[Numpy-discussion] converting list of int16 values to bitmask and back to bitmask and back to list of int32\float values

Thomas Jollans tjol at tjol.eu
Sun Oct 8 16:58:41 EDT 2017


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 = '<I' # Big endian
> < is little endian. Make sure you're getting out the right values!
>>     ParameterFormat = 'f' # float32
>>     RawDataList = [17252, 26334, 16141, 58057,17252, 15478, 16144, 43257] # list of int32 registers
>>     NumOfParametersInRawData = int(len(RawDataList)/2)
>>     Result = []
>>     for i in range(NumOfParametersInRawData):
> Iterating over indices is not very Pythonic, and there's usually a
> better way. In this case: for int1, int2 in zip(RawDataList[::2],
> RawDataList[1::2])
> 
>>         # 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('<I',RawDataList[i*2])[0:2])))[0])
> 
> 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('<HH', i1 & 0xffff, i2 & 0xffff))[0] for
> i1, i2 in zip(raw_data[::2], raw_data[1::2])]
> 
> Numpy can also do it. You can get your array of little-endian shorts with
> 
> 
> le_shorts = np.array(raw_data, dtype='<u2')
> 
> 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='<u2').data,
dtype='f4')
6.45 µs ± 60.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [85]: %timeit np.frombuffer(np.array(raw_data2, dtype='<u2').data,
dtype='f4')
854 µs ± 37.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [86]: %timeit [struct.unpack('f', struct.pack('<HH', i1 & 0xffff, i2
& 0xffff))[0] for i1, i2 in zip(raw_data[::2], raw_data[1::2])]
4.87 µs ± 17.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [87]: %timeit [struct.unpack('f', struct.pack('<HH', i1 & 0xffff, i2
& 0xffff))[0] for i1, i2 in zip(raw_data2[::2], raw_data2[1::2])]
3.6 s ± 9.78 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

> 
> 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='<u2').data, dtype='f4') 6.45 µs
> ± 60.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each) In
> [85]: %timeit np.frombuffer(np.array(raw_data2, dtype='<u2').data,
> dtype='f4') 854 µs ± 37.3 µs per loop (mean ± std. dev. of 7 runs, 1000
> loops each) In [86]: %timeit [struct.unpack('f', struct.pack('<HH', i1 &
> 0xffff, i2 & 0xffff))[0] for i1, i2 in zip(raw_data[::2],
> raw_data[1::2])] 4.87 µs ± 17.3 ns per loop (mean ± std. dev. of 7 runs,
> 100000 loops each) In [87]: %timeit [struct.unpack('f',
> struct.pack('<HH', i1 & 0xffff, i2 & 0xffff))[0] for i1, i2 in
> zip(raw_data2[::2], raw_data2[1::2])] 3.6 s ± 9.78 ms per loop (mean ±
> std. dev. of 7 runs, 1 loop each)
> 
> -- Thomas
> 
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