Command and Control vectors
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I am having trouble trying to make this work. I want it so that whenever a value in t is 1, it will output five zeros onto a row vector. Whenever the value of t is 2, I want it to output 5 ones to the same row vector. When the value of t is 3, I want it to output 5 twos onto the row vector. I want it so that my ending vector will be [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]. Right now, it is only showing [0 1 0 1 0 2], so basically it is only outputting the value that I want once. How can I make it output the same number that I want 5 times instead of just once?
t = [1, 2, 1, 2, 1, 3];
d = 0;
for n = 1:length(t)
if (t(n) == 1)
for k = 0:4
k = k+1;
d(n) = 0;
end
elseif (t(n) == 2)
for k = 0:4
k = k+1;
d(n) = 1;
end
elseif (t(n) == 3)
for k = 0:4
k = k+1;
d(n) = 2;
end
end
end
display(d)
采纳的回答
Guillaume
2018-11-23
A 'correct' version of your code, with all the pointless loops could be:
t = [1, 2, 1, 2, 1, 3];
d = zeros(1, numel(t)*5); %at least predeclar d with the correct size.
idx = 1; %to keep track of where we are in d
for n = 1 : numel(t)
if t(n) == 1 %note that using switch ... case statements would be more elegant than if.. elseif...
for k = 1:5
d(idx) = 0;
idx = idx + 1;
end
elseif t(n) == 2
for k = 1:5
d(idx) = 1;
idx = idx + 1;
end
elseif ((n) == 3
for k = 1:5
d(idx) = 2;
idx = idx + 1;
end
end
end
display(d)
Now of course, the k loops are completely pointless, so we could simplify that to:
t = [1, 2, 1, 2, 1, 3];
d = zeros(1, numel(t)*5); %at least predeclar d with the correct size.
idx = 1; %to keep track of where we are in d
for n = 1 : numel(t)
if t(n) == 1 %note that using switch ... case statements would be more elegant than if.. elseif...
d(idx:idx+4) = 0;
idx = idx + 5;
elseif t(n) == 2
d(idx:idx+4) = 1;
idx = idx + 5;
elseif ((n) == 3
d(idx:idx+4) = 2;
idx = idx + 5;
end
end
display(d)
Now of course, we're just repeating the same code in each of the if with the only difference being the value assigned, so they're also completely pointless:
t = [1, 2, 1, 2, 1, 3];
d = zeros(1, numel(t)*5); %at least predeclar d with the correct size.
idx = 1; %to keep track of where we are in d
for n = 1 : numel(t)
d(idx:idx+4) = t(n)-1;
idx = idx + 5;
end
display(d)
An of course, even the for loop is pointless. As per Stephen's answer you can use kron, or you can use repelem:
t = [1, 2, 1, 2, 1, 3];
d = repelem(t-1, 5)
更多回答（1 个）
Stephen23
2018-11-23
0 个投票
>> t = [1,2,1,2,1,3];
>> kron(t-1,ones(1,5))
ans =
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
>>
4 个评论
Chase Lee
2018-11-23
Walter Roberson
2018-11-23
write the same value to 5 rows and one column of aa matrix . reshape it to aa row vector afterwards
Is there any other way besides the use of kron or repelem? I would prefer just relying on a for loop within the if statement.
Yes, sure, why use one line of code when you can use 20 lines of code? You could waste even more time by adding even more lines of code to do the same thing but why?
Even ignoring the use of loops and ifs there's already plenty of pointless lines in your code, for example:
for k = 0:4
k = k + 1;
%...
end
is simply a more complicated version of
for k = 1:5
%...
end
Walter Roberson
2018-11-23
d(n*5-4:n*5) = value
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