0,i)/data(0,0)
endfor
for i=0,n-1 do dff(i)=min(df)+i*(max(df)-min(df))/float(n-1.0)
print,dff
; calculate the number distribution (dc), I want the counter to reset to zero after every point and start to count again
for i=0,n-2 do begin
c=0.0
for j=0,nx-2 do begin
IF (df(j) < dff(i+1)) or (df(j) > dff(i)) THEN begin
c=c+1
dc(i)=c
endif
endfor
print, dc(i)
endfor
end
when I run the code all the value of dc is 4116 . I think the way I calculated dc is wrong. Any suggestion to do this in proper way?
I'm pretty sure the problem is this line:
IF (df(j) < dff(i+1)) or (df(j) > dff(i)) THEN begin
In IDL, a < b is shorthand for "the lesser of a and b", and a > b is likewise "the greater of a and b". So right now, your IF statement is actually evaluated as:
IF (df(j) or dff(i+1), whichever is less) OR (df(j) or dff(i), whichever is more) THEN begin
and since non-zero floats evaluate as TRUE in IDL, it's ultimately this:
IF TRUE or TRUE THEN begin
Because the IF statement is always true, c is always incremented, and every value of dc ends up being 4116.
To fix this, you want to use the relational operators LT and GT, which return TRUE or FALSE. In other words, your code should read:
IF (df(j) LT dff(i+1)) or (df(j) GT dff(i)) THEN begin
Related
Hi¡ I have a doubt and I hope someone can help me please, I have a dataframe in R and it makes a double cicle for and an if, the data frame has some values and then if the condition is True, it makes some operations, the problem is I can't understand neither the cicle and the operation the code makes under the condition.
I reply the code I have in a simpler one but the idea is the same. And if someone can explain me the whole operation please.
w<-c(2,5,4,3,5,6,8,2,4,6,8)
x<-c(2,5,6,7,1,1,4,9,8,8,2)
y<-c(2,5,6,3,2,4,5,6,7,3,5)
z<-c(2,5,4,5,6,3,2,5,6,4,6)
letras<-data.frame(w,x,y,z)
l=1
o=1
v=nrow(letras)
letras$op1<-c(1)
letras$op2<-c(0)
for (l in 1:v) {
for (o in 1:v) {
if(letras$x[o]==letras$y[l] & letras$z[l]==letras$z[o] & letras$w[l]){
letras$op1<-letras$op1+1
letras$op2<-letras$x*letras$y
}
}
}
The result is the following:
Thanks¡¡¡¡¡
This segment of code is storing values into vectors labeled w,x,y,z.
w<-c(2,5,4,3,5,6,8,2,4,6,8)
x<-c(2,5,6,7,1,1,4,9,8,8,2)
y<-c(2,5,6,3,2,4,5,6,7,3,5)
z<-c(2,5,4,5,6,3,2,5,6,4,6)
It then transforms the 4 vectors into a data frame
letras<-data.frame(w,x,y,z)
This bit of code isn't doing anything as far as I can tell.
l=1 #???
o=1 #???
This counts how many rows is in the letras data frame and stores to v, in this case 11 rows.
v=nrow(letras)
This creates new columns in letras dataframe with all ones in op1 and all zeros in op2
letras$op1<-c(1)
letras$op2<-c(0)
Here each for loop is acting as a counter, and will run the code beneath it iteratively from 1 to v (11), so 11 iterations. Each iteration the value of l will increase by 1. So first iteration l = 1, second l=2... etc.
for (l in 1:v) {
You then have a second counter, which is running within the first counter. So this will iterate over 1 to 11, exactly the same way as above. But the difference is, this counter will need to complete it's 1 to 11 cycle before the top level counter can move onto the next number. So o will effectively cycle from 1 to 11, for each 1 count of 1l. So with the two together, the inside for loop will count from 1 to 11, 11 times.
for (o in 1:v) {
You then have a logical statement which will run the code beneath if the column x and column y values are the same. Remember they will be calling different index values so it could be 1st x value vs the 2nd y value. There is an AND statement so it also needs the two z position values to be equal. and the last part letras$w[l] is always true in this particular example, so could possibly be removed.
