Performing computation using contents stored in a table
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Hi,
I've the following table
t = [1 2 3 4 5 6 7 8 9]';
h = [2 3 4 5 6 7 8 9 10]';
value = [1 2 1 2 1 2 1 2 1]';
tbl = table(t,h,value);
tbl = mergevars(tbl,[1 2]);
I'm filtering the contents of the table using a search value. The contents present in the resulting table is used to perform some operations.
searchvalues = 1:10;
tic
for i = 1:length(searchvalues)
searchval = searchvalues(i);
newtable = tbl(any(tbl.Var1 == searchval, 2), :)
tochange = newtable.Var1(:, 2) == searchval;
newtable.Var1(tochange,:) = fliplr(newtable.Var1(tochange,:));
result(i) = sum((newtable.Var1(:,1) - newtable.Var1(:,2)).*newtable.value);
end
toc
result
I'd like to ask for suggestions on how to speed up the steps carried out in for loop.
I tried the following,
searchvalues = 1:10;
tic
i = searchvalues;
newtable = tbl(any(tbl.Var1 == i, 2), :)
tochange = newtable.Var1(:, 2) == i;
newtable.Var1(tochange,:) = fliplr(newtable.Var1(tochange,:));
result(i) = sum((newtable.Var1(:,1) - newtable.Var1(:,2)).*newtable.value);
end
toc
result
But, this didn't work.
EDIT: what happens in the for loop?
The following contents are stored in a table, in variable 'tbl',
Var1 value
______________
1 2 1
2 3 2
3 4 1
4 5 2
I would like to do the following,
i = 2,
If the variable i is present in tbl.Var1, I would like to retain only those rows and delete the remining rows.
Example,
Var1 value
______________
1 2 1
2 3 2
Also, if i is present in column 2 of Multico, the value in column 2 of tbl.Var1 , the columns are flipped
Th expected output is,
Var1 value
______________
2 1 1
2 3 2
Then, I substract tbl(:,1) - tbl(:,2) , this results in a column vector. Sum of the dot product of this column vector and tbl.value is obtained, stored in variabl result.
I repeat the same for different values of i.
4 件のコメント
darova
2020 年 2 月 25 日
Yes, it's clear now. Thank you
Do you always operate on numbers? Why do you need table?
回答 (1 件)
darova
2020 年 2 月 26 日
Version without tables
var11 = [1 2 3 4 5 6 7 8 9]';
var12 = [2 3 4 5 6 7 8 9 10]';
value = [1 2 1 2 1 2 1 2 1]';
i = 2;
ind = (var11==i) | (var12==i);
% (without replacing columns) tbl(:,1) - tbl(:,2) the same as
result = sum( (2*i-var11-var12).*value.*ind );
2 件のコメント
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