转自Stackoverflow.备忘用. Question In Python 2 I could do the following: import numpy as np f = lambda x: x**2 seq = map(f, xrange(5)) seq = np.array(seq) print seq # prints: [ 0 1 4 9 16] In Python 3 it does not work anymore: import numpy as np f = lambd
对象的引用 看例子: a = np.array([0, 1, 2, 3]) b = a a[0] = 5 print("b=", b) # 判断a和b是否是同样的地址 print(b is a) 运行结果: b= [5 1 2 3] True 上面的例子中,我们改变了a的值,但打印出来b中的值也被修改了,原因是a和b指向相同的对象. 复制 如果我们想要解决修改了a的值不会影响到b中的值,该如何做到呢? import numpy as np a = np.array([0, 1, 2,