Python (AoC第5天)使用二进制搜索解码字符串与二进制转换一样快?
这个问题是关于解决问题的(上半年) 为了得到相同的结果,我编写了两个不同的函数,即将登机牌字符串解码为行、列坐标。在第一种情况下,我根据字符串中的每个字符进行了二进制搜索:Python (AoC第5天)使用二进制搜索解码字符串与二进制转换一样快?,python,performance,optimization,binary,binary-search,Python,Performance,Optimization,Binary,Binary Search,这个问题是关于解决问题的(上半年) 为了得到相同的结果,我编写了两个不同的函数,即将登机牌字符串解码为行、列坐标。在第一种情况下,我根据字符串中的每个字符进行了二进制搜索: def decode(bp): row = bp[:7] col = bp[-3:] row_lower, row_upper = 0, 127 col_lower, col_upper = 0, 7 for char in row: if char=='F': row_u
def decode(bp):
row = bp[:7]
col = bp[-3:]
row_lower, row_upper = 0, 127
col_lower, col_upper = 0, 7
for char in row:
if char=='F':
row_upper = ((row_upper+row_lower))//2
else:
row_lower = ((row_upper+row_lower)+1)//2
for char in col:
if char=='L':
col_upper = ((col_upper+col_lower))//2
else:
col_lower = ((col_upper+col_lower)+1)//2
sid = (row_lower*8)+col_lower
return (row_lower, col_lower, sid)
我意识到,如果把字符串看成二进制数,每个行、列坐标都有1∶1的映射,所以我也写了这个问题的另一种解决方案:
def alternative_decode(bp):
bp = bp.replace('F', '0').replace('L', '0').replace('B', '1').replace('R', '1')
return(int(bp[:7], 2), int(bp[-3:], 2), (int(bp[:7], 2)*8)+int(bp[-3:], 2))
我写第二个解决方案是因为我希望它比第一个快得多,因为它是一个简单的二进制转换,而不是二进制搜索。然而,当对这两种方法计时时,我注意到它们都有相同的运行时间,即大约在0.0000020和0.0000025秒之间
这是什么原因?是否有一些Python魔术在幕后发生,使这两种解决方案都同样有效,或者我编写它们的方式使它们同样无效?在我的计算机上,如果您重用行/列结果,而不是重复对
sid
返回的int()调用,则您的替代方法会显著加快:
def alternative_decode_reuse(bp):
bp = bp.replace('F', '0').replace('L', '0').replace('B', '1').replace('R', '1')
row = int(bp[:7], 2)
col = int(bp[-3:], 2)
sid = row*8 + col
return (row, col, sid)
将所有三个函数放入一个文件decoding.py
,这将为我们提供:
tmp$ python -m timeit -s 'import decoding' -- 'decoding.decode("FFBBFBFLLR")'
1000000 loops, best of 3: 1.75 usec per loop
tmp$ python -m timeit -s 'import decoding' -- 'decoding.alternative_decode("FFBBFBFLLR")'
1000000 loops, best of 3: 1.62 usec per loop
tmp$ python -m timeit -s 'import decoding' -- 'decoding.alternative_decode_reuse("FFBBFBFLLR")'
1000000 loops, best of 3: 1.32 usec per loop
此时,大部分时间用于执行replace()
行。如果我们没有呢
def third_decode(bp):
row = 0
for c in bp[:7]:
row <<= 1
if c == 'B':
row += 1
col = 0
for c in bp[7:]:
col <<= 1
if c == 'R':
col += 1
sid = row * 8 + col
return (row, col, sid)
稍微差一点,或者至少没有明显好转。如果我们还使用所需的sid
相当于行/列编号的(二进制)串联这一事实,会怎么样
def fourth_decode(bp):
sid = 0
for c in bp:
sid <<= 1
if c in 'BR':
sid += 1
row = sid >> 3
col = sid & 7
return (row, col, sid)
现在,我已经厌倦了编辑cmdline参数来重新运行所有内容,所以让我们将其添加到decoding.py
的底部:
if __name__ == '__main__':
loops = 1000000
funcs = (
decode, alternative_decode, alternative_decode_reuse, replace_only,
third_decode, fourth_decode,
)
from timeit import Timer
for fun in funcs:
cmd = 'decoding.%s("FFBBFBFLLR")' % fun.__name__
timer = Timer(cmd, setup='import decoding')
totaltime = min(timer.repeat(5, loops))
fmt = '%25s returned %14r -- %8d loops, best of 5: %6d ns per loop'
arg = (fun.__name__, fun('FFBBFBFLLR'), loops, totaltime*1000000000/loops)
print(fmt % arg)
这让我们可以轻松地运行所有功能:
tmp$ python decoding.py
decode returned (26, 1, 209) -- 1000000 loops, best of 5: 2090 ns per loop
alternative_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 1829 ns per loop
alternative_decode_reuse returned (26, 1, 209) -- 1000000 loops, best of 5: 1414 ns per loop
replace_only returned None -- 1000000 loops, best of 5: 700 ns per loop
third_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 1368 ns per loop
fourth_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 1123 ns per loop
在那之后,我就不知道如何让它跑得更快了。但去年,一位代码解决熟人告诉我,他正在使用pypypy
