using linspace in a matrix incrementing by a certain number

2023 1 19
2 个回答
更新时间：2023 1 19
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in a row of a matrix I need start with 51 and subtract 7.25 each time, 4 times, so the row would look like
51 43.75 36.5 29.25 22
I have
A=[2 4; 6 8]
B=[A(1,2:-1:1) A(1,2:-1:1) ; A(2,2:-1:1) A(2,2:-1:1);linspace(51,22,4)]
but the result is
A =
2 4
6 8
B =
4.0000 2.0000 4.0000 2.0000
8.0000 6.0000 8.0000 6.0000
51.0000 41.3333 31.6667 22.0000
and is incorrect

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回答（2 个）

Walter Roberson
Walter Roberson 2023-1-19

0 个投票

that would be a row of 5 values inside a matrix that has 4 columns not 5. That is not possible
Steven Lord
Steven Lord 2023-1-19
Ran in:

0 个投票

Ran in:
There are several functions for creating regularly spaced vectors of given lengths depending on which three of the four parameters (start point, end point, increment, and number of points) you know.
  • For start point, end point, increment use colon.
startPoint = 1;
endPoint = 10;
increment = 0.5;
numPoints = 19;
x1 = startPoint:increment:endPoint
x1 = 1×19
1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000 4.5000 5.0000 5.5000 6.0000 6.5000 7.0000 7.5000 8.0000 8.5000 9.0000 9.5000 10.0000
  • For start point, end point, number of points use linspace.
x2 = linspace(startPoint, endPoint, numPoints)
x2 = 1×19
1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000 4.5000 5.0000 5.5000 6.0000 6.5000 7.0000 7.5000 8.0000 8.5000 9.0000 9.5000 10.0000
  • For start point, increment, number of points use colon.
x3 = startPoint + increment*(0:numPoints-1)
x3 = 1×19
1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000 4.5000 5.0000 5.5000 6.0000 6.5000 7.0000 7.5000 8.0000 8.5000 9.0000 9.5000 10.0000
  • For end point, increment, number of points use colon. This one essentially flips the sign of the increment and the vector from the start point, increment, number of points case. Or you could negate the sign of the increment, use the (start point, increment, number of points) case, then flip the resulting vector.
x4 = endPoint + increment*((1-numPoints):0)
x4 = 1×19
1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000 4.5000 5.0000 5.5000 6.0000 6.5000 7.0000 7.5000 8.0000 8.5000 9.0000 9.5000 10.0000
startPoint2 = endPoint;
x5 = flip(startPoint2 + (-increment)*(0:numPoints-1))
x5 = 1×19
1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000 4.5000 5.0000 5.5000 6.0000 6.5000 7.0000 7.5000 8.0000 8.5000 9.0000 9.5000 10.0000
Are they all the same?
[isequal(x1, x2), isequal(x2, x3), isequal(x3, x4), isequal(x4, x5)]
ans = 1×4 logical array
1 1 1 1

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