if(letras$x[o]==letras$y[l] & letras$z[l]==letras$z[o] & letras$w[l]){
Lastly, is the bit that happens if the above statement is true.
op1 get's 1 added (remember this was starting from 1 anyway), and op2 multiplies x*y columns together. This multiplication is perhaps a little bit inefficient, because x and y do not change, so the answer will calculate the same result each time the the if statement evaluates TRUE.
letras$op1<-letras$op1+1
letras$op2<-letras$x*letras$y
}
}
}
Hope this helps.
I am creating a scheduling model for a blending facility in Mini Zinc. I have asked a similar question earlier but I have since progressed. I will summarize what I think my existing model should do, it would be super helpful if someone could correct any logical or syntax errors I have made. The model currently errors, with several instances of "expecting end of file". It seems way to simplistic in comparison to several other sequence dependent scheduling models I have found. Below you will find the model code commented with my understanding of each line.
Besides an overview of the logic and syntax, I am looking for help with the "missing constraint" in this model, which needs to require that the array of blends [y] contain no more than the declared integer quantity of each blend.
Notable future goals for this model include automatic generation of the blendcost matrix, output the schedule array given a starting day into a 5 column matrix representing weekdays, and showing the blend name as opposed to blend number.
enum Blends = { A, B, C, D, E, F};
%Establish names for each blend and link them to their order number.
int: nb = count([Blends])
%Count the number of blends, plan to use later.
int: qA; %Error: syntax error, unexpected int, expecting end of file
int: qB;
int: qC;
int: qD;
int: qE;
int: qF;
int: sb;
%Call for inputs of the quantity of each of the blends needed, as well as the number/letter of the blend currently in the machine.
int: mc = qA + qB + qC + qD + qE + qF;
%Sum the blend quantities to determine total number of blends
[Blendcost] : [|1,2,2,2,2,2,|1,1,1,1,1,1,|1,1,1,1,1,1,|2,2,2,1,2,2,|1,1,1,1,1,1,|1,1,1,1,1,1,|]; %Error: syntax error, unexpected [|, expecting identifier
%Establishes a blend cost matrix, 6X6 depicting the transition costs from any blend A-F to any other blend A-F
array [Blends] of int: 1..6;
%Is this line needed to establish the link between A/1, B/2, C/3 etc;? Or is that taken care of when Blends is enumerated?
array [0..mc] of var 1..6: y;
%Create an array from 0 to the number of blends with potential values from 1-6, corresponding to the blend numbers.
%Missing constraint: [y] can contain no more than the quantity of each blend declared above, except for the blend declared in the starting blend, which will be allowed that blend quantity + 1
constraint y(0) = sb
%y(0) is set equal to the starting blend Letter/Number Defined earlier, used to determine the first transitionary cost.
array [1..mc] of int: x(i); %Error: syntax error, unexpected array, expecting end of file
%Create an array from 1 to number of blends, which will be filled with the transition costs in response to variations in y
constraint forall(i in x)(x(i) = Blendcost(y(i-1),y(i)))
%For each space in x, x will equal the blend cost value obtained from the previous blend in the y array vs the next blend in the y array
solve minimize sum (x); %Error: syntax error, unexpected solve, expecting end of file
%Solves this model attempting to minimize the sum of the x array, which should be filled with the transition costs.
show(y):
%Print the final array of blend numbers that has minimized blend cost transition.
%Error: unexpected end of file, expecting identifier.
