如何使用来自另一个字典的聚合值创建一个新字典

Question

I have a dictionary with time of day as the key and stock price as the value. There is a key for each minute in a day:

original_dict = {'09:30': '2170.8601', '09:31': '2180.21', '09:32': '2181.76'... '16:15': '2179.98'}

I am trying to create a new dictionary which will have times as keys and values to be a list consisting of open, high, low and close stock prices (values from the original dictionary) using rolling 10 minute increments:

new_dict = {'09:40': [open, high, low, close], '09:41': [open, high, low, close]...}

( new_dict starting at 9:40 because 10 minutes is required for the first key/list to be calculated)

For example, the list of values for the 9:40 key in the new_dict should be calculated using the values of 9:30-9:39 in the original_dict . The open, high, low and close are the 9:30 , max of 9:30-9:39 , min of 9:30-9:39 , and 9:39 values from original_dict , respectively. The 9:41 key will have open, high, low, close values calculated using 9:31-9:40 values from original_dict and so on.

I am not sure how to easily do a for statement, for example, using original_dict since it is not a list. Should I create an intermediate list of keys from original_dict and use that to iterate through times?

What would be the best way to create new_dict ?

Thank you!

Answer 1

It would require a fair bit of coding so I'm going to instead write out steps that should help you get the job done.

1. Convert the String keys ('09:30') to numbers('09:30' is 9*60 + 30 = 2130 hours).
2. Iterate from the earliest minute in your dictionary to the last minute, and store the values associated with each minute in a list. You should have a list of values sorted chronologically now.
3. Maintain a sliding window maximum queue and sliding window minimum queue, with the window size in your case being 10.
4. Then at every step, fetch the 'high price' from the sliding window maximum queue, and likewise for the 'low price'. For the 'start price', you just need to look back 10 indexes behind in the sorted list of values we earlier formed. For the 'end price', we look behind just a single index.

And that should be it. Perhaps the tricky part here would be implementing the Sliding Window Maximum/Minimum, you could take the following implementation of a function that takes in a list of numbers, and returns another list of of the maximum of every contiguous window of size k in our list. Runs in O(N) time where N is the size of the list.

def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
    deque = collections.deque()
    res = []
    for i, num in enumerate(nums):
        while(deque and nums[deque[-1]] < num):
            deque.pop()
        if(deque and i - deque[0] >= k):
            deque.popleft()
        deque.append(i)
        res.append(nums[deque[0]])
    return res[k-1:]

You'd similarly obtain Sliding Window Minimum, and with that you should have hopefully enough to get it working.