直接使用Python来实现向量的相加
# -*-coding:utf-8-*- #向量相加 def pythonsum(n): a = range(n) b = range(n) c = [] for i in range(len(a)): a[i] = i**2 b[i] = i**3 c.append(a[i]+b[i]) return a,b,c print pythonsum(4),type(pythonsum(4)) for arg in pythonsum(4): print arg
从这里这个输出结果可以看得出来，return多个值时，是以列表的形式返回的
([0, 1, 4, 9], [0, 1, 8, 27], [0, 2, 12, 36]) <type 'tuple'> [0, 1, 4, 9] [0, 1, 8, 27] [0, 2, 12, 36]
使用numpy包实现两个向量的相加
def numpysum(n): a = np.arange(n) ** 2 b = np.arange(n) ** 3 c = a + b return a,b,c
(array([0, 1, 4, 9]), array([ 0, 1, 8, 27]), array([ 0, 2, 12, 36])) <type 'function'> [0 1 4 9] [ 0 1 8 27] [ 0 2 12 36]
比较用Python实现两个向量相加和用numpy实现两个向量相加的情况
size = 1000 start = datetime.now() c = pythonsum(size) delta = datetime.now() - start # print 'The last 2 elements of the sum',c[-2:] print 'pythonSum elapsed time in microseconds',delta.microseconds size = 1000 start1 = datetime.now() c1 = numpysum(size) delta1 = datetime.now() - start1 # print 'The last 2 elements of the sum',c1[-2:] print 'numpySum elapsed time in microseconds',delta1.microseconds
从下面程序运行结果我们可以看到在处理向量是numpy要比Python计算高出不知道多少倍
pythonSum elapsed time in microseconds 1000 numpySum elapsed time in microseconds 0
以上这篇关于Python中的向量相加和numpy中的向量相加效率对比就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持。