Python实现来提高速度。也许能帮上忙
tmp$ pypy decoding.py
decode returned (26, 1, 209) -- 1000000 loops, best of 5: 151 ns per loop
alternative_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 4 ns per loop
alternative_decode_reuse returned (26, 1, 209) -- 1000000 loops, best of 5: 3 ns per loop
replace_only returned None -- 1000000 loops, best of 5: 3 ns per loop
third_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 141 ns per loop
fourth_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 138 ns per loop
好吧,我所有的努力都白费了!:)
看起来pypy的
replace()
和int()
函数要快得多。此外,虽然它的JIT确实使我们的各种循环函数更快,但在可能的情况下还是最好使用内置函数。在我的计算机上,如果您重用行/列结果,而不是重复对sid
返回的int()调用,那么您的选择速度会显著加快:
def alternative_decode_reuse(bp):
bp = bp.replace('F', '0').replace('L', '0').replace('B', '1').replace('R', '1')
row = int(bp[:7], 2)
col = int(bp[-3:], 2)
sid = row*8 + col
return (row, col, sid)
将所有三个函数放入一个文件decoding.py
,这将为我们提供:
tmp$ python -m timeit -s 'import decoding' -- 'decoding.decode("FFBBFBFLLR")'
1000000 loops, best of 3: 1.75 usec per loop
tmp$ python -m timeit -s 'import decoding' -- 'decoding.alternative_decode("FFBBFBFLLR")'
1000000 loops, best of 3: 1.62 usec per loop
tmp$ python -m timeit -s 'import decoding' -- 'decoding.alternative_decode_reuse("FFBBFBFLLR")'
1000000 loops, best of 3: 1.32 usec per loop
此时,大部分时间用于执行replace()
行。如果我们没有呢
def third_decode(bp):
row = 0
for c in bp[:7]:
row <<= 1
if c == 'B':
row += 1
col = 0
for c in bp[7:]:
col <<= 1
if c == 'R':
col += 1
sid = row * 8 + col
return (row, col, sid)
稍微差一点,或者至少没有明显好转。如果我们还使用所需的sid
相当于行/列编号的(二进制)串联这一事实,会怎么样
def fourth_decode(bp):
sid = 0
for c in bp:
sid <<= 1
if c in 'BR':
sid += 1
row = sid >> 3
col = sid & 7
return (row, col, sid)
现在,我已经厌倦了编辑cmdline参数来重新运行所有内容,所以让我们将其添加到decoding.py
的底部:
if __name__ == '__main__':
loops = 1000000
funcs = (
decode, alternative_decode, alternative_decode_reuse, replace_only,
third_decode, fourth_decode,
)
from timeit import Timer
for fun in funcs:
cmd = 'decoding.%s("FFBBFBFLLR")' % fun.__name__
timer = Timer(cmd, setup='import decoding')
totaltime = min(timer.repeat(5, loops))
fmt = '%25s returned %14r -- %8d loops, best of 5: %6d ns per loop'
arg = (fun.__name__, fun('FFBBFBFLLR'), loops, totaltime*1000000000/loops)
print(fmt % arg)
这让我们可以轻松地运行所有功能:
tmp$ python decoding.py
decode returned (26, 1, 209) -- 1000000 loops, best of 5: 2090 ns per loop
alternative_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 1829 ns per loop
alternative_decode_reuse returned (26, 1, 209) -- 1000000 loops, best of 5: 1414 ns per loop
replace_only returned None -- 1000000 loops, best of 5: 700 ns per loop
third_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 1368 ns per loop
fourth_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 1123 ns per loop
在那之后,我就不知道如何让它跑得更快了。但去年,一位代码解决熟人告诉我,他正在使用pypypy
Python实现来提高速度。也许能帮上忙
tmp$ pypy decoding.py
decode returned (26, 1, 209) -- 1000000 loops, best of 5: 151 ns per loop
alternative_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 4 ns per loop
alternative_decode_reuse returned (26, 1, 209) -- 1000000 loops, best of 5: 3 ns per loop
replace_only returned None -- 1000000 loops, best of 5: 3 ns per loop
third_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 141 ns per loop
fourth_decode returned (26, 1, 209) -- 1000000 loops, best of 5: 138 ns per loop
好吧,我所有的努力都白费了!:)
看起来pypy的
replace()
和int()
函数要快得多。此外,虽然它的JIT确实使我们的各种循环函数更快,但如果可能,最好还是使用内置函数。感谢您提供了令人惊讶的深入答案!我不知道pypy在这里的优势,我将有一些跟进工作:)感谢您提供了令人惊讶的深入回答!我不知道pypy在这里的优势,我将有一些事情要做:)