以下的纯Python代码可以解决上述问题： def pythonsum(n): a = range(n) b = range(n) c = [] for i in range(len...datetime import datetime import numpy as np #省略上面两处代码 size = int(sys.argv[1]) start = datetime.now() c = pythonsum...(size) delta = datetime.now() - start print "The last 2 elements of the sum", c[-2:] print "PythonSum
#3次幂太大会溢出，注意(后面细说) b = np.arange(n, dtype = 'int64')**3 c = a + b return c #普通方法 def pythonSum...c.append(a[i] + b[i]) return c size = int(sys.argv[1]) start = datetime.now() c = pythonSum...(size) delta = datetime.now() - start print ("最后两个元素的和：", c[-2:]) print ("pythonSum花费的时间（微秒）", delta.microseconds
实现方式 实现：实现了两个向量的相加 # -*- coding: utf-8 -*- # 此处两种操作方式： # 第一种对于每一个元素的操作，第二种是对于整体的操作 # 向量相加-Python def pythonsum...import sys from datetime import datetime import numpy as np size = 1000 start = datetime.now() c = pythonsum...(size) delta = datetime.now() - start print "The last 2 elements of the sum", c[-2:] print "PythonSum...time in microseconds", delta.microseconds res: The last 2 elements of the sum [995007996, 998001000] PythonSum
实现方式 实现：实现了两个向量的相加 # -*- coding: utf-8 -*- # 此处两种操作方式： # 第一种对于每一个元素的操作，第二种是对于整体的操作 # 向量相加-Python def pythonsum...import sys from datetime import datetime import numpy as np size = 1000 start = datetime.now() c = pythonsum...(size) delta = datetime.now() - start print "The last 2 elements of the sum", c[-2:] print "PythonSum...time in microseconds", delta.microseconds res: The last 2 elements of the sum 995007996, 998001000 PythonSum
以下函数使用不带 NumPy 的纯 Python 解决了向量加法问题： def pythonsum(n): a = range(n) b = range(n) c = [] for...以下程序将通过测量numpysum和pythonsum函数的经过时间（以微秒为单位）向我们展示。 它还打印向量和的最后两个元素。...def numpysum(n): a = np.arange(n) ** 2 b = np.arange(n) ** 3 c = a + b return c def pythonsum...** 3 c.append(a[i] + b[i]) return c size = int(sys.argv[1]) start = datetime.now() c = pythonsum...(size) delta = datetime.now() - start print "The last 2 elements of the sum", c[-2:] print "PythonSum
使用纯 Python 相加向量：以下函数使用不带 NumPy 的纯 Python 解决了向量相加问题： def pythonsum(n): a = range(n) b = range(n)...以下程序将通过为numpysum()和pythonsum()函数测量经过的时间（以微秒为单位）向我们展示。 它还打印向量和的最后两个元素。...def numpysum(n): a = np.arange(n) ** 2 b = np.arange(n) ** 3 c = a + b return c def pythonsum...** 3 c.append(a[i] + b[i]) return c size = int(sys.argv[1]) start = datetime.now() c = pythonsum...(size) delta = datetime.now() - start print("The last 2 elements of the sum", c[-2:]) print("PythonSum
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