how i can eliminate the row which equal those if conditions - arrays

Question

function prealloc()
situation=zeros(Int64,3^5,5);
i=1;
for north=0:2, south=0:2, east=0:2, west=0:2, current=0:2
situation[i,:]=[north, south, east, west, current]
if situation[i,:]=[2, 2, 2, 2, 2]
elseif situation[i,:]=[2, 2, 2, 2, 1]
elseif situation[i,:]=[2, 2, 2, 2, 0]`enter code here`
end
i+=1
end
situation
end
How can I eliminate the row which equal those if conditions from the array which called situation

Answer 1

First things first: the code in your question doesn't run (for several reasons). When posting code in questions, it is good form to put it in a "working example" form, where users can copy and paste it into their editor of choice and it will work without the user having to make educated guesses as to what you are actually trying to do. This is probably one reason the question has received down-votes.
With that out of the way, there are two approaches to accomplish what you are trying to do:
1) Construct your matrix without the indicated rows in the first step. Then you don't need to worry about "deleting the rows" later on. For situations as simple as the one in the question, you could just do something like this:
function prealloc()
x = zeros(Int, 3^5 - 3, 5)
i = 1
for n=0:2, s=0:2, ea=0:2, w=0:2, cur=0:2
if !([n, s, ea, w, cur] == [2, 2, 2, 2, 2] || [n, s, ea, w, cur] == [2, 2, 2, 2, 1] || [n, s, ea, w, cur] == [2, 2, 2, 2, 0])
x[i, :] = [n, s, ea, w, cur]
i += 1
end
end
return(x)
end
Notice I'm using Int, not Int64. This will not affect performance, and it means your code will run on both 32-bit and 64-bit architectures.
Another style tip. Don't use semi-colons to end lines. This is a Matlab quirk, and it is not needed in Julia.
2) As other users have suggested, you could construct the entire matrix (including the undesirable rows), and then remove them at a later point. Of course, this necessitates re-allocating the entire matrix, and so is somewhat inefficient (note, you can remove elements of vectors in place, i.e. without re-allocation, but not any arrays of dimension 2 or greater). In this case, to encourage code re-use, it makes sense to break the routine down into three separate functions. First, we allocate the entire matrix:
function prealloc1()
x = zeros(Int64,3^5,5)
i = 1
for north=0:2, south=0:2, east=0:2, west=0:2, current=0:2
x[i,:]=[north, south, east, west, current]
i += 1
end
return(x)
end
Next, we obtain a vector of indices that we wish to remove. We do this as its own step because we only want to re-allocate the matrix once, rather than re-allocating every time we find a new row we want to delete. For your situation, you could use a function like this:
function findCondition(x::Matrix{Int})
inds = Array(Int, 0)
for i = 1:size(x, 1)
if x[i, :] == [2 2 2 2 2]
push!(inds, i)
elseif x[i, :] == [2 2 2 2 1]
push!(inds, i)
elseif x[i, :] == [2 2 2 2 0]
push!(inds, i)
end
end
return(inds)
end
Notice that in my comparison statements in this function I use [2 2 2 2 2] instead of [2, 2, 2, 2, 2]. This is because the first construct is a 2-dimensional array (type Matrix) while the second is 1-dimensional (type Vector). Since x[i, :] is of type Matrix, the difference is important.
Finally, we need to re-allocate the matrix without the offending rows. As user #Matt B. suggests, this can be done with the following one-liner function:
removeIndices(x::Matrix{Int}, inds::Vector{Int}) = x[setdiff(IntSet(1:size(x, 1)), IntSet(inds)), :]
Note, applying setdiff to IntSet here is fast because by construction inds will already be sorted in ascending order.

Answer 2

You cannot just delete a row in Julia, the only way to do it, is to create a copy of the array without the row you want to delete. And I think that's not internally implemented and it's intentional.
So you will have to do to it manually, something like this will create a copy of situation without the row i (which is not the same as saying that it will delete row i).
situation = vcat(situation[1:i-1,:],situation[i+1:end,:])
also, this will actually change the dimensions of situation in each iteration, so be careful with that...
Also2, your loop will finish in a bounds error since eventually it will be off limits of your array, maybe you could write something like this to end your loop.
if i = length(situation)
break
else
i += 1
end
Ultimately, you can make a function delrow and call it from within your loop:
function delrow(array,row)
return vcat(array[1:row-1,:],array[row+1:end,:])
end
then call situation = delrow(situation,i)