倒数

如何解决倒数

| 我有一个这样组织的清单：

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

最好的方式是“向后比较？”，基本上，如果我们得到一系列2次“下降”后跟一个“”，则我想返回“是”（或其他任何一种。）。并且第二个值小于0.0095。 我希望他有道理..     

解决方法

干得好：

def frob(l):
    downcount = 0
    for ele in l:
        if downcount >= 2 and ele[0] == \'up\' and ele[1] < 0.0095:
                return True
        downcount = (downcount + 1) if ele[0] == \'down\' else 0
    return False

    ,创建一个滑动窗口，并对此进行测试：

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

    ,

for index in xrange(0,len(list) - 2):
    if list[index][0] == \'down\' and list[index + 1][0] == \'down\' and list[index + 2][0] == \'up\' and list[index + 1][1] < 0.0095:
         return True

    ,这是我的尝试：

def test(data):
  for x in xrange(2,len(data)):
    if data[x-2][0] is \'down\' and data[x][x-1] is \'down\' and data[x][0] is \'up\' and data[x][1] < 0.0095:
      return True
  return False

    ,我的建议（尽管现在有了第三部分，这已经不再是那么漂亮了）：

def compback(l):
    return any(i1[0] == i2[0] == \"down\" 
               and i2[1] < 0.0095
               and i3[0] == \"up\"
               for i1,i2,i3 in zip(l,l[1:],l[2:]))