indexing a numbering in a string.

This is two questions in one. Firstly, I am trying to add labels to an array, but for the life of me I cannot figure out how to, so I decided to use a cheaty way using the statistics tool dataset.

In the program I have indexed a variable number of loops. These loops are stored in the array, and I want to be able to number the labels on the array to display something like:

Division 1, Division 2, Division 3,.....Division n, Total Cells, Lin A, Lin B

So basically, I want the first set of labels to be able to be indexed, and then there will be a few labels that are not indexed.

The first thought was that I need a string array that was indexed. I thought I could do something like:

for i=1:n

char(i)=('Help' num2str(i));

end

but this is not working. It will to accept the indexed number and the string. together, although the code without the text works fine.

    for i=1:n

    char(i)=(num2str(i));

    end

The second issue, is that when i try to use dataset, with an array created as the above char, it will not accept it.

B=dataset({results, char,'Lineage 1','Lineage 2','Total Cells'})

where "results" is the data that will go into the table. And "char" right now is simply ['1' '2' '3'...'n'] built as above (without the text I want). If you have any suggestions please let me know!

Answer 1

[Walter Roberson]

Do not name a variable "char": that overrides the char() function.

for i = 1 : n
  labels{i} = sprintf('Division %d', i);
end

B = dataset({results, labels{:},'Lineage 1','Lineage 2','Total Cells'})

[JJ]

Thanks! but for some reason the dataset function won't accept the "labels" array. It does not seem to accept an array as the labels.

[Walter Roberson]

To confirm, you have the {:} right after labels ? If so then it should be equivalent to you having typed out each of the individual labels, exactly the same as

B  dataset({results, 'Division 1', 'Division 2', [...] 'Lineage 1' etc

What error are you getting?

[JJ]

The error I am getting is:

??? Error using ==> dataset.dataset>dataset.dataset at 129
Must have one variable name for each column when creating multiple variables from an array.

[Walter Roberson]

How many columns are there in "results" ? How does that compare to "n" ? If the number of columns in "results" is not n+3 then you would get that error. The "+3" being for 'Linage 1', 'Linage 2', 'Total Cells'

Answer 2

[JJ]

the function 'dataset' does not seem to want to accept a vector for the column titles. even if i write a program as follows;

results=ones(1,6);

for i=1:6
label{i} = transpose(sprintf('Division %d', i)); %had to transpose it because otherwise this becomes a string column vector.

end

B=dataset({results,label});

Any thoughts anyone? Is there another better way to title these columns that will accept a string vector?

[Walter Roberson]

Correct, dataset() does not take a vector for the column titles. Which is why the code I gave above

B = dataset({results, labels{:},'Lineage 1','Lineage 2','Total Cells'})

does not use a vector for the column titles. The construct labels{:} does not mean 'pass in the array "labels" as an array at that point': it means 'at this point create new arguments just as if you had originally typed them in as part of the call, with each successive argument taking a value from the next element of the cell array "labels"'

Did you try

B=dataset({results,label{:}});

I notice you did not answer my previous question about how many columns there were in "results" and how that compares to "n".

[Jan]

Have you tried your code? The transposition is not useful here, because sprintf does not create a column vector, but as expected a row vector.

Answer 3

[Jan]

"I want the first set of labels to be able to be indexed"

Then using an index seems to be obvious:

Division{1} = ...
Division{2} = ...
...

This is much more convenient than hiding the index in the name of the variable. But if you really have good reasons to do this, dynamic fieldnames are a solution also:

for ii = 1:10
  data.(sprintf('Division %d', ii)) = rand(1,10);
end

Answer 4

[JJ]

thanks Walter! You were right, I was not using the right bracketing. Sorry! I had to read up about the bracketing.

Jan, thanks! something like that is also working for me!