How to use an array as an indexing table to add search result as a new row to the querying array ?

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Tamura Kentai 2019-7-24

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评论： Tamura Kentai 2019-7-25

采纳的回答： Andrei Bobrov

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Hi, :

I thank you at first.

My question is , 'How to use an array as an indexing table to add search result as a new row to the querying array ?'

For example, the Q matrix is 1000 x 1, the Idx matrix is 5 x 3, as below:

Then use Q( : ) every cell to search Idx by the logical expression, " if Idx(1 , 1 , : ) < Q(:,1) < Idx(1, 2 , :) "

,to search which row will the Q(:) be located, then return the columns 3 of the Idx(,,3) respectively to store every it in Q by added a new column 2.

For example Q(1) = 3.5, then Idx(3, 1) = 3, Idx(3, 2) = 4, so that " Idx(3, 1) < Q(1) < Idx(3, 2)", so the Q(1) should have the returned result from Idx's column3 = 30, then store this '30', to a new column2 to Q(1,2) = '30'

Is it possible to avoid using loop, or any existing similar function to simplify it ?

Thank you very much.

Best regards.

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采纳的回答

Andrei Bobrov 2019-7-24

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编辑：Andrei Bobrov 2019-7-24

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Q_new = [Q,Idx(discretize(Q,[Idx(:,1);Idx(end,2)]),3)];

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Tamura Kentai 2019-7-24

Hi, Andrei:

Thank you very much.

This is exactly what I want. I appreciate your help.

Best regards.

Tamura Kentai 2019-7-24

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Hi, Andrei:

I beg your pardon.

I discovered it after I tried some successfully.

Is it possible, to get an interpolating value between 2 index ?

For example, if the indexing value of Q(1) is 3.5, which is the middle value of the row4 & row5, then return the value (40 -30) * ( (3.5 -3) / (4-3) ) + 30 = 35 .

Thank you very much.

Best regards.

>> [Inew(:,1) Inew(:,3)]
ans =
  -Inf   NaN
     1    10
     2    20
     3    30
     4    40
     5    50
     6   NaN

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Tamura Kentai 2019-7-24

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Hi, Andrei:

I'm sorry, after some hours, I think I have one solution. It's the following:

% index_array_search.m
load('Idx.mat');
load('Q.mat');
Idx = [(1:5)',(2:6)',10*(1:5)'];
Inew = [-inf,1,nan;Idx;6,inf,nan];
[~,ii] = histc(Q,[Inew(:,1);Inew(end,2)]);
out = [Q,Inew(ii,3),Inew(ii+1,3), Inew(ii,1), Inew(ii+1, 1)];
out = [out,( ( out(:,1)-out(:,4) ).*( out(:,3) - out(:,2) ) ./ (out(:,5)-out(:,4)) + out(:,2) )];

I add those necessary for computing into the array.

It looks pretty fine.

Thank you for the ideas.

Best regards.

>> index_array_search
>> out
out =
    1.1000   10.0000   20.0000    1.0000    2.0000   11.0000
    2.2000   20.0000   30.0000    2.0000    3.0000   22.0000
    3.3000   30.0000   40.0000    3.0000    4.0000   33.0000
    4.4000   40.0000   50.0000    4.0000    5.0000   44.0000
    5.5000   50.0000       NaN    5.0000    6.0000       NaN

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Tamura Kentai 2019-7-25

Hi, Andrei:

Thank you for the suggestions. These are very helpful.

Best regards.

Tamura Kentai 2019-7-25

Hi, Andrei:

In fact, I am thinking if there are other options other than 'linear', actually I mean it's 'linear' distribution in log-log plane.

For example, if plot the 'y=-x' using 'loglog(x,y)', the line will be straight, but if add some constant offset, eg: 'y=-x + 500', the plot 'loglog(x,y)', it won't be straight line in log-log plane.

If you think it's interesting. In fact I have tried survey ,but no good result except using manually adjusting such math equations.

Thank you very much.

Best regards.

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