I'm not sure what you mean by clear, the array will always have some values stored in it, but you can set those values to something, for example:
>>> A = numpy.array([[1, 2], [3, 4], [5, 6]], dtype=numpy.float)
>>> A
array([[ 1., 2.],
[ 3., 4.],
[ 5., 6.]])
>>> A.fill(0)
>>> A
array([[ 0., 0.],
[ 0., 0.],
[ 0., 0.]])
>>> A[:] = 1.
>>> A
array([[ 1., 1.],
[ 1., 1.],
[ 1., 1.]])
Update
First, your question is very unclear. The more effort you put into writing a good question the better answers you'll get. A good question should make it clear to us what you're trying to do and why. Also example data is very helpful, just a small amount, so we can see exactly what you're trying to do.
That being said. It seems like maybe you should just create a new array for each iteration. Creating arrays is pretty fast and it's not clear why you would want to reuse an array when the size and contents need to change. If you're trying to reuse it for performance reasons, you're probably not going to see any measurable difference, resizing arrays is not noticeably faster than creating a new array. You can create a new array by calling numpy.zeros((X, 6))
Also in your question you say:
the columns are filled from lists constructed from the tuples
If your data is already housed as a list of tuples you use numpy.array
to convert it to an array. You don't need to go the the trouble of creating an array and filling it. For example if I wanted to get a (2, 3) array from a list of tuples I would do:
data = [(0, 0, 1), (0, 0, 2)]
A = numpy.array(data)
# or if the data is stored like this
data = [(0, 0), (0, 0), (1, 2)]
A = numpy.array(data).T
Hope that helps.