I can't understand what my mistakes are

[Manikanta Aditya]

Can you share the entire code file?

[Alexandre Da Silva]

Changin the () to {} worked but now it's complaining about the multiplications

Answer 1

[Star Strider]

Using parentheses () to index into a table indexes the table itself rather than the elements of the table. To index the elements, use curly brackets {} instead. See the documentation section on Access Data in Tables for details.

You can also use the prod function to make this easier —

t = array2table(rand(100, 5))

t = 100x5 table
      Var1         Var2        Var3       Var4         Var5   
    _________    ________    ________    _______    __________

      0.36021     0.78416     0.19064    0.77828       0.33928
      0.86373     0.75798     0.06804    0.70605       0.70315
      0.13617     0.11643     0.71102    0.38285       0.69832
      0.72425     0.28794     0.34274    0.93101       0.28691
      0.91087     0.68707     0.10527    0.20181      0.018597
      0.25297      0.6527     0.12287    0.56841       0.35464
      0.76311      0.6136     0.15964    0.61596       0.95528
      0.69494     0.23664     0.67301    0.44096       0.82481
    0.0093195     0.74846     0.14387    0.31925       0.41236
      0.61274     0.42674     0.32932    0.10132       0.60915
      0.34004     0.71392    0.078349     0.7659        0.5931
     0.073266     0.33247      0.9104    0.34555       0.72223
      0.48368    0.059958     0.75113     0.6843      0.069615
      0.59842     0.37785     0.86768    0.97629       0.88959
      0.46492    0.063566     0.59707    0.97527    3.9778e-05
       0.8089    0.052195     0.85856    0.77135       0.24753

h = linspace(1, 100, 100);

l1 = prod(h-t{2:5,1}) ./ prod(t{1,1}-t{2:5,1})

l1 = 1×100

1.0e+09 * -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0001 -0.0001 -0.0002 -0.0003 -0.0005 -0.0007 -0.0010 -0.0014 -0.0019 -0.0024 -0.0032 -0.0040 -0.0050 -0.0062 -0.0076 -0.0092 -0.0110 -0.0131 -0.0155 -0.0182 -0.0213 -0.0247 -0.0285 -0.0328

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l1_original = ((h-t{2,1}).*(h-t{3,1}).*(h-t{4,1}).*(h-t{5,1})) ./ ((t{1,1}-t{2,1}).*(t{1,1}-t{3,1}).*(t{1,1}-t{4,1}).*(t{1,1}-t{5,1}))

l1_original = 1×100

1.0e+09 * -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0001 -0.0001 -0.0002 -0.0003 -0.0005 -0.0007 -0.0010 -0.0014 -0.0019 -0.0024 -0.0032 -0.0040 -0.0050 -0.0062 -0.0076 -0.0092 -0.0110 -0.0131 -0.0155 -0.0182 -0.0213 -0.0247 -0.0285 -0.0328

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Check = all(l1 == l1_original)                      % ‘1’ (true) If All Elements Are Equal

Check = logical
   1

Checking that this version of ‘l1’ produces the same result as your hard-coded version. (It should, and it does here.)

EDIT — (27 Dec 2024 at 16:25)

EDIT — (27 Dec 2024 at 16:31)

Now having the code and data —

%1 without overcomplicating
t = readtable('TabelaValores1');
h = linspace(1,100,100);

%em comum

l1=(((h-t{2,1}).*(h-t{3,1}).*(h-t{4,1}).*(h-t{5,1})) ...
    ./((t{1,1}-t{2,1}).*(t{1,1}-t{3,1}).*(t{1,1}-t{4,1}).*t{1,1}-t{5,1}));

l2=(((h-t{1,1}).*(h-t{3,1}).*(h-t{4,1}).*(h-t{5,1})) ...
    ./((t{2,1}-t{1,1}).*(t{2,1}-t{3,1}).*(t{1,1}-t{4,1}).*t{1,1}-t{5,1}));

l3=(((h-t{1,1}).*(h-t{2,1}).*(h-t{4,1}).*(h-t{5,1})) ...
    ./((t{3,1}-t{1,1}).*(t{3,1}-t{2,1}).*(t{3,1}-t{4,1}).*t{3,1}-t{5,1}));

l4=(((h-t{1,1}).*(h-t{2,1}).*(h-t{3,1}).*(h-t{5,1})) ...
    ./((t{4,1}-t{1,1}).*(t{4,1}-t{2,1}).*(t{4,1}-t{3,1}).*t{4,1}-t{5,1}));

l5=(((h-t{1,1}).*(h-t{2,1}).*(h-t{3,1}).*(h-t{4,1})) ...
    ./((t{5,1}-t{1,1}).*(t{5,1}-t{2,1}).*(t{5,1}-t{3,1}).*t{5,1}-t{4,1}));


