easy way of getting number of bits from a numpy type? [duplicate]

Question

Is there an easy way to get the number of 8-bit bytes that corresponds to a particular numpy type?

I know I can do this:

lookup = {np.uint8:   1, np.int8:   1,
          np.uint16:  2, np.int16:  2, 
          np.uint32:  4, np.int32:  4,
          np.uint64:  8, np.int64:  8}
def getByteSize(dtype):
   return lookup[dtype]

but this seems slightly kludgey and it seems like there ought to be a built-in method of retrieving this info.

Answer 1

You can use the nbytes attribute of an instance of the type:

In [8]: np.uint8(0).nbytes
Out[8]: 1

In [9]: np.uint16(0).nbytes
Out[9]: 2

In [10]: np.uint32(0).nbytes
Out[10]: 4

etc.

You can drop the argument (0 is assumed), and you can also use the itemsize attribute instead of nbytes:

In [13]: np.int32().itemsize
Out[13]: 4

A method that (apparently) avoids creating an instance of the numeric type is to create a dtype object, and then access its itemsize attribute (but I doubt this is going to be more memory or time efficient than simply creating an instance of the numeric type):

In [14]: dt = np.dtype(np.int32)

In [15]: dt.itemsize
Out[15]: 4

(I said "apparently" because I haven't looked at the source code for dtype. For all I know, it might create an instance of its argument.)