Question

我的python3程序以下列格式从其他地方接收数据作为字符串（...表示我需要输入更多数据）：

data = "0,12,145,234;1,0,0,128;2,255,255,255;...;909,100,100,100;"

我想将此转换为压缩二进制数据，而忽略,和;个字符。目前，我正在做以下事情：

splitData = data.split(';')[:-1] # ignore the last ';'
buff = []
for item in splitData:
    addr, R, G, B = item.split(',')
    addr = int(addr) # two bytes
    R    = int(R)    # one byte
    G    = int(G)    # one byte
    B    = int(B)    # one byte
    packed = struct.pack('HBBB', addr, R, G, B)
    buff.append(packed)
dataBytes = b''.join(buff)

对于上面的示例数据，此过程为我提供了以下内容：

dataBytes = b'\x00\x00\x0c\x91\xea\x01\x00\x00\x00\x80...\x8d\x03ddd'

这就是我想要的（大约是原始字符串大小的三分之一）。

但是，此过程大约需要0.002秒。我需要每帧执行33次此过程，这导致计算大约0.05秒，达到每秒约20帧。如果可能的话，我想加快速度。

有没有办法将字符串数据转换为比上述方法更快的字节数据？

Answer 1

使用itertools，进行替换然后拆分，映射到int，最后用四肢拉伸速度提高约25％：

 In [82]: data = "0,12,145,234;1,0,0,128;2,255,255,255;909,100,100,100;" * 1000
 In [83]: from itertools import  imap, izip
 [84]: %%timeit  
splitData = data.split(';')[:-1] # ignore the last ';'
buff = []
for item in splitData:
    addr, R, G, B = item.split(',')
    addr = int(addr) # two bytes
    R    = int(R)    # one byte
    G    = int(G)    # one byte
    B    = int(B)    # one byte
    packed = struct.pack('HBBB', addr, R, G, B)
    buff.append(packed)
dataBytes = b''.join(buff)
   ....: 
100 loops, best of 3: 8.61 ms per loop

In [85]: %%timeit     
mapped = imap(int, data[:-1].replace(";", ",").split(","))
b"".join([struct.pack('HBBB', *sub) for sub in izip(mapped, mapped, mapped, mapped)])
   ....: 
100 loops, best of 3: 6.27 ms per loop

使用python3，只需使用map和zip：

In [4]: %%timeit
mapped = map(int, data[:-1].replace(";", ",").split(","))
b"".join([struct.pack('HBBB', *sub) for sub in zip(mapped, mapped, mapped, mapped)])
   ...: 
100 loops, best of 3: 3.61 ms per loop

In [5]: %%timeit        
splitData = data.split(';')[:-1] # ignore the last ';'
buff = []                                                                  for item in splitData:
    addr, R, G, B = item.split(',')
    addr = int(addr) # two bytes
    R    = int(R)    # one byte
    G    = int(G)    # one byte
    B    = int(B)    # one byte
    packed = struct.pack('HBBB', addr, R, G, B)
    buff.append(packed)
dataBytes = b''.join(buff)
   ...: 
100 loops, best of 3: 4.89 ms per loop

有效地将逗号分隔值字符串转换为字节

1 个答案: