Convert to a function

14 ビュー (過去 30 日間)
jones matthew
jones matthew 2017 年 10 月 10 日
編集済み: jones matthew 2017 年 10 月 11 日
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

採用された回答

Andrei Bobrov
Andrei Bobrov 2017 年 10 月 10 日
編集済み: Andrei Bobrov 2017 年 10 月 10 日
>> 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 件のコメント
jones matthew
jones matthew 2017 年 10 月 10 日
Would you be able to make the code in terms of statements such as switch, else, else if, and if?

サインインしてコメントする。

その他の回答 (1 件)

Jan
Jan 2017 年 10 月 10 日
編集済み: Jan 2017 年 10 月 10 日
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 件のコメント
Jan
Jan 2017 年 10 月 10 日
編集済み: Jan 2017 年 10 月 10 日
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 2017 年 10 月 10 日
Thank you so much!

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeDates and Time についてさらに検索

タグ

タグが未入力です。

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by