Algebra function with table and vector

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Ana Castanheiro
Ana Castanheiro 2016-10-3
编辑： Jan Orwat 2016-10-3
Hi guys! I have 1 table (Composition) and 1 vector (S), and I want to compute the following function: Composition(i,j)=Composition(i,j)./S(i)*100.
For this, I tried to use a loop (a and b are the size of Composition and S, respectively):
i = 1;
j = 1;
for i = i:a
for j = j:b
Composition(i,j)= Composition(i,j)./S(i)*100;
j = j+1;
end
i = i+1;
end
I get the following error: >> import_excel Undefined operator './' for input arguments of type 'table'.
Error in import_excel (line 73) Composition(i,j)= Composition(i,j)./S(i)*100;
I'm at begginer level, so any help is most appreciated! Thanks!
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  2 个评论
KSSV
KSSV 2016-10-3
How you have read the excel data into MATLAB?
Ana Castanheiro
Ana Castanheiro 2016-10-3
I'm not sure if I understand your question. I import the excel data using this:
[FileIn,PathIn] = uigetfile('C:\Users\location\*.xlsx',...
'Multiselect','off');
FilePath = strcat(PathIn,FileIn);

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采纳的回答

Guillaume
Guillaume 2016-10-3
First, some points not really related to your question:
1.
i = 1;
for i = i:a
While the above will work and do what you want, using the same variable name for the indexing variable and for the start value is really not good practice. Maybe:
istart = 1;
for i = istart:a
but really
for i = 1:a
is simpler.
2.
for i = 1:n
%... do something
i = i+1;
end
The i = i+1 is completely unnecessary as the for loop will increase i anyway (and you should actually have a warning in the editor). When matlab reaches the end of the loop, it completely ignores any change you've made to the indexing variable and proceeds with the next indexing value regardless.
Please read the documentation about for loops, particularly the examples.
Now, to answer your question. The proper way to access the elements of a table is with {} indexing, not () which returns a smaller table. So, if you do want to use a loop:
for row = 1:height(Composition) %probably the safest way to write the loop
for col = 1:width(Composition)
Composition{row, col} = Composition{row, col} ./ S(row) * 100; %Assuming S is NOT a table
end
end
But note that you actually don't need the loop. If using matlab before R2016b:
Composition{:, :} = bsxfun(@rdivide, Composition{:, :}, S(:)) * 100;
And if using R2016b or later, simply:
Composition{:, :} = Composition{:, :} ./ S(:) * 100;

更多回答（2 个）

Jan Orwat
Jan Orwat 2016-10-3
编辑：Jan Orwat 2016-10-3
If you want to work on tables in MATLAB functions varfun and rowfun come handy.
See following example:
>> a = [1;2;3;4;5;6];
>> b = a+1;
>> c = a*0;
>> T = table(a,b,c)
T =
a b c
_ _ _
1 2 0
2 3 0
3 4 0
4 5 0
5 6 0
6 7 0
>> bsxfun(@plus,T,a)
Undefined function 'bsxfun' for input arguments of type 'table'.
>> varfun(@(col)plus(a,col),T)
ans =
Fun_a Fun_b Fun_c
_____ _____ _____
2 3 1
4 5 2
6 7 3
8 9 4
10 11 5
12 13 6
For your problem, it would look like:
varnames = Composition.Properties.VariableNames; % Store original variable names.
Composition = varfun(@(column)column./S*100, Composition);
Composition.Properties.VariableNames = varnames; % Restore original variable names.
KSSV
KSSV 2016-10-3
doc xlsread......
Read the data of the excel file using xlsread...then the data will be stored as numbers. You can do all the algebraic operations.
  2 个评论
Ana Castanheiro
Ana Castanheiro 2016-10-3
I have something like that:
[title,myheader] = xlsread(FilePath, 'Sheet1','A1:ZZ1');
Guillaume
Guillaume 2016-10-3
That is really not a good answer!
The data is stored as numbers, just as a table. The issue has nothing to do with the way the data is read. It's a simple matter of indexing a table.
Since this is linked to the previous question, I would add that rather than using xlsread followed by celltotable, it would be simpler to read the file with:
Composition = readtable(FilePath, 'Sheet1', 'A1:ZZ1');
which will create the table in one go, and use row 1 as the table column name.
readtable is a lot more powerful and flexible than xlsread, so there's really no reason to use xlsread anymore.

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