序列中连续值之间的差值列表

标签：job   next   print   bsp   OLE   tuples   .com   一个   ble   

 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)
]

序列中连续值之间的差值列表

标签：job   next   print   bsp   OLE   tuples   .com   一个   ble   

原文地址：https://www.cnblogs.com/qingsheng/p/9191623.html