How would I replace a string of numbers with more numbers using for and if statements?
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Here is my code so far. I'm trying to make it so that if the number is "1" in the string, it replaces it with five zeros. If the number is "2" in the string it replaces that with 5 ones, and if the number is "3" in the string, it replaces it with 5 twos and make that all a different string of numbers called d. The end should look like this: d = [0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,2,2,2,2,2]. I really need help, I am awful with matlab.
if true
t = [1,2,1,2,1,3];
% For Loop
for n = 1:length(t)
if t(n) == 1
d(n) = 0,0,0,0,0;
elseif t(n) == 2
d(n) = 1,1,1,1,1;
else t(n) == 3
d = 2,2,2,2,2;
end
end
end
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Réponse acceptée
madhan ravi
le 3 Nov 2018
Modifié(e) : madhan ravi
le 3 Nov 2018
t = string([1,2,1,2,1,3]);
d=cell(1,numel(t)); %pre-allocation
for i = 1:numel(t)
if t(i)=='1'
d{i}=[ones(1,5)*0];
elseif t(i)=='2'
d{i}=[ones(1,5)*1];
elseif t(i)=='3'
d{i}=[ones(1,5)*2];
end
end
d = horzcat(d{:})
command window:
>> COMMUNITY
d =
Columns 1 through 13
0 0 0 0 0 1 1 1 1 1 0 0 0
Columns 14 through 26
0 0 1 1 1 1 1 0 0 0 0 0 2
Columns 27 through 30
2 2 2 2
>>
2 commentaires
madhan ravi
le 3 Nov 2018
1)just created 1 by 5 cell array;
2) made a loop until the number of cell array;
3)made if and else if conditions to store elements in each cell when the condition is satisfied;
4)finally concatenated them horizontally.
What's the advantage?
Cells are huge containers they can compress how much ever you feed inside them unlike the normal arrays
Plus de réponses (1)
Tyler Johns
le 3 Nov 2018
Keep in mind when programming there is usually more than one way to accomplish your goal. Here is an additional solution. This may not be the most efficient way since the size of vector d changes every loop iteration but it is still a solution nonetheless.
t = [1 2 1 2 1 3];
d=[]; % create empty vector d
for n = 1:length(t)
if t(n)==1
d = [d 0 0 0 0 0]; %add new numbers to end of vector d
elseif t(n)==2
d = [d 1 1 1 1 1];
elseif t(n)==3
d = [d 2 2 2 2 2];
end
end
disp(d)
4 commentaires
Stephen23
le 3 Nov 2018
Follow madhan ravi's answer: it shows good practice: a preallocated cell array to store the intermediate arrays.
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