Here is a basic version of your CP-model that runs (assuming some demand q):
enum BLEND = { A, B, C, D, E, F};
array[BLEND] of int: q = [1, 2, 4, 6, 8, 1];
array[BLEND, BLEND] of int: Blendcost =
[|1,2,2,2,2,2|1,1,1,1,1,1|1,1,1,1,1,1|2,2,2,1,2,2|1,1,1,1,1,1|1,1,1,1,1,1|];
int: mc = sum(q);
array[1..mc] of var BLEND: y;
include "global_cardinality.mzn";
constraint global_cardinality(y, BLEND, q);
var int: obj = sum(p in 1..mc-1)(Blendcost[y[p],y[p+1]]) + 1;
array[int] of var opt BLEND: day = [y[p] | p in 1..mc-1, q in 1..max(Blendcost) where q <= Blendcost[y[p],y[p+1]]] ++ [y[mc]];
array[int] of var opt int: wash = [bool2int(q > 1) | p in 1..mc-1, q in 1..max(Blendcost) where q <= Blendcost[y[p],y[p+1]]] ++ [0];
solve minimize obj;
output ["obj=\(obj)\n"] ++
["day=\n"] ++ [
show(day[d]) ++ if fix(wash[d]) > 0 then "W" else "" endif ++ " " ++
if d mod 5 = 0 then "\n" else "" endif | d in 1..length(day)
] ++ ["\nmc=\(mc)\n"] ++ ["y=\(y)\n"] ++ ["wash=\(wash)\n"]
Have a look at https://www.minizinc.org/doc-2.2.3/en/lib-globals.html#counting-constraints for alternative versions of the counting constraint.
For larger instances a MIP-model might show better performance.
I've been working on a hex calculator for a while, but seem to be stuck on the subtraction portion, particularly when B>A. I'm trying to simply subtract two positive integers and display the result. It works fine for A>B and A=B. So far I'm able use two 7-segment displays to show the integers to be subtracted and I get the proper difference as long as A>=B
When B>A I see a pattern that I'm not able to debug because of my limited knowledge in Verilog case/if-else statements. Forgive me if I'm not explaining the best way but what I'm observing is that once the first number, A, "reaches" 0 (after being subtracted from) it loops back to F. The remainder of B is then subtracted from F rather than 0.
For example: If A=1, B=3
A - B =
1 - 1 = 0
0 - 1 = F
F - 1 = E
Another example could be 4-8=C
Below are the important snippets of code I've put together thus far.
First, my subtraction statement
always#*
begin
begin
Cout1 = 7'b1000000; //0
end
case(PrintDifference[3:0])
4'b0000 : Cout0 = 7'b1000000; //0
4'b0001 : Cout0 = 7'b1111001; //1
...
4'b1110 : Cout0 = 7'b0000110; //E
4'b1111 : Cout0 = 7'b0001110; //F
endcase
end
My subtraction is pretty straightforward
output [4:0]Difference;
output [4:0] PrintDifference;
assign PrintDifference = A-B;
I was thinking I could just do something like
if A>=B, Difference = B-A
else, Difference = A-B
Thank you everyone in advance!
This is expected behaviour of twos complement addition / subtraction which I would recommend reading up on since it is so essential.
The result obtained can be changed back into an unsigned form by inverting all the bits and adding one. Checking the most significant bit will tell you if the number is negative or not.
I tried implementing Lloyd's algorithm and it seemed good until I ran it multiple times. Sometimes it gives the results I want, sometimes it gives strange centres.
I tried to change the condition so it stops when it has converged, but it doesn't help. Sorry for not translating comments to English, I hope it's clear enough.
The only randomness I have in the code is in the situation where my cluster empties so I replace it with a random point. I have no other idea what to do when this happens.
I can't see the problem. Can you give me an idea what might be the problem from the result figures?
