Creating a table by extracting rows from a table based off their values in relation to another table of a different size.

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Low Kay Tong 2022-9-15

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https://ww2.mathworks.cn/matlabcentral/answers/1805860-creating-a-table-by-extracting-rows-from-a-table-based-off-their-values-in-relation-to-another-table

编辑： Low Kay Tong 2022-9-15

I have two tables of different size, for example, I have a 16x4 table with headers a,b,c and d and I have another 6x2 table with headers c and d. I have constructed both tables in matlab

a = [11;33;55;77;99;23;45;67;89;87;65;43;10;1;3;5];
b = [22;44;66;88;12;34;56;78;98;76;54;32;21;2;4;6];
c = [2;2;2;2;4;4;4;4;6;6;6;6;8;8;8;8];
d = [2;4;6;8;2;4;6;8;2;4;6;8;2;4;6;8];
T1 = table(a,b,c,d)
cc = [1.2;1.6;5.5;6.7;5.9;2.3];
dd = [2.5;4.6;3.5;3.2;1.5;1.9];
T2 = table(cc,dd)
T3 = T1(abs(T1.c-T2.cc)<=1 & abs(T1.d-T2.dd)<=1)

I wish to construct another table by filtering out rows based on values of abs(c - cc) <= 1 and abs(d - dd)<= 1 as the conditions and construct a new table using the extracted rows. looking like this in due to the order of the previous table t2:

.

But I couldn't do it as the number of variables are different. Any help will be appreciated!

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Answer 1

Stephen23 2022-9-15

在 MATLAB Online 中打开

a = [11;33;55;77;99;23;45;67;89;87;65;43;10;1;3;5];
b = [22;44;66;88;12;34;56;78;98;76;54;32;21;2;4;6];
c = [2;2;2;2;4;4;4;4;6;6;6;6;8;8;8;8];
d = [2;4;6;8;2;4;6;8;2;4;6;8;2;4;6;8];
T1 = table(a,b,c,d)
T1 = 16×4 table
    a     b     c    d
    __    __    _    _

    11    22    2    2
    33    44    2    4
    55    66    2    6
    77    88    2    8
    99    12    4    2
    23    34    4    4
    45    56    4    6
    67    78    4    8
    89    98    6    2
    87    76    6    4
    65    54    6    6
    43    32    6    8
    10    21    8    2
     1     2    8    4
     3     4    8    6
     5     6    8    8
cc = [1.2;1.6;5.5;6.7;5.9;2.3];
dd = [2.5;4.6;3.5;3.2;1.5;1.9];
T2 = table(cc,dd)
T2 = 6×2 table
    cc     dd 
    ___    ___

    1.2    2.5
    1.6    4.6
    5.5    3.5
    6.7    3.2
    5.9    1.5
    2.3    1.9
ixc = any((T1.c-reshape(T2.cc,1,1,[]))<=1,3);
ixd = any((T1.d-reshape(T2.dd,1,1,[]))<=1,3);
T3 = T1(ixc&ixd,:)
T3 = 6×4 table
    a     b     c    d
    __    __    _    _

    11    22    2    2
    33    44    2    4
    99    12    4    2
    23    34    4    4
    89    98    6    2
    87    76    6    4

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Low Kay Tong 2022-9-15

I am sorry, but your output is different from my intended output. is it due to the lack of the absolute sign? I'd appreciate it if you could explain the code!

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