为什么使用zip对象的列表理解导致空列表？

Question

f = lambda x : 2*x
g = lambda x : x ** 2
h = lambda x : x ** x
funcTriple = ( f, g, h )
myZip = ( zip ( funcTriple, (1, 3, 5) ) )
k = lambda pair : pair[0](pair[1])

# Why do Output # 1 (2, 9, 3125) and Output # 2 ( [ ] ) differ?

print ("\n\nOutput # 1:  for pair in myZip: k(pair) ...")
for pair in myZip :
    print ( k(pair) )

print ("\n\nOutput # 2:  [ k(pair) for pair in myZip ] ...")
print ( [ k(pair) for pair in myZip ] )

# script output is ...
# Output # 1:  for pair in myZip: k(pair) ...
# 2
# 9
# 3125
# 
# Output # 2:  [ k(pair) for pair in myZip ] ...
# []

Answer 1

在Python 2.6中完美地工作但在Python 3.0中失败,因为zip返回一个生成器样式的对象,并且第一个循环耗尽它.改为列出一个清单:

myZip = list( zip ( funcTriple, (1, 3, 5) ) )

它适用于Python 3.0