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I seem to have developed the right reduce operation to find the union of intervals, only to realize that reduce gives you a final result. So I looked up the documentation and figured out that what I should be using is in fact accumulate.

I need someone to help me convert this reduce to accumulate so I have the intermediate intervals

The code below is an example of how I used reduce. I'm assuming that the intermediate values can be stored using accumulate. I'm not sure if this is even possible.. But I looked at examples how accumulate gives you a list of items where each item is an intermediate calculated result.

example_interval = [[1,3],[2,6],[6,10],[15,18]]

def main():

    def function(item1, item2):


        if item1[1] >= item2[0]:

            return item1[0], max(item1[1], item2[1])

        else:

            return item2

    return reduce(function, example_interval)

To understand the problem, [1, 3], [2, 6] can be reduced to [1, 6] since item1[1] >= item2[0], [1, 6] is then taken in as item1 and then compared to [6,10] which is item2, to give [1, 10]. [1, 10] is then compared with the final item [15, 18], in this case, it is not merged, so the final result is [1, 10], [15, 18].

I do know how to do this problem without reduce and accumulate. I just have a general interest in understanding how I can use accumulate to replicate this task where intermediate values are stored.

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1 回答 1

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from itertools import accumulate

def function(item1, item2):
    if item1[1] >= item2[0]:
        return item1[0], max(item1[1], item2[1])
    return item2

example_interval = [(1,3),(2,6),(6,10),(15,18)]
print(list(accumulate(example_interval, function)))

结果是：

[(1, 3), (1, 6), (1, 10), (15, 18)]

请注意，我将项目example_interval从列表更改为元组。如果你不这样做，when item1[1] < item2[0]，返回的值是item2 一个列表对象，但是 if item[1] >= item2[0]，返回的表达式是item1[0], max(item1[1], item2[1])，它被转换为一个元组：

example_interval = [[1,3],[2,6],[6,10],[15,18]]
print(list(accumulate(example_interval, function)))

现在输出是：

[[1, 3], (1, 6), (1, 10), [15, 18]]
于 2019-07-26T20:14:49.630 回答