Making an array using loop
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A=[2;3;4;5;12;13;14;15;16;17;24;25;26;27;28;29;36;37;38;39;40;41;48;49;50;51;52;53;60;61;62;63;64;65;72;73;74;75;76;77;84;85;86;87;88;89;96;97;98;99;100;101;108;109;110;111;112;113;120;121];
How to make an array like A using loop. Thank you.
3 个评论
Cris LaPierre
2020-12-30
is there a pattern to the numbers? If not, a for loop isn't going to help you much.
SWARNENDU PAL
2020-12-31
回答（1 个）
Paul Hoffrichter
2020-12-31
编辑：Paul Hoffrichter
2021-1-1
Change ORIGINAL_POST to false to get slightly different result.
clearvars; clc; % remove previous debug runs
lenA = 60; % assumes you know the length of A
A = zeros(lenA,1); % pre-allocate A
maxSeqLen = 6;
ORIGINAL_POST = true;
if ORIGINAL_POST
start = 2;
seqLength = 4;
else
% rng(123);
start = randi(maxSeqLen,1);
seqLength = randi(maxSeqLen);
end
nominalSequence = 1:maxSeqLen;
skip = 7;
% Initialize
A(1:seqLength) = start:(start + seqLength - 1);
startLoc = seqLength + 1;
k = seqLength + 1;
while startLoc + maxSeqLen - 1 <= lenA
start = A(startLoc - 1) + skip;
A(startLoc:startLoc + maxSeqLen - 1) = start: start + maxSeqLen - 1;
start = start + maxSeqLen - 1 + skip;
startLoc = startLoc + maxSeqLen;
end
if startLoc <= lenA
maxSeqLen = lenA - startLoc + 1;
A(startLoc:startLoc + maxSeqLen - 1) = start: start + maxSeqLen - 1;
% if you cannot pre-allocate A, and have to enter values one at a time,
% then you can add elements like this: A = [A; new-value];
end
2 个评论
Rik
2020-12-31
You don't print anything to the command window, so what is the clc doing? And why are you suggesting clear all? mlint is clearly warning you not to use this.
Paul Hoffrichter
2021-1-1
编辑：Paul Hoffrichter
2021-1-1
clc - used to remove previous debug runs where printing was done.
clear all - good point. Edited to say clearvars.
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2020-12-30
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