倒計數

Question

我有這樣的組織的清單：「比較落後」

[('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), ..... 

什麼將是最好的方式，基本上我想返回「是」（或者其他）。如果我們得到了一系列2個「下降」，然後是一個「上升」，第二個值低於0.0095。

我希望他是有道理的..

Answer 1

這裏你去：

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 

Answer 2

這裏是我的嘗試：

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 

Answer 3

我的建議（雖然現在與第三片，這是不是真的漂亮了）：

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:])) 

Answer 4

上創建一個滑動窗口，並測試：

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 

Answer 5

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