how to make a function that return cell array of the month

Hi every one;

I am going to attempt that query:

Write a function called June2015 that returns a cell array of dimensions 30-by-3, whose rows correspond to the days of June, 2015. The three elements of each row must be set as follows:

• The first element refers to the string 'June' (uppercase ‘J’). • The second element refers to a scalar of type double that equals the date (1 through 30). • The third element refers to the three-letter abbreviation of the day chosen from this list: 'Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun'.

For example, here is a call of the function followed by a command that shows the eleventh element of the cell array that is returned by the function:

>> m = June2015;
>> m(11,:)
ans =
'June' [11] 'Thu'

I am using that code

function m = June2015
A=cell(30,3)%declar cell array of 30 by 3
for i = 1:30
    [DateNumber, DateName] = weekday(datenum([2015 6 i]));% making the value of DateNume for DateNumber 
    %and loop run number of rows in cell array
    for j=1:3% loop run for column of cell array
    A(i,:)={'June', i, 'DateName'} 
    end
end
end

but in testing i am getting that error

Your solution is _not_ correct.

Guide me about my corrections.. thanks in advance

[Walter Roberson]

A(i,:)={'June', i, DateName(i,:)}

[Muhammad Usman Saleem]

@walter i use that:

function m = June2015
A=cell(30,3)%declar cell array of 30 by 3
for i = 1:30
    [DateNumber, DateName] = weekday(datenum([2015 6 i]));% making the value of DateNume for DateNumber 
    %and loop run number of rows in cell array
    for j=1:3% loop run for column of cell array
    A(i,:)={'June', i, 'DateName(i,:)'} 
    end
end
end

but error is same???

[Walter Roberson]

'DateName(i,:)' means to put in the literal string 'DateName(i,:)', not what is contained in DateName(i,:)

Oh I see, I thought you were pre-calculating all of the date names. In that case, just

   A(i,:) = {'June', i, DateName};

and your "for j" loop is useless

[Christos Vyzantios]

I made some small changes but the grader didnt accept :(

function m = June2015 %#ok<STOUT>
A=cell(30,3); 
for i = 1:30
    [~, DateName] = weekday(datenum([2015 6 i]));% making the value of DateNume for DateNumber 
    %and loop run number of rows in cell array

      A(i,:) = {'June', i, DateName};
      end
  end

[Walter Roberson]

Your "function" defines the output as going to the variable named "m" but you never assign anything to that variable.

[Muhammad Usman Saleem]

@walter see my comment below...

[Christos Vyzantios]

I transform liitle the code but the problem exist

function m = June2015 %#ok<STOUT>
A=cell(30,3);%declar cell array of 30 by 3
for i = 1:30 %#ok<ALIGN>
    [~, DateName] = weekday(datenum([2015 6 i]));
      A(i,:) = {'June', i, DateName};
    end
 end

[Muhammad Usman Saleem]

@Christos Vyzantios i have formatted your code but i am getting below mentioned error

function m = June2015 %#ok<STOUT> 
A=cell(30,3);%declar cell array of 30 by 3 
for i = 1:30 %#ok<ALIGN> 
    [~, DateName] = weekday(datenum([2015 6 i])); 
    A(i,:) = {'June', i, 'DateName'}; 
end
end

error i get

Your solution is _not_ correct.

Please correct my code. Thanks in advance for that...

Answer 1

[Walter Roberson]

In your line

A(i,:)={'June', i, 'DateName'}

that would attempt to put the literal string 'DateName' into the array, rather than trying to put any content from the variable DateName into the array.

You probably want to select a portion of DateName rather than the whole thing.

[Muhammad Usman Saleem]

@Walter thanks for contributions.. I do not getting your guide. I only understand , you are saying me about correction of 'DateNume', but when i used without quote i still gets same error. looking for more understanding of your assistance

[Muhammad Usman Saleem]

@Walter???????

[Muhammad Usman Saleem]

@walter see my comment on @jan post please..

Answer 2

[Jan]

The loop over j is not required. Simply omit it.

The 3rd column of A should not be the string 'DateName(i,:)', but the contents of the variable DateName. So use:

A(i,:) = {'June', i, DateName} ;

[Muhammad Usman Saleem]

@Jan Simon and @walter last time when i test my code it was correct but now when i again testing my code its shown an error. That code i was using

function m = June2015
A=cell(30,3)%declar cell array of 30 by 3
for i = 1:30
    [DateNumber, DateName] = weekday(datenum([2015 6 i]));% making the value of DateNume for DateNumber 
    %and loop run number of rows in cell array
  A(i,:) = {'June', i, 'DateName'} ;
    end
end

guide me as my paper is tomorrow. Thanks in advance for assistance

[Muhammad Usman Saleem]

if i use

A(i,:) = {'June', i, DateName} ;

then again error is same....

[Jan]

Then please be so kind and post, which error message you see. The code runs fine on my machine.

[Muhammad Usman Saleem]

@Jan thanks for contributions. I am checking that code in a grader.. which has output

Your solution is _not_ correct.

We are doing some major mistake in the code?

[Muhammad Usman Saleem]

function A = June2015
A=cell(30,3)%declar cell array of 30 by 3
for i = 1:30
    [DateNumber, DateName] = weekday(datenum([2015 6 i]));
 % making the value of DateNume for DateNumber and loop run number of rows in cell array
  A(i,:) = {'June', i, 'DateName'} ;
    end
end

that code i using...

[Walter Roberson]

function A = June2015
A=cell(30,3)%declar cell array of 30 by 3
for i = 1:30
    [DateNumber, DateName] = weekday(datenum([2015 6 i]));
 % making the value of DateNume for DateNumber and loop run number of rows in cell array
  [A{i,:}] = deal('June', i, DateName) ;
    end
end

[Muhammad Usman Saleem]

@Walter , i have solved this yesterday..

Answer 3

[abdo desoki]

change that variable A with m and it will run correctly .

[Muhammad Usman Saleem]

@abdo , i use that but error remains same

function m = June2015
m=cell(30,3)%declar cell array of 30 by 3
for i = 1:30
    [DateNumber, DateName] = weekday(datenum([2015 6 i]));
 % making the value of DateNume for DateNumber and loop run number of rows in cell array
  m(i,:) = {'June', i, 'DateName'} ;
    end
end

that error is come the code

   Your solution is _not_ correct.

Answer 4

[Luxman Maheswaran]

You have to convert the number of DateName to a string such as 'Thu' I have checked it with a grader and it is correct

[Walter Roberson]

The second output of weekday() is a string.