On 12 April 2010 11:59, John Jack itsmilesda...@gmail.com wrote:
Hello all.
I am (relatively) new to python, and 100% new to numpy.
I need a way to store arrays of booleans and compare the arrays for
equality.
I assume I want arrays of dtype Boolean, and I should compare the arrays
with array_equal
tmp.all_states
array([False, True, False], dtype=bool)
tmp1.all_states
array([False, False, False], dtype=bool)
tmp1.all_states[1]=True
tmp1.all_states
array([False, True, False], dtype=bool)
array_equal(tmp.all_states,tmp1.all_states)
True
any(tmp.all_states)
True
Would this be (a) the cheapest way (w.r.t. memory) to store Booleans and (b)
the most efficient way to compare two lists of Booleans?
The short answer is yes and yes.
The longer answer is, that uses one byte per Boolean, which is a
tradeoff. In some sense, modern machines are happier working with 32-
or 64-bit quantities, so loading a one-byte Boolean requires a small
amount of byte-shuffling. On the other hand, if you're really short of
memory, 8 bits for a Boolean is really wasteful. In fact, since modern
machines are almost always limited by memory bandwidth, a packed
Boolean data structure would probably be much faster for almost all
operations in spite of the bit-fiddling required. But such a
representation is incompatible with the whole infrastructure of numpy
and so would require a great deal of messy code to support.
So yes, it's the best representation of Booleans available, unless
you're dealing with mind-bogglingly large arrays of them, in which
case some sort of packed-Boolean representation would be better. This
can even be partially supported by numpy, using uint8s, bitwise
operations, and manually-specified bitmasks. There are probably not
many applications for which this is worth the pain.
Anne
P.S. There's actually at least one python package for bit vectors,
outside numpy; I can't speak for how good it is, though. -A
Thanks for your advice.
-JJ.
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