计算后总值不变(total value is not changing after calculation)

 第一部分代码实际上是误导性的，并且依赖于lena是方形图像的事实：发生的事情等同于调用zip(range(xmax), range(ymax)) ，然后将每个结果元组设置为0 。 你可以看到这里可能出现的问题：如果xmax != ymax ，那么事情将无效： 
 
>>> test = lena[:,:-3]
>>> test.shape
(512, 509)
>>> xmax, ymax = test.shape
>>> test[range(xmax), range(ymax)] = 0
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: shape mismatch: objects cannot be broadcast to a single shape
 
 定义diag_max = min(xmax, ymax) ，然后设置lena[range(diag_max), range(diag_max)] = 0可能会更好。 
 
 第二个问题的答案更容易： range(from, to, step)是对range的一般调用： 
 
>>> range(1, 10, 2)
[1, 3, 5, 7, 9]
>>> range(1, 10, -2)
[]
>>> range(10, 1, -2)
[10, 8, 6, 4, 2]
>>> range(10, 0, -1)
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
 
 特别是，这会反转先前的列表，因此从右到左而不是从左到右抓取对角线。 

The first bit of code is actually misleading, and relies on the fact that lena is a square image: what happens is equivalent to calling zip(range(xmax), range(ymax)), and then setting each of the resulting tuples to 0. You can see what could go wrong here: if xmax != ymax, then things won't work:
 
>>> test = lena[:,:-3]
>>> test.shape
(512, 509)
>>> xmax, ymax = test.shape
>>> test[range(xmax), range(ymax)] = 0
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: shape mismatch: objects cannot be broadcast to a single shape
 
It would probably be better to define diag_max = min(xmax, ymax), and then set lena[range(diag_max), range(diag_max)] = 0. 
 
The answer to your second question is easier: range(from, to, step) is the general call to range:
 
>>> range(1, 10, 2)
[1, 3, 5, 7, 9]
>>> range(1, 10, -2)
[]
>>> range(10, 1, -2)
[10, 8, 6, 4, 2]
>>> range(10, 0, -1)
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
 
In particular, this reverses the previous list, and so grabs the diagonal from right to left instead of left to right.