How to get values ​​from a column based on the values ​​of other columns?

Hi,
I ordered a set of data as in the example below:
A B C D E F G J
,1985,04,10,1800,4,140,1, 2.0,
,1985,04,11,0900,4,999,1, 0.0,
,1985,07,09,0300,4,999,1, 0.0,
,1985,04,11,2100,4,020,1, 5.0,
,1987,03,03,1000,4,360,1, 10.0,
,1987,03,03,1100,4,360,1, 10.0,
,1987,01,03,0700,4,360,1, 8.0,
,1987,01,03,0900,4,360,1, 7.0,
,1987,03,03,1200,4,360,1, 10.0,
,1987,03,03,1300,4,360,1, 10.0,
,1985,07,09,1500,4,180,1, 10.0,
,1985,07,09,1800,4,180,1, 11.0,
,1985,07,10,0600,4,160,1, 3.0,
,1985,07,10,1000,4,180,1, 4.0,
,1985,04,11,1500,4,360,1, 9.0,
,1985,04,11,2000,4,050,1, 5.0,
,1985,07,10,1200,4,180,1, 5.0,
,1985,07,10,1500,4,180,1, 6.0,
,1987,01,03,0600,4,360,1, 8.0,
,1987,01,03,1100,4,360,1, 6.0,
,1985,07,09,0600,4,999,1, 0.0,
,1985,07,09,1200,4,180,1, 6.0,
,1987,01,03,1300,4,360,1, 6.0,
,1987,01,03,1600,4,360,1, 7.0,
,1987,01,03,1900,4,360,1, 6.0,
,1987,03,03,0500,4,020,1, 8.0,
,1987,03,03,0900,4,360,1, 10.0,
,1987,01,03,1400,4,320,1, 5.0,
,1987,03,03,1500,4,360,1, 10.0,
I want to create an array with the values in column "J" based on the values in column "B". . For example the vector Z should contain only the values of "J" that are on the same row with the value "01" in column "B", vector X only the "03" and so on ...
Thanks in advanace for the help and suggestions.

 Akzeptierte Antwort

Are A...J cell arrays with strings or numeric arrays?
If you have numbers, this should do the job:
Z=J(B==1);
If it's cell arrays with strings, you need:
Z=J(strcmp(B,'01'));
Best regards,
Michael

2 Kommentare

Thank You, I tried but I dont get the correct values... I'l try some more
In the end the problem was with the dataset now I have the correct answer. Thanks for the help.
As a side note I tried the other answers once I realized the error and are right now... Thank you all

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Weitere Antworten (1)

Joseph Cheng
Joseph Cheng am 27 Aug. 2014
Bearbeitet: Joseph Cheng am 27 Aug. 2014
you can use == to find the rows for each B value. here is an example of an implementation
uB = unique(MatAnsExample(:,2));
for ind = 1:length(uB)
Result(ind).B = uB(ind);
Result(ind).J = MatAnsExample(MatAnsExample(:,2)==uB(ind),8);
end
here i find each unique instance of values within column B. then with the for loop i store the current B in the Result structure as well as the values of J that correspond with that B. to do this i generate the logical mask of when column B is equal to the B in question.

4 Kommentare

Nym
Nym am 27 Aug. 2014
Bearbeitet: Nym am 27 Aug. 2014
Seems that i failed to make myself understood, sorry. I need a new vector with J values only that are on the same row with a specific value from B, not entire row. In the example above:
Z = [8.0 7.0 8.0 6.0 6.0 7.0 6.0 5.0]
Thank you for support.
Joseph's answer is correct and creates all the possible vectors out of your matrix. If you just want one vector, you can simplify it to:
Z = m(m(:, 2) == 1), 8);
Thank you but I fail to adapt to my needs suggested code, my knowledge of MATLAB is poor.
so if you run my code lets say with an adaptation
uB = unique(MatAnsExample(:,2));
for ind = 1:length(uB)
Result(ind).B = uB(ind);
Result(ind).J = MatAnsExample(MatAnsExample(:,2)==uB(ind),8);
disp(['For a B of: ' num2str(uB(ind)) ' the J column are: ' num2str(Result(ind).J')]);
end
you get the results
For a B of: 1 the J column are: 8 7 8 6 6 7 6 5
For a B of: 3 the J column are: 10 10 10 10 8 10 10
For a B of: 4 the J column are: 2 0 5 9 5
For a B of: 7 the J column are: 0 10 11 3 4 5 6 0 6
you get the examples you just wrote. so instead of using individual variables Z, X, Y, etc. for each value of B. I am storing it as Result(N) where N from 1 to the number of unique instances within the B column. So to get the Z, X, etc. variable you can simply just type in Result(1).J instead. I also just put it in this form as i don't know what you are doing. The Right side of the equation is what you should be interested in and not the left side. The left side is for you to determine.

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Gefragt:

Nym
am 27 Aug. 2014

Kommentiert:

Nym
am 29 Aug. 2014

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