On 08/10/17 22:50, Thomas Jollans wrote:

On 08/10/17 09:12, Nissim Derdiger wrote:

...

Hi again, I realize that my question was not clear enough, so I've refined it into one runnable function (attached below) My question is basically - is there a way to perform the same operation, but faster using NumPy (or even just by using Python better..) Thanks again and sorry for the unclearness.. Nissim.

import struct

def Convert(): Endian = '

...

# pack every 2 registers, take only the first 2 bytes from each one, change their endianess than unpack them back to the Parameter format Result.append((struct.unpack(ParameterFormat,(struct.pack(Endian,RawDataList[(i*2)+1])[0:2] + struct.pack('

You can do this a little more elegantly (and probably faster) with struct by putting it in a list comprehension:

[struct.unpack('f', struct.pack('

Numpy can also do it. You can get your array of little-endian shorts with

le_shorts = np.array(raw_data, dtype='

and then reinterpret the bytes backing it as float32 with np.frombuffer:

np.frombuffer(le_shorts.data, dtype='f4')

For small lists like the one in your example, the two approaches are equally fast. For long ones, numpy is much faster:

*sigh* let's try that again: In [82]: raw_data Out[82]: [17252, 26334, 16141, 58057, 17252, 15478, 16144, 43257] In [83]: raw_data2 = np.random.randint(0, 2**32, size=10**6, dtype='u4') In [84]: %timeit np.frombuffer(np.array(raw_data, dtype='

In [82]: raw_data Out[82]: [17252, 26334, 16141, 58057, 17252, 15478, 16144, 43257] In [83]: raw_data2 = np.random.randint(0, 2**32, size=10**6, dtype='u4') # 1 million random integers In [84]: %timeit np.frombuffer(np.array(raw_data, dtype='

-- Thomas

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