Convert to a function

Question

jones matthew 2017-10-10

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https://ww2.mathworks.cn/matlabcentral/answers/360552-convert-to-a-function

编辑： jones matthew 2017-10-11

I would like this code to be converted to a function.

% Convert to function
year=input('Enter specified year(yyyy):');
if year>0
  if mod(year,400)==0
      leap_day=1;
  else if mod(year,100)==0
          leap_day=0;
      else if mod(year,4)==0
              leap_day=1;
          else 
              leap_day=0;
          end
      end
  end
end
          month =input('Enter specified month(1-12):');
          if month>=1 && month <=12
              switch (month)
                  case {1,3,5,7,8,10,12}
                      max_day=31;
                  case {4,6,9,11}
                      max_day=30;
                  case {2}
                      max_day=28+leap_day;
              end
              fprintf('Enter specified day (1-%d):',max_day);
              day=input('');
              if day>=1 && day <=max_day
                  day_of_year=day;
                  for ii=1:month-1
                      switch(ii)
                          case {1,3,5,7,8,10,12}
                              day_of_year=day_of_year+31;
                          case {4,6,9,11}
                              day_of_year=day_of_year+30;
                          case {2}
                              day_of_year=day_of_year+28+leap_day;
                      end
                  end

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Answer 1

Andrei Bobrov 2017-10-10

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https://ww2.mathworks.cn/matlabcentral/answers/360552-convert-to-a-function#answer_285046

编辑：Andrei Bobrov 2017-10-10

在 MATLAB Online 中打开

>> calender1 = @(day1,month1,year1)datenum(year1,month1,day1) - datenum(year1 - 1,12,31);
>> calender1(3,1,2017)
ans =
     3
>> calender1(3,3,2020)
ans =
    63
>> calender1(3,3,2017)
ans =
    62
>>

or create m - file: calcDaysOfYear.m

function num_of_day = calcDaysOfYear(day1,month1,year1)
    dm = 30*ones(12,1);
    x = rem(year1,[4,100,400]);
    dm(2) = dm(2) - 1 - (x(:,1) | x(:,2)== 0 & x(:,3));
    dm([1:2:7,8:2:end]) = dm([1:2:7,8:2:end]) + 1;
    num_of_day = day1 + sum(dm(1:month1-1));
end

use

>>calcDaysOfYear(3,1,2017)
  ans =
       3
>>calcDaysOfYear(3,3,2016)
  ans =
       63
>>calcDaysOfYear(3,3,2017)
  ans =
       62

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jones matthew 2017-10-10

Would you be able to make the code in terms of statements such as switch, else, else if, and if?

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Answer 2

Jan 2017-10-10

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/360552-convert-to-a-function#answer_285049

编辑：Jan 2017-10-10

在 MATLAB Online 中打开

If you really want to use this code instead of the built-in function of the Matlab toolbx (see Andrei's answer), start with:

function [day] = calender(day,month,year)

Now add the code and remove the fprintf and input lines.

Currently your code checks only if year > 0, but what should happen otherwise? Either add an else or define e.g. day=NaN as default values on top of the code.

Note that there is a function called "calender" already, so it is better to use a unique name.

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jones matthew 2017-10-10

在 MATLAB Online 中打开

I currently have this code below. But I have errors on MATLAB. Would you be able to correct my code, and maybe look at the image I just posted? Thank you.

    % code
function [day] = calender(day,month,year)
day=NaN;
  if mod(year,400)==0
      leap_day=1;
  else if mod(year,100)==0
          leap_day=0;
      else if mod(year,4)==0
              leap_day=1;
          else 
              leap_day=0;
          end
      end
  end
end
          if month>=1 && month <=12
              switch (month)
                  case {1,3,5,7,8,10,12}
                      max_day=31;
                  case {4,6,9,11}
                      max_day=30;
                  case {2}
                      max_day=28+leap_day;
              end
              day=input('');
              if day>=1 && day <=max_day
                  day_of_year=day;
                  for ii=1:month-1
                      switch(ii)
                          case {1,3,5,7,8,10,12}
                              day_of_year=day_of_year+31;
                          case {4,6,9,11}
                              day_of_year=day_of_year+30;
                          case {2}
                              day_of_year=day_of_year+28+leap_day;
                      end
                  end

end

Jan 2017-10-10

编辑：Jan 2017-10-10

在 MATLAB Online 中打开

There is no image. If you mention an error, please post a complete copy of it also. It is much easier to fix a problem, than ti guess, what the problem is. Thanks.

Start with:

function day_of_year = calender(day,month,year)
day_of_year = NaN;

to avoid overwriting the input "day".

Care for a proper indentation: Mark all code with Ctrl-A and press Ctrl-I.

It will be nicer to replace the "else if" by "elseif":

if mod(year,400)==0
  leap_day=1;
elseif mod(year,100)==0
  leap_day=0;
elseif mod(year,4)==0
  leap_day=1;
else 
  leap_day=0;
end

Or shorter:

leap_day = (mod(year,4) == 0 && mod(year,100) ~= 0) || ...
           (mod(year,400) == 0);

Remove the "day=input('');" also.

jones matthew 2017-10-10

Thank you so much!

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Convert to a function

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Convert to a function

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