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 commentaires
Rik
le 31 Déc 2020
Modifié(e) : Rik
le 31 Déc 2020
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.
Réponses (1)
Paul Hoffrichter
le 31 Déc 2020
Modifié(e) : Paul Hoffrichter
le 1 Jan 2021
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 commentaires
Rik
le 31 Déc 2020
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
le 1 Jan 2021
Modifié(e) : Paul Hoffrichter
le 1 Jan 2021
clc - used to remove previous debug runs where printing was done.
clear all - good point. Edited to say clearvars.
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