Filling out an empty column in an existing table based on conditions matching other columns

I have a table. If column 3 is "fast" AND column 4 contains any of these characters: boat, car, or airplane. then column 5 become those strings on column 4. If it meets just the "fast" in column 3, then column 5 becomes "fast". Otherwise it's none. Do I need an if else statement? Or how would I do this?

column 3
fast
fast
slow
medium
fast

column 4
mike boat day
pax airplane oscar
ship
car
train

column 5
boat
airplane
none
none
fast

[Sole]

I tried this and it didn't work. It still gives me "fast" even though the allnames contains a 1 as the logic for detecting "boat" in the columns

t1 = table();

str = table2cell(lut(:,3));

allnames = contains(str,"boat");

if allnames == 1

t1.secondnames(allnames,:) = "boat";

else

t1.secondnames(allnames,:) = "fast";

end

Answer 1

[Mohammad Sami]

% t1 = yourtable
logic1 = startsWith(t1.column3varname,'fast') & startsWith(t1.column4varname,{'boat' 'car' 'airplane'});
logic2 = startsWith(t1.column3varname,'fast') & ~startsWith(t1.column4varname,{'boat' 'car' 'airplane'});
t1.column5varname(logic1) = t1.column4varname(logic1);
t1.column5varname(logic2) = t1.column3varname(logic2);
t1.column5varname(~logic1 & ~logic2) = {'none'};

[Sole]

when running t1.column5varname(logic1) = t1.column4varname(logic1), I get a "conversion to double from cell is not possible" error. What does that mean?

[Mohammad Sami]

Maybe initialise the column 5 as cell array first

t1.column5varname = cell(height(t1),1);