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Mathematics Wolfram Language Numerical Computation

Calculate the outer product of the third layer's of two lists?

Shaohui Yin

Posted 5 years ago

When there is only one layer in the lists: x1 = Range[-1., 1., 0.5]; y1 = Range[1., -1., -0.5]; Sqrt@N@Outer[Plus, x1^2, y1^2] // MatrixForm The result is a {5,5} dimensions list. If the original lists are changed into two {5,5,3} dimensions lists, how to calculate the most inner layer's outer product? xx1 = Table[x1, 5]; yy1 = Table[y1, 5]; kx={-0.005208333333333333, 0.10^-18, 0.00520833333333333}; ky={0.005208333333333333, 0.10^-18, -0.00520833333333333}; dx[x_] := x + kx; SetAttributes[dx, Listable]; dy[y_] := y + ky; SetAttributes[dy, Listable]; Sqrt@N@Outer[Plus, dx[xx1]^2, Transpose@dy[yy1]^2, 3] // MatrixForm The corresponding result should be a {5,5,3,3} dimensions list, but it is a {5, 5, 3, 5, 5, 3} dimensions list.

When there is only one layer in the lists:

    x1 = Range[-1., 1., 0.5];
    y1 = Range[1., -1., -0.5];
    Sqrt@N@Outer[Plus, x1^2, y1^2] // MatrixForm

The result is a {5,5} dimensions list. If the original lists are changed into two {5,5,3} dimensions lists, how to calculate the most inner layer's outer product?

    xx1 = Table[x1, 5];
    yy1 = Table[y1, 5];
    kx={-0.005208333333333333, 0.*10^-18, 0.00520833333333333};
    ky={0.005208333333333333, 0.*10^-18, -0.00520833333333333};
    dx[x_] := x + kx;
    SetAttributes[dx, Listable];
    dy[y_] := y + ky;
    SetAttributes[dy, Listable];
    Sqrt@N@Outer[Plus, dx[xx1]^2, Transpose@dy[yy1]^2, 3] // MatrixForm

The corresponding result should be a {5,5,3,3} dimensions list, but it is a {5, 5, 3, 5, 5, 3} dimensions list.

POSTED BY: Shaohui Yin

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Estevao Teixeira

Posted 5 years ago

Hey Shaohui, how are you? Do you want to do basically that? you could do that Outer[Plus, dx[xx1][[1, 1]]^2, dy[yy1][[1, 1]]^2, 3] Result: you will have to run over any i and j in dx and dy. Is it what you wanted? Bye.

POSTED BY: Estevao Teixeira

Shaohui Yin

Posted 5 years ago

Hi, Estevao. I have worked it out. Anyway, thank you. Here is my code. k=3; kTimes[x_] := Table[x, k]; SetAttributes[kTimes, Listable]; xx2 = kTimes[xx1]; yy2 = kTimes[yy1]; xx3 = dx[xx22]; yy3 =Transpose[dy[yy2], 3 <-> 4]; Sqrt@N@Plus[xx3^2, yy3^2]; In this way, I do not need to use Part, which would be faster.

POSTED BY: Shaohui Yin

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