python中的Numpy二维数组遍历与二维数组切片后遍历效率比较
在python-numpy
使用中,可以用双层 for循环对数组元素进行访问,也可以切片成每一行后进行一维数组的遍历。
代码如下:
import numpy as np import time NUM = 160 a=np.random.random((NUM,NUM)) start = time.time() for i in range(NUM): for j in range(NUM): if a[i][j] == 1.0: pass end1 = time.time() for ii in range(NUM): b = a[ii,:] for jj in range(NUM): if b[jj] == 1.0: pass end2 = time.time() print("end1",end1-start) print("end2",end2-end1)
由于生成的是[0,1)中的数,因此两种操作会遍历所有的元素。多轮测试后,耗时如下:
当NUM为160时:
end1 0.006983518600463867
end2 0.003988742828369141
当NUM为1600时:
end1 0.71415114402771
end2 0.45178747177124023
结论:切片后遍历更快
原因:
楼主还暂不明确
一个想法:
b=a[ii,:]
在numpy中,为了提高效率,这种切片出来的子矩阵其实都是原矩阵的引用而已,所以改变子矩阵,原矩阵还是会变的
所以在内层循环中,第二种方法是在那一行元素所在的内存进行寻找。而第一种方法是先定位到行,再定位到列,所以比较慢?
大家是怎么想的呢?
关于numba
在小数据量下的速度慢于普通操作
什么是numba?
numba
实验比较:
import numpy as np import time NUM = 160 from numba import jit a=np.random.random((NUM,NUM)) @jit(nopython=True) def fun1(a): for i in range(NUM): for j in range(NUM): if a[i][j] == 1.0: pass def fun2(a): for i in range(NUM): for j in range(NUM): if a[i][j] == 1.0: pass @jit(nopython=True) def fun3(a): for ii in range(NUM): b = a[ii,:] for jj in range(NUM): if b[jj] == 1.0: pass def fun4(a): for iii in range(NUM): b = a[iii,:] for jjj in range(NUM): if b[jjj] == 1.0: pass start = time.time() fun1(a) end1 = time.time() fun2(a) end2 = time.time() fun3(a) end3 = time.time() fun4(a) end4 = time.time() print("end1",end1-start) print("end2",end2-end1) print("end3",end3-end2) print("end4",end4-end3)
首先,当NUM为1600时,结果如下:
end1 0.2991981506347656 #无切片,有加速
end2 0.6372940540313721 #无切片,无加速
end3 0.08377814292907715 #有切片,有加速
end4 0.358079195022583 #有切片,无加速
其他条件相同的情况下,有切片的速度更快。同样,有numba加速的也比没加速的快。
但当NUM =160时,结果如下:
end1 0.29620814323425293 #无切片,有加速
end2 0.006980180740356445 #无切片,无加速
end3 0.08580684661865234 #有切片,有加速
end4 0.0029993057250976562 #有切片,无加速
有切片依旧比无切片的快。但是有numba加速的却比没有numba加速的慢。
原来@jit(nopython=True)只是对函数进行修饰,第一次调用会进行编译,编译成机器码,之后速度就会很快。
实验代码如下:
import numpy as np import time NUM = 160 from numba import jit a=np.random.random((NUM,NUM)) @jit(nopython=True) def fun1(a): for i in range(NUM): for j in range(NUM): if a[i][j] == 1.0: pass def fun2(a): for i in range(NUM): for j in range(NUM): if a[i][j] == 1.0: pass @jit(nopython=True) def fun3(a): for ii in range(NUM): b = a[ii,:] for jj in range(NUM): if b[jj] == 1.0: pass def fun4(a): for iii in range(NUM): b = a[iii,:] for jjj in range(NUM): if b[jjj] == 1.0: pass for b in range(4): start = time.time() fun1(a) end1 = time.time() fun2(a) end2 = time.time() fun3(a) end3 = time.time() fun4(a) end4 = time.time() print("end1",end1-start) print("end2",end2-end1) print("end3",end3-end2) print("end4",end4-end3) print("---")
结果如下:
end1 0.29421305656433105
end2 0.0059833526611328125
end3 0.08181905746459961
end4 0.0029909610748291016
---
end1 0.0
end2 0.005949735641479492
end3 0.0
end4 0.004008769989013672
---
end1 0.0
end2 0.006977558135986328
end3 0.0
end4 0.00399017333984375
---
end1 0.0
end2 0.005974292755126953
end3 0.0
end4 0.003837108612060547
---
结论:
numba
加速时,第一次需要编译,需要耗时。之后调用就不需要了。
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