这是一些代码来说明问题的大小(截至 2015 年 5 月 11 日)以及如何“修复”它。
import numpy as np
import bisect
import timeit
from random import randint
dtype = np.dtype([ ('pos','<I'),('sig','<H') ]) # my data is unsigned 32bit, and unsigned 16bit
data1 = np.fromfile('./all2/840d.0a9b45e8c5344abf6ac761017e93b5bb.2.1bp.binary', dtype)
dtype2 = np.dtype([('pos',np.uint32),('sig',np.uint32)]) # convert data to both unsigned 32bit
data2 = data1.astype(dtype2)
data3 = data2.view(('uint32', len(data2.dtype.names))) # convert above to a normal array (not structured array)
print data1.dtype.descr # [('pos', '<u4'), ('sig', '<u2')]
print data2.dtype.descr # [('pos', '<u4'), ('sig', '<u4')]
print data3.dtype.descr # [('', '<u4')]
print data1.nbytes # 50344494
print data2.nbytes # 67125992
print data3.nbytes # 67125992
print data1['pos'].max() # 2099257376
print data2['pos'].max() # 2099257376
print data3[:,0].max() # 2099257376
def b1(): return bisect.bisect_left(data1['pos'], randint(100000000,200000000))
def b2(): return bisect.bisect_left(data2['pos'], randint(100000000,200000000))
def b3(): return bisect.bisect_left(data3[:,0], randint(100000000,200000000))
def ss1(): return np.searchsorted(data1['pos'], randint(100000000,200000000))
def ss2(): return np.searchsorted(data2['pos'], randint(100000000,200000000))
def ss3(): return np.searchsorted(data3[:,0], randint(100000000,200000000))
def ricob1(): return bisect.bisect_left(data1['pos'], np.uint32(randint(100000000,200000000)))
def ricob2(): return bisect.bisect_left(data2['pos'], np.uint32(randint(100000000,200000000)))
def ricob3(): return bisect.bisect_left(data3[:,0], np.uint32(randint(100000000,200000000)))
def ricoss1(): return np.searchsorted(data1['pos'], np.uint32(randint(100000000,200000000)))
def ricoss2(): return np.searchsorted(data2['pos'], np.uint32(randint(100000000,200000000)))
def ricoss3(): return np.searchsorted(data3[:,0], np.uint32(randint(100000000,200000000)))
print timeit.timeit(b1,number=300) # 0.0085117816925
print timeit.timeit(b2,number=300) # 0.00826191902161
print timeit.timeit(b3,number=300) # 0.00828003883362
print timeit.timeit(ss1,number=300) # 6.57477498055
print timeit.timeit(ss2,number=300) # 5.95308589935
print timeit.timeit(ss3,number=300) # 5.92483091354
print timeit.timeit(ricob1,number=300) # 0.00120902061462
print timeit.timeit(ricob2,number=300) # 0.00120401382446
print timeit.timeit(ricob3,number=300) # 0.00120711326599
print timeit.timeit(ricoss1,number=300) # 4.39265394211
print timeit.timeit(ricoss2,number=300) # 0.00116586685181
print timeit.timeit(ricoss3,number=300) # 0.00108909606934
更新!
因此,多亏了 Rico 的评论,似乎为要搜索的数字设置类型排序/对分确实很重要!但是,在具有 32 位和 16 位整数的结构化数组上,它仍然很慢(尽管没有以前那么慢了)