In a table, when I try assigning a value to a new column based on some criteria, I get error that "assignment to elements using simple assignment statement is not supported"
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I have a table for which I create an index where the values in a particular column are not NaN. I then try to use this index to create an entry into a new array and replace the values at the corresponding index (preloaded with '0000') with '0001'. I will then add this array to the table as a new column.
My intention is to fill the corresponding indx in the column with '0001', if the value in the newTable.PEM1_GA_Current_Estimate_Error a valid number (not NaN).
PEM1b ='0001';
PEM2b ='0010';
PEM3b = '0100';
PEM4 = '1000';
temp={};
indx=(find(newTable.PEM1_GA_Current_Estimate_Error ~= NaN))
temp ={repmat('0000',size(newTable,1),1)}
temp{indx} = PEM1b
%...and add this array as a new column to the table
the last assigment in the code above generates this error:
Assigning to 2609 elements using a simple assignment statement is not supported. Consider using comma-separated list assignment.
I don't think I should need a for loop to iterate through each row and replace the value at the "indx" location --what am I doing wrong?
I included the data....thank you.
3 commentaires
Don't use == or ~= to compare to NaN, because NaN is not equal to NaN.
Consider this vector x:
x = [1 NaN 3 NaN 5];
Checking that each element of x is not equal to (~=) NaN indicates that all elements of x are non-NaN, which is obviously not the case.
x ~= NaN
That happens because NaNs are considered not equal to each other (so NaN ~= NaN is true):
NaN ~= NaN
Instead, use the isnan() function to determine whether each element of an array is NaN, as in:
~isnan(x)
newTable = load('data.mat').newTable
PEM1b = '0001';
indx = ~isnan(newTable.PEM1_GA_Current_Estimate_Error);
temp = repmat({'0000'},size(newTable,1),1);
temp(indx) = {PEM1b}
%...and add this array as a new column to the table
newTable.newColumn = temp
Jorge
le 16 Sep 2024
Réponse acceptée
"I don't think I should need a for loop to iterate through each row and replace the value at the "indx" location"
You do not need a loop, just assign a scalar cell array to the existing cell array using parentheses:
temp(indx) = {PEM1b};
"what am I doing wrong?"
You are trying to assign one array to multiple cells of the cell array. You could think of it something like this:
[a,b,c,d,e,f,..] = pi
While this might be a defined syntax in some languages, with MATLAB this does not assign one array to lots of cells (or variables or fields or ...). In general it is an error. You could use DEAL for some of those situations, but your usecase is actually much simpler: you can use the special case that applies for any array type: a scalar on the RHS will always be expanded to fit the indexing on the LHS into an array of the same type. So all you need is a scalar cell array on the RHS. Which means that the LHS must use parentheses to assign to the cell array itself, not curly braces to replace its content. Then you assign a scalar array element on the RHS to as many elements on the LHS as the indexing specifies (note how I did not write cell, because this concept is exactly the same for every array class).
8 commentaires
C = num2cell('a':'i')
C([2,8]) = {'xxx'} % assign scalar to array
[C{[3,7]}] = deal('yyy') % comma-separated list
C{[4,6]} = 'zzz' % error
Jorge
le 16 Sep 2024
Jorge
le 16 Sep 2024
Jorge
le 16 Sep 2024
What you showed in your question:
temp={};
temp ={repmat('0000',size(newTable,1),1)}
What you showed in your comment:
temp=[];
temp =[repmat('0000',size(newTable,1),1)]
They are completely different things:
- in your question you create a cell array (which is a container array) and put things into it.
- in your comment you create a character matrix (which is NOT a container array).
I could give you a box with an apple in it. Or I could give you an apple. But they are not the same thing: you cannot have a command which says "replace the content of the apple with a banana". But you can do that with the box.
A table is a container array, it may contain a character matrix. So from that point of view, you could use either.
But if you use a character matrix then you need to use parentheses indexing to refer to those characters themselves (not some inappropriate curly-brace indexing as if those characters were stored inside containers, which in your comment they are not). Using the character matrix will be more effort because you are not allocating a scalar array on the RHS, you are now attempting to allocate a vector (i.e. multiple elements) which means that you will need the number of elements on the LHS and the RHS to match exactly, which will mean that you will need to use e.g. REPMAT or similar on the RHS. You will most likely need to use subscript indexing or perform some conversion to linear indexing, which would take a few lines of code... and so your code will be more complex and more liable to bugs...
It would be simpler to stick with a cell array of character vectors, just like you showed in your question.
Or perhaps even better, a string array.
Also: if you use a character matrix get rid of those superfluous square brackets:
temp=[]; % <- this does nothing, get rid of it.
temp =[repmat('0000',size(newTable,1),1)]
% ^ superfluous, get rid of them ^
"which seemed to work. But I didn't understand why it would...."
Unlike some other languages, type coercion in MATLAB was in the past basically limited to the numeric types and char. For better or for worse, modern users apparently expect type coercion between everything, so the (relatively new) string class can coerce numeric, char, and cell-of-char into strings:
S = "this" % string
S(2) = 'assigns' % char
S(3) = 3 % numeric
S(4) = {'types'} % cell of char
"which to me makes more sense. can you explain what I "did" here?"
You created a string array and then assigned a character vector to it, which got coerced into a string:
T = ["hello";"x"]
T(2) = 'world'
Using a scalar string on the RHS would not use any coercion.
Jorge
le 17 Sep 2024
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