Constructing a string with several index requirements
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Hello, I have a vector of numbers r2 that I need to send over a communciation to an array (called ScanArray). The comms is such that I can only send upto 50 elements in each send (Im using writeline)
Heres the 1st 8 values I need to send
r2 =
0 30 60 90 120 150 180 210
and heres my string that I construct that I send with writeline function
command=['ScanArray(0)(0)=',num2str(r2(1,1)),'ScanArray(0)(1)=',num2str(r2(1,2)),'ScanArray(0)(2)=',num2str(r2(1,3)),'ScanArray(0)(3)=',num2str(r2(1,4))]
Rather than write this out for 50 elements ScanArray(0)(0) -> ScanArray(0)(49), is there a more effcient way to construct this. This was my attemp:
command=[];
for i=1:50
commandnew=['ScanArray(0)(',num2str(i),')=',num2str(r2(1,i))]
command=[command,commandnew]
end
2 commentaires
Stephen23
le 8 Nov 2024
"is there a more effcient way to construct this."
Do not fight MATLAB with loops. Use e.g. strings or COMPOSE:
r2 = 0:30:210;
ix = 1:numel(r2);
tx = "ScanArray(0)(" + ix(:) + ")=" + r2(:)
tx = compose('ScanArray(0)(%u) = %u',ix(:),r2(:))
Voss
le 8 Nov 2024
r2 = 0:30:210;
ix = 0:numel(r2)-1;
tx = "ScanArray(0)(" + ix(:) + ")=" + r2(:);
tx = [tx{:}]
tx = compose('ScanArray(0)(%u)=%u',ix(:),r2(:));
tx = [tx{:}]
Réponse acceptée
Star Strider
le 8 Nov 2024
I’m not certain what you need, however this is one option —
r2 = [0 30 60 90 120 150 180 210];
command=['ScanArray(0)(0)=',num2str(r2(1,1)),'ScanArray(0)(1)=',num2str(r2(1,2)),'ScanArray(0)(2)=',num2str(r2(1,3)),'ScanArray(0)(3)=',num2str(r2(1,4))]
for k = 1:numel(r2)-3
command = ["ScanArray(0)("+(k-1)+")="+r2(1,k)+"ScanArray(0)("+k+")="+r2(1,k+1)+"ScanArray(0)("+(k+1)+")="+r2(1,k+2)+"ScanArray(0)("+(k+2)+")="+r2(1,k+3)]
end
It may be necessary to expand on that, perhaps with a nested loop, for a multi-row ‘r2’.
.
1 commentaire
Voss
le 8 Nov 2024
The square brackets are unnecessary when using string concatenation:
"S"+1
["S"+1]
Plus de réponses (3)
Voss
le 8 Nov 2024
r2 = [0 30 60 90 120 150 180 210];
command = sprintf('ScanArray(0)(%d)=%g',[0:numel(r2)-1; r2])
The %g is to handle cases where elements of r2 are not integers. If they are always integers you can use %d there instead.
And I don't know but you may need a delimiter between adjacent ScanArray(0) assignments, as in
command = sprintf('ScanArray(0)(%d)=%g ',[0:numel(r2)-1; r2])
or
command = sprintf('ScanArray(0)(%d)=%g;',[0:numel(r2)-1; r2])
etc.
0 commentaires
Steven Lord
le 8 Nov 2024
r2 = [0 30 60 90 120 150 180 210];
ind = 0:numel(r2)-1;
result = join("ScanArray(0)(" + ind + ")=" + r2, newline)
Replace newline with ";" to join the substrings with semicolons rather than newlines (which I used so you can easily see the individual substrings included in result.)
4 commentaires
Voss
le 8 Nov 2024
Or:
r2 = (0:47)*30;
N = numel(r2);
n = 4;
assert(mod(N,n) == 0)
result = strjoin("ScanArray("+(0:N/n-1)+")("+(0:n-1).'+")="+reshape(r2,n,[]),newline())
Voir également
Catégories
En savoir plus sur Matrix Indexing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!