vec = randi(10,10,1)
vec(vec < 5) = 0
func = @(x) x(x < 5) = 0 % This isn't valid
How am I supposed to translate the second line of code into a function handle that I can use in conjunction with cellfun?
You can use multiplication, since if your condition is satisfied you have 1 and 0 otherwise.
Multiplying by the inverse of the condition therefore gives you either an unchanged value (if condition is not satisfied) or your desired substitution of 0!
func = @(x) x .* (~(x < 5)) % Replace values less than 5 with 0
If you had a different substitution, you could expand the same logic
func = @(x) x .* (~(x < 5)) + 10 * (x < 5) % replace values less than 5 with 10