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.
댓글 수: 3
Rik
2020년 12월 31일
편집: Rik
2020년 12월 31일
Help us help you: explain the pattern and we might be able to help you write the code to create it.
The pattern is not obvious to me (nor, apperently, to Cris). It is also missing from the OEIS, which is generally an indication that the sequence is fairly obscure.
답변 (1개)
Paul Hoffrichter
2020년 12월 31일
편집: Paul Hoffrichter
2021년 1월 1일
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
댓글 수: 2
Rik
2020년 12월 31일
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
2021년 1월 1일
편집: Paul Hoffrichter
2021년 1월 1일
clc - used to remove previous debug runs where printing was done.
clear all - good point. Edited to say clearvars.
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