Compensate the vector with the last entry

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mingcheng nie
mingcheng nie 2023년 4월 25일
답변: Steven Lord 2024년 7월 31일
I have a length L vector contains some numbers, I want to compensate this vector to length K, where K > L, with repeating the last entry of the vector. For example, the vector is [2 4 7 3], after compensate, it will be [2 4 7 3 3 3 3 3]. I hope there is an efficient way to do so because I actually have more than 10^4 vectors to compensate.
Thanks,
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Stephen23
Stephen23 2023년 4월 25일
"I hope there is an efficient way to do so because I actually have more than 10^4 vectors to compensate."
Do you really have 1e4 separate vectors stored in the workspace? How did you get them all there?
mingcheng nie
mingcheng nie 2024년 7월 31일
sorry for the ambiguity. I have a loop around 10^4 times, where within each loop I will need compensate the vector :)

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채택된 답변

Stephen23
Stephen23 2023년 4월 25일
V = [2,4,7,3];
K = 8;
V(end+1:K) = V(end)
V = 1×8
2 4 7 3 3 3 3 3

추가 답변 (1개)

Steven Lord
Steven Lord 2024년 7월 31일
If you were using release R2023b or later, you could use the paddata function with the Side name-value argument and either the FillValue name-value argument or the Pattern name-value argument with the 'edge' pattern.
x = [2 4 7 3]
x = 1x4
2 4 7 3
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
y = paddata(x, 7, Side = 'trailing', Pattern = 'edge')
y = 1x7
2 4 7 3 3 3 3
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Or to show FillValue with a different value:
y = paddata(x, 7, Side = 'trailing', FillValue = -999)
y = 1x7
2 4 7 3 -999 -999 -999
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

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