Thomas, Thanks for your answers!
Just for the closer of this issue, here are the 2 np solutions that I used with
Thomas help:
(both are MUCH faster than my original solution, func2 slightly more)
def func1():
Endian = '<I' # Little endian
ParameterFormat = 'f' # float32
RawDataList = np.array([17252, 26334, 16141, 58057,17252, 15478, 16144,
43257]) # list of int32 registers
w1 = np.array(RawDataList, dtype='<u2').reshape((-1, 2))
w2 = w1[:, ::-1].ravel()
w3 = np.frombuffer(w2.data,dtype='f4')
print(w3)
def func2():
Endian = '<I' # Little endian
ParameterFormat = 'f' # float32
RawDataList = np.array([17252, 26334, 16141, 58057,17252, 15478, 16144,
43257]) # list of int32 registers
x1 = np.array(RawDataList, dtype='>u2')
x2 = np.frombuffer(x1.data, dtype='>u4')
x3 = np.array(x2, dtype='<u4')
x4 = np.frombuffer(x3.data, dtype='f4')
print(x4)
Nissim.
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Contents of NumPy-Discussion digest..."
Today's Topics:
1. Re: converting list of int16 values to bitmask and back to
bitmask and back to list of int32\float values (Thomas Jollans)
2. Re: converting list of int16 values to bitmask and back to
bitmask and back to list of int32\float values (Thomas Jollans)
----------------------------------------------------------------------
Message: 1
Date: Sun, 8 Oct 2017 22:50:02 +0200
From: Thomas Jollans <t...@tjol.eu>
To: numpy-discussion@python.org
Subject: Re: [Numpy-discussion] converting list of int16 values to
bitmask and back to bitmask and back to list of int32\float values
Message-ID: <78197fb1-3f0f-06a6-1385-a310fd556...@tjol.eu>
Content-Type: text/plain; charset=utf-8
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,RawDa
> taList[(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:
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 ?s ? 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 ?s ? 37.3 ?s 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 ?s ? 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
------------------------------
Message: 2
Date: Sun, 8 Oct 2017 22:58:41 +0200
From: Thomas Jollans <t...@tjol.eu>
To: numpy-discussion@python.org
Subject: Re: [Numpy-discussion] converting list of int16 values to
bitmask and back to bitmask and back to list of int32\float values
Message-ID: <a1ac7020-857a-366f-43cb-6ea2b4f24...@tjol.eu>
Content-Type: text/plain; charset=utf-8
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,RawD
>> ataList[(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 ?s ? 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 ?s ? 37.3 ?s 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 ?s ? 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
> ?s ? 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 ?s ? 37.3 ?s 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 ?s ? 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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> NumPy-Discussion@python.org
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