for loop returns NaN

6 Ansichten (letzte 30 Tage)
negin tebyani
negin tebyani am 10 Feb. 2018
Beantwortet: Walter Roberson am 10 Feb. 2018
I am trying to do this in my code:
for i=1:c+1
load(row) = load(row)+(ro(row,i)* MinRateCurrentUserRRH(row,i))/(log(1+SINR(row,i)));
end
ro is a matrix of 1 and 0 s and MinRateCurrentUserRRH and SINR s are other matrixes. before running the for loop, load (row) is fine and for example 0.3, after first loop, it turns to NaN, so it gives errors, I can't figure out the problem. can anyone help me with it?
  3 Kommentare
negin tebyani
negin tebyani am 10 Feb. 2018
Bearbeitet: negin tebyani am 10 Feb. 2018
SINR and currnetminrate are zero matrices that are filled in the code in other loops, in the time of running this, they look like:
and
and c is 1.
dpb
dpb am 10 Feb. 2018
As other poster says, nothing anybody here can do without actual data but you've got all the information you need; just use the debugger and see where your logic fails.
While not the specific problem, you can probably replace all the loops with Matlab array operations; "the MATLAB way". Look at the "dot" operators under
>> help punct
Punctuation.
. Decimal point. 325/100, 3.25 and .325e1 are all the same.
. Array operations. Element-by-element multiplicative operations
are obtained using .* , .^ , ./ , .\ or .'. For example,
C = A ./ B is the matrix with elements c(i,j) = a(i,j)/b(i,j).
...

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 10 Feb. 2018
Your row is a vector, so (log(1+SINR(row,i)) is a vector, so your / operator is mrdivide, the matrix right division operator, with A/B being approximately equivalent to A * pinv(B) where the * indicates mtimes, the algebraic matrix multiplication operator. For vector row you need the ./ operator instead of /
You will also need to watch out for the * operator earlier in the line: you probably want it to be .*

Weitere Antworten (0)

Kategorien

Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by