1 readings=[1,8,3,4,9,6,7] 2 current=readings[0] 3 defferences=[] 4 for next_item in readings[1:]:#注意next_item的值 5 defferences.append(next_item-current) 6 current=next_item 7 print(current) 8 9 for i in defferences: 10 print(i) 11 12 #辅助函数 13 def with_next(iterable): 14 '''yield(current,next_item) tuples for each item in iterable.''' 15 iterator =iter(iterable) 16 current=next(iterator) 17 for next_item in iterator: 18 yield current,next_item 19 current=next_item 20 21 differences=[] 22 for current,next_item in with_next(readings): 23 differences.append(next_item-current) 24 25 defferences2=[ 26 27 (next_item-current) for current,next_item in with_next(readings) 28 ]
迭代器每迭代一次就会消耗一个元素，是一次性的。所以会得到对迭代器对象求和结果为0
http://python.jobbole.com/89181/
readings=[1,8,3,4,9,6,7]
current=readings[0]
defferences=[]
for next_item in readings[1:]:#注意next_item的值
defferences.append(next_item-current)
current=next_item
print(current)
for i in defferences:
print(i)
#辅助函数
def with_next(iterable):
'''yield(current,next_item) tuples for each item in iterable.'''
iterator =iter(iterable)
current=next(iterator)
for next_item in iterator:
yield current,next_item
current=next_item
differences=[]
for current,next_item in with_next(readings):
differences.append(next_item-current)
defferences2=[
(next_item-current) for current,next_item in with_next(readings)
]