%para o x

x1=t{1,2};
x2=t{2,2};
x3=t{3,2};
x4=t{4,2};
x5=t{5,2};

interpolx=(x1*l1+x2*l2+x3*l3+x4*l4+x5*l5);

%para o y

y1=t{1,3};
y2=t{2,3};
y3=t{3,3};
y4=t{4,3};
y5=t{5,3};

interpoly=(y1*l1+y2*l2+y3*l3+y4*l4+y5*l5);

figure
plot(interpolx, interpoly)
grid

.

This works after replacing the appropriate parentheses with curly brackets, and using element-wise multiplication.

.

[Alexandre Da Silva]

Thank you so much! You helped a lot!!

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.

Answer 2

[Image Analyst]

Looks like line 6 is really not empty. Maybe scroll to the right to see if there is a - on that line.

Also t(2,1) I think is a table, not the contents of the table at that location. Try using {} instead of () for all your t references, like t{2,1}.

If you have any more questions, then attach your files (Time.m and TabelaValores1.xlsx) and code to read it in with the paperclip icon after you read this:

TUTORIAL: How to ask a question (on Answers) and get a fast answer

[Alexandre Da Silva]

Thank you!! Switching the () for {} worked for the subtraction but now it's complaining about the multiplications even after removing the . from .*

this are the files:

[Image Analyst]

I'm glad my suggestion of using braces instead of parentheses got you further. You don't want to remove the . from .* because they are multi-element arrays, not scalars. So I put them back in and replaced the remaining parentheses with braces and it works fine. It usually doesn't hurt if you use .* instead of * but you need to make sure you want element-by-element multiplication, not matrix multiplication, and the two arrays are compatible sizes.

%1 without overcomplicating
t = readtable('TabelaValores1');
h = linspace(1,100,100);
%em comum
l1=(((h-t{2,1}).*(h-t{3,1}).*(h-t{4,1}).*(h-t{5,1})) ...
    ./((t{1,1}-t{2,1}).*(t{1,1}-t{3,1}).*(t{1,1}-t{4,1}).*t{1,1}-t{5,1}));
l2=(((h-t{1,1}).*(h-t{3,1}).*(h-t{4,1}).*(h-t{5,1})) ...
    ./((t{2,1}-t{1,1}).*(t{2,1}-t{3,1}).*(t{1,1}-t{4,1}).*t{1,1}-t{5,1}));
l3=(((h-t{1,1}).*(h-t{2,1}).*(h-t{4,1}).*(h-t{5,1})) ...
    ./((t{3,1}-t{1,1}).*(t{3,1}-t{2,1}).*(t{3,1}-t{4,1}).*t{3,1}-t{5,1}));
l4=(((h-t{1,1}).*(h-t{2,1}).*(h-t{3,1}).*(h-t{5,1})) ...
    ./((t{4,1}-t{1,1}).*(t{4,1}-t{2,1}).*(t{4,1}-t{3,1}).*t{4,1}-t{5,1}));
l5=(((h-t{1,1}).*(h-t{2,1}).*(h-t{3,1}).*(h-t{4,1})) ...
    ./((t{5,1}-t{1,1}).*(t{5,1}-t{2,1}).*(t{5,1}-t{3,1}).*t{5,1}-t{4,1}));
%para o x
x1=t{1,2};
x2=t{2,2};
x3=t{3,2};
x4=t{4,2};
x5=t{5,2};
interpolx=(x1*l1+x2*l2+x3*l3+x4*l4+x5*l5);
%para o y
y1=t{1,3};
y2=t{2,3};
y3=t{3,3};
y4=t{4,3};
y5=t{5,3};
interpoly=(y1*l1+y2*l2+y3*l3+y4*l4+y5*l5);
fprintf('DONE!  It worked (or at least it finished)\n');

DONE!  It worked (or at least it finished)

See this link for help in understanding when to use braces, parentheses, or brackets:

What is a cell array

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