基于类的方向指示器 [英] Class based directional indicator
问题描述
我创建了一个给定的天数基于类方向指示器（ n_days ）和号码的列表，它给号码的（数出来的最近 n_days 上的数量比previous数量减去 n_days 出$的对更高$ pvious n_days 上的号码去了）。所以，如果我想它返回列表的数量增加 +1 ，如果它减少我希望它返回 1 ，否则应 0 所以第一个数字应始终是 0 ，因为你不能把它比作任何东西。然后根据 n_days 我基本上要采取的近期n_days的总和，因此，例如在列表[1,2,2 ，1,2,1] 的变化应 [0，+ 1,0，1,1，-1] ，如果我希望每个日子，所以它应该是 [0，+ 1，-1,0，+ 1,0] ，因为对在2号最近变化的总和第一天只有0，第二天你把最新的2天 0 +（+ 1）= 1 ，第三天<$ C $的总和C>（+ 1）+ 0 = + 1 ，第四天 0 +（ - 1）= - 1 等等。这里是我的code，它不工作：
I'm creating a class based directional indicator that given a number of days (n_days) and a list of numbers, it gives the (number of numbers out of the most recent n_days on which the number was higher than the previous number minus the n_days out of the previous n_days on which the number went down). So if the number in the list increases I want it to return +1, if it decreases I want it to return -1, otherwise it should be 0 so the first number should always be 0 since you can't compare it to anything. Then based on n_days I basically want to take the sum of the of the recent n_days, so for example in a list of [1,2,2,1,2,1] the change should be [0,+1,0,-1,1,-1] and if I want the sum of the change in the 2 recent numbers for each day so it should be [0,+1,-1,0,+1,0] because on the first day there is only 0, on the second day you take the sum of the most recent two days 0+(+1)=1, the third day (+1)+0=+1, the fourth day 0+(-1)=-1 and so forth. Here is my code that's not working:
class Directionalindicator():
def __init__(self, n_days, list_of_prices):
self.n_days = n_days
self.list_of_prices = list_of_prices
def calculate(self):
change = []
for i in range(len(self.list_of_prices)):
if self.list_of_prices[i-1] < self.list_of_prices[i]:
change.append(1)
elif self.list_of_prices[i-1] > self.list_of_prices[i]:
change.append(-1)
else:
change.append(0)
directional = []
for i in range(len(change)):
directional.append(sum(change[i+1-self.n_days:i+1]))
return directional
与测试它：
y = Directionalindicator(2,[1,2,2,1,2,1])
y.calculate()
应该返回：
[0,+1,+1,-1,0,0]
和它。
但随着测试它：
y = Directionalindicator(3, [1,2,3,4,5,6,7,8,9,10])
y.calculate()
应该返回
[0, 0, 2, 3, 3, 3, 3, 3, 3, 3]
但它返回
[0, 0, 1, 3, 3, 3, 3, 3, 3, 3]
我打印的变化，看它是做什么，第一个值是-1，而不是0。此外，在回答一个code工作使用拉链，但我不明白为什么我的没有按T该列表1-10工作。
I printed change to see what it was doing and the first value is a -1 instead of a 0. Also, the code in one of the answers works using zip, but I don't understand why mine doesn't work for that list from 1-10.
推荐答案
您比较
i > i-1
将总是的是真。你正在减一，这永远是较小每个价格比较本身。相反，你应该比较对的价格。 拉链是有用的：
will always be True. You are comparing each price to itself minus one, which will always be smaller. Instead, you should be comparing pairs of prices. zip is useful for this:
change = [0] # first price always zero change
for a, b in zip(self.list_of_prices, self.list_of_prices[1:]):
if a < b: # price went up
change.append(1)
elif a > b: # price went down
change.append(-1)
else: # price stayed the same
change.append(0)
在此插入到你的code和使用您的例子
When you plug this into your code and use your example
Directionalindicator(2, [1, 2, 2, 1, 2, 1])
您可以：
change == [0, 1, 0, -1, 1, -1]
directional == [0, 1, 1, -1, 0, 0]
这似乎根据你的规则最初说法是正确的，但由于某些原因不符合你的期望的输出 [0，1，-1，0，1，0] 从你的问题的结束。
This seems to be correct according to your initial statement of the rules, but for some reason doesn't match your "expected output" [0, 1, -1, 0, 1, 0] from the end of your question.
您的编辑不工作的原因是您正在使用的列表上的指数 I 。当我== 0 ， I-1 == -1 。当作为索引使用 list_of_prices [-1] ，这让你在列表的最后一个元素。因此，变更包含 [ - 1，1，1，1，1，1，1，1，1，1] ，因为它比较 1 与 10 ，而不是 [0，1，1，1 ，1个，1个，1个，1个，1个，1] 如您所愿。
The reason your edit doesn't work is that you are using an index i on the list. When i == 0, i-1 == -1. When used as an index list_of_prices[-1], this gives you the last element in the list. Therefore change contains [-1, 1, 1, 1, 1, 1, 1, 1, 1, 1], as it compares 1 with 10, not [0, 1, 1, 1, 1, 1, 1, 1, 1, 1] as you expected.
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