This is my code:
(A is a matrix whose rows are my points)
% initialization of centroids; further-first method
n=size(A,1);
dim=size(A,2);
centri=zeros(k,dim); %matrix of centroids
for i=1:n
centri(1,:)=centri(1,:)+A(i,:);
end
centri(1,:)=centri(1,:)/n;
for j=2:k %u svakom koraku postavljamo za centar onu tocku koja je najdalje od centra 1,..j-1
maks=zeros(1,n);
%maks(i) je najveca udaljenost te tocke do centra =max d(x(i),c), c centri
for i=1:n
dist=zeros(1,j-1);
for l=1:j-1
dist(l)=norm(A(i,:)-centri(l,:));
end
if(size(dist,2)==1) maks(i)=dist;
else
maks(i)=max(dist);
end
%maks(i)=0;
%for l=1:j-1
% if(maks(i)<dist(l)) maks(i)=dist(l);
% end
%end
end
[maksi, ind]=max(maks);
centri(j,:)=A(ind(1),:);
end
indeksi=zeros(1,n);
for i=1:n
indeksi(i)=randi(k,1);
end
% u centrima je postavljena pocetna inicijalizacija
br_iter=0;
tic
while br_iter<=1000
br_iter=br_iter+1;
for i=1:n
dist=zeros(1,k); % udaljenosti od tocke x do centra j
for j=1:k
dist(j)=norm(A(i,:)-centri(j,:));
end
[mini, ind]=min(dist); % ind je vektor za koji se poprima minimalna vrijednost
indeksi(i)=ind(1); % uzmemo prvi po redu
end
% sad radimo nove centroide koji su aritmetička sredina svih vektora koji mu pripadaju
for j=1:k
centri(j,:)=zeros(1,dim);
brojac=0;
for i=1:n
if indeksi(i)==j
centri(j,:)=centri(j,:)+A(i,:);
brojac=brojac+1;
end
end
if brojac
centri(j,:)=centri(j,:)/brojac;
else
ind=randi(n, 1);
centri(j,:)=A(ind,:);
end
end
end
toc
for i=1:n
plot(A(i,1), A(i,2), '.b');
if(i==1) hold on;
end
end
for i=1:k
plot(centri(i,1), centri(i,2), '+r');
end
hold off
Starting with centers all zero is not a recommended approach.
After the first iteration, all but one of these centers will be empty. So randomness does have an effect on your result.
The results you show are typical for k-means. It does not guarantee to fond the optimum, but it can get stuck in a "local optimum".
So I don't think there is an error in your code. Just the starting condition is not chosen very wisely & you are mistaken to expect k-means to always give good results.
I have a problem to understand the role of to-report and report in NetLogo, even it seems pretty useful and I can't really find a help written in "human style" language.
In NetLogo dictionnary http://ccl.northwestern.edu/netlogo/docs/dictionary.html#report I can find definitions for to-report :
to-report procedure-name
to-report procedure-name [input1 ...]
Used to begin a reporter procedure.
The body of the procedure should use report to report a value for the procedure. See report.
and for report:
report value
Immediately exits from the current to-report procedure and reports value as the result of that procedure. report and to-report are always used in conjunction with each other. See to-report for a discussion of how to use them.
So, it seems to-report and report calculate some value and report it.
Thus, when I try add
to-report average [a b c]
report (a + b + c) / 2
end
to my code, and then use the average variable somewhere in my code p.e.:
to go
...
print average
tick
end
I've got an error: AVERAGE expected 3 inputs. When I try to create my variables [a b c] in globals [a b c] I've got an error There is already a global variable called A.
If I define my variables [a b c] within to-report procedure:
to-report average [a b c]
set a 1
set b 2
set c 3
report (a + b + c) / 2
end
My error is again AVERAGE expected 3 inputs.
Thus, how can I simply test the usefulness of to-report procedure? And where to place it correctly in my code to see what it is really doing? From Urban Suite - Economic Disparity (http://ccl.northwestern.edu/netlogo/models/UrbanSuite-EconomicDisparity) I see that to-report is used to calculate values related to each patch:
to-report patch-utility-for-poor
report ( ( 1 / (sddist / 100 + 0.1) ) ^ ( 1 - poor-price-priority ) ) * ( ( 1 / price ) ^ ( 1 + poor-price-priority ) )
end
however this reported value is not directly defined as patch variable which increase my confusion...
Thank you !
A function can take some input (usually one or more variables or values) and return some output (usually a single value). You can specify that a function returns a value using to-report in your function header and report returns the actual value.
Your error is due to the fact that you never passed in arguments to your average function
to go
...
print average
tick
end
should be
to go
...
print average 5 2 3 ;;a = 5, b = 2, c =3
tick
end
Inside your average function, you should not reassign values of a,b,and c.
You should use report whenever you want to return a result from a function.