Simplify code with nested if

4 vues (au cours des 30 derniers jours)
Gaetano Pavone
Gaetano Pavone le 21 Oct 2019
Commenté : the cyclist le 21 Oct 2019
I have a matrix C in which every row contains dates and times. In the fourth column months are indicated as Jun, Feb, Mar, etc... I would like to replace such abbreviations with their corresponding numeric values: Jan->1, Feb->2, etc... My first attempt is:
for i=1:size(C,1)
if C(i,4)=='Jen' C(i,4)==1;
elseif C(i,4)=='Feb' C(i,4)==2; ... etc
end
end
Is there any way for simplify this code?
  2 commentaires
the cyclist
the cyclist le 21 Oct 2019
What data type is C?
Gaetano Pavone
Gaetano Pavone le 21 Oct 2019
C is a 153435x34 cell

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Réponses (1)

the cyclist
the cyclist le 21 Oct 2019
Modifié(e) : the cyclist le 21 Oct 2019
% A little pretend data, where I only fill in the 4th column with a few months
C = cell(5,6);
C(:,4) = {'Jan';'Jan';'Dec';'Mar';'Jul'};
% Replace the 4th column with the numeric indices
C(:,4) = num2cell(month(datetime(C(:,4),'Format','MMM')));
  2 commentaires
Gaetano Pavone
Gaetano Pavone le 21 Oct 2019
C(:,4) actually is like:
C(:,4) = {"Jan";"Jan";"Dec";"Mar";"Jul"};
your code doesn't work for this format
the cyclist
the cyclist le 21 Oct 2019
Try
month(datetime([C{:,4}],"Format","MMM"))
instead. That creates a string array from the cell array of strings, which is an allowed input to the datetime function.

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