If and elseif with Or

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kathy
kathy el 3 de Nov. de 2017
Comentada: kathy el 3 de Nov. de 2017
I would like to distinguish id2 if we encounter a leap year. However, this current code sets id2 to true for values(days) 32 through 60 for nonleap years as well, so the elseif isn't working properly.
function mmean_Feb_year=mmean_Feb_(year,file)
T = readtable(file);
M =table2array(T);
id1 = M(:,1)==year;
dmean_year = M(id1,:);
if year==2008||2012||2016
id2 = dmean_year(:,2)>=32 & dmean_year(:,2)<=60;
elseif year==2007||2009||2010||2011
id2 = dmean_year(:,2)>=32 & dmean_year(:,2)<=59;
end
dmean_Feb_year = dmean_year(id2,:);
mmean_Feb_year = mean(dmean_Feb_year(:,3:end),'omitnan');
end
I also tried the following, but this did not save the leap year id2, and made id2 true for days 32-59 for all years.
function mmean_Feb_year=mmean_Feb_(year,file)
T = readtable(file);
M =table2array(T);
id1 = M(:,1)==year;
dmean_year = M(id1,:);
if year==2008||2012||2016
id2 = dmean_year(:,2)>=32 & dmean_year(:,2)<=60;
end
if year==2007||2009||2010||2011
id2 = dmean_year(:,2)>=32 & dmean_year(:,2)<=59;
end
dmean_Feb_year = dmean_year(id2,:);
mmean_Feb_year = mean(dmean_Feb_year(:,3:end),'omitnan');
end

Respuesta aceptada

KL
KL el 3 de Nov. de 2017
Editada: KL el 3 de Nov. de 2017
when you use | |, you should use it like,
if year == 2012 || year == 2016 % and so on
you need have "an expression" on both side of the | |.
when you write
if year == 2012 || 2016
it checks the first part of the condition (year==2012) but the second part is simply a number which is not zero, so the condition becomes true. So based on what you wrote, the condition becomes always true.
  1 comentario
kathy
kathy el 3 de Nov. de 2017
Oh! My apologies for the silly mistake. Thank you!

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