Convert to a function

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jones matthew
jones matthew el 10 de Oct. de 2017
Editada: jones matthew el 11 de Oct. de 2017
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

Respuesta aceptada

Andrei Bobrov
Andrei Bobrov el 10 de Oct. de 2017
Editada: Andrei Bobrov el 10 de Oct. de 2017
>> 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
  1 comentario
jones matthew
jones matthew el 10 de Oct. de 2017
Would you be able to make the code in terms of statements such as switch, else, else if, and if?

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Más respuestas (1)

Jan
Jan el 10 de Oct. de 2017
Editada: Jan el 10 de Oct. de 2017
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.
  3 comentarios
Jan
Jan el 10 de Oct. de 2017
Editada: Jan el 10 de Oct. de 2017
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
jones matthew el 10 de Oct. de 2017
Thank you so much!

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