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python之倒数

2025年05月04日11oomusou

我有一个这样组织的列表:

[('down', 0.0098000000000000309),  
('up', 0.0015000000000000568),  
('down', 0.008900000000000019),  
('down', 0.023300000000000098),  
('down', 0.011599999999999944),  
('down', 0.0027000000000000357),  
('up', 0.0023999999999999577),  
('up', 0.0065000000000000613),  
('down', 0.0057000000000000384),  
('down', 0.018400000000000083),  
('up', 0.009300000000000086),  
('down', 0.0038000000000000256),  
('down', 0.00050000000000005596),  
('up', 0.0082000000000000961), ..... 

“向后比较”的最佳方式是什么? ,基本上我想返回"is"(或其他什么......)如果我们有一系列的 2 个“下降”然后是一个“上升”并且第二个值低于 0.0095。

我希望他说得有道理..

请您参考如下方法:

创建一个滑动窗口,并对其进行测试:

def slidingwindow(iterable): 
    iterator = iter(iterable) 
    first, second = iterator.next(), iterator.next() 
    for next in iterator: 
        yield (first, second, next) 
        first, second = second, next 
 
def testforcondition(data): 
    for window in slidingwindow(data): 
        direction = [w[0] for w in window] 
        if direction == ['down', 'down', 'up'] and window[2][1] < 0.0095: 
            return True 
    return False