function for if loop
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Kacper Witasinski
am 19 Feb. 2022
Beantwortet: Voss
am 19 Feb. 2022
Hello guys,
I would like to create a function that I could use it in elsewhere script in the future. Here is what I've came up with:
function output = name()
output = [?????? ??????]
for k = 1:size(table)
if table.("My Column1")(k) == "Manchester City"
ChampionsYears(k) = 2012;
elseif table.("My Column2")(k) = "Manchester City"
ChampionYears(k) = 2012;
end
.
.
.
end
end
I have table with 10 columns, and I want the function (and loop) to check through the table (row-wise) if any string in the cell is "Manchester".
If so, assign the value of 2012 to "Champion Years". I want the function to have no inputs, and to be like so: displayed message (any) in command window and assign value of 2012 to Champion Years. Could you please modify this piece of code?
Regards
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Voss
am 19 Feb. 2022
name(); % call the function defined below, as specified
function name() % no input, as specified, and no output
% table has to come from somehwere (can't be an input), so I make one up
% here. yours may be loaded from a file or whatever.
% table of empty strings with 5 rows and 10 columns:
my_table = cell2table(repmat({""},5,10),'VariableNames',sprintfc('My_Column%d',1:10));
% put some non-empty strings in some cells of the table:
my_table{1,2} = "Manchester City";
my_table{1,6} = "Manchester City";
my_table{2,6} = "Manchester City";
my_table{3,8} = "Pasadena, CA";
my_table{4,5} = "Austin, TX";
my_table{4,9} = "Manchester City";
my_table{5,10} = "Manchester City";
% display table for visual reference:
disp(my_table);
% build ChampionYears column vector, which is 2012 when any cell in
% the corresponding row of my_table is "Manchester City" and NaN otherwise:
ChampionYears = NaN(size(my_table,1),1);
for k = 1:size(my_table,1)
if any(my_table{k,:} == "Manchester City")
ChampionYears(k) = 2012;
end
end
% display ChampionYears to command line:
disp(ChampionYears);
end
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