组合大量字典键重叠间隔的值 [英] combining values for a large number of overlapping intervals of dictionary keys

问题描述

我有一个字典的字典有这样的项目

I have a dictionary of dictionaries that has items like this

all={
    1:{ ('a',123,145):20, ('a',155,170):12, ('b',234,345): 34},
    2:{ ('a',121,135):10, ('a',155,175):28, ('b',230,345): 16},
    3:{ ('a',130,140):20, ('a',150,170):10, ('b',234,345): 30}, 
    ...
    n: {...}
}

编辑：字典名称是由我根据从初始数据读取的文件名任意给出的，我可以使用任何值来命名这些字典。
我想获得每个重叠区域的这些值的总和。显示如何重叠应该是这样的输出是这样的

edit: The dictionary names are arbitrarily given by me according to the file names the initial data is read from, I can use any value I want to name these dictionaries. I would like to get the sum of these values for each overlapping region. The output showing how the overlaps should be like is this

 { ('a',121,122):10, ('a',123,130):30, ('a',131,135):50, 
   ('a',136,140):40,('a',141,145):20, ...}

编辑：每个字典都有不重叠的间隔，所以没有（'a'，2,10）和（'a' ，3,12），但由于开始和结束位置不一样（即字典之间的键不相同），字典之间的间隔重叠。

edit: Each dictionary has non-overlapping intervals so there never is ('a',2,10) and ('a',3,12) in a given dictionary but the intervals overlap between dictionaries as the start and end positions are not the same (i.e keys are not the same between dictionaries).

我不必使用字典数据结构，因为我首先创建了这个字典，如果这更容易做到列表，集合等，我可以得到这些结构之一的数据，我可以工作另一种解决方案也是基于不同的数据结构。

I don't have to use the dictionary data structure and since I have created this dictionary in the first place, if this is more easy to do with lists, sets etc I can get the data in one of those structures, I can work with another solution based on a different data structure as well.

感谢您的帮助。

推荐答案

认为我得到它：基本上你有一堆重叠的间隔，由具有给定厚度的某个位置的条形表示。你会把这些酒吧放在彼此之下，看看他们在任何一点上的浓度如何。

Ok, now i think i get it: Basically you have a bunch of overlapping intervals, represented by bars at a certain position with a given thickness. You would draw these bars below each other and see how thick they are together at any given point.

我认为滥用你的整数位置的事实是最简单/最快的这样做：

I think it's easiest/fastest to abuse the fact that you have integer positions to do this:

all={
    1:{ ('a',123,145):20, ('a',155,170):12, ('b',234,345): 34},
    2:{ ('a',121,135):10, ('a',155,175):28, ('b',230,345): 16},
    3:{ ('a',130,140):20, ('a',150,170):10, ('b',234,345): 30}
}

from collections import defaultdict
summer = defaultdict(int)
mini, maxi = 0,0
for d in all.values():
    for (name, start, stop), value in d.iteritems(): 
        # im completely ignoring the `name` here, not sure if that's what you want
        # else just separate the data before doing this ...
        if mini == 0:
            mini = start
        mini, maxi = min(mini, start), max(maxi, stop)
        for i in range(start, stop+1):
            summer[i]+=value

# now we have the values at each point, very redundant but very fast so  far
print summer

# now we can find the intervals:
def get_intervals(points, start, stop):
    cstart = start
    for i in range(start, stop+1):
        if points[cstart] != points[i]: # did the value change ?
            yield cstart, i-1, points[cstart]
            cstart = i

    if cstart != i:
        yield cstart, i, points[cstart]


print list(get_intervals(summer, mini, maxi))

当仅使用它给出的'a'项目时：

When using only the 'a' items it give:

[(121, 122, 10), (123, 129, 30), (130, 135, 50), (136, 140, 40), (141, 145, 20), (146, 149, 0), (150, 154, 10), (155, 170, 50), (171, 175, 28)]

编辑：它只是打我怎么做这个真的简单：

It just hit me how to do this really simple:

from collections import defaultdict
from heapq import heappush, heappop

class Summer(object):
    def __init__(self):
        # its a priority queue, kind of like a sorted list
        self.hq = []

    def additem(self, start, stop, value):
        # at `start` add it as a positive value
        heappush(self.hq, (start, value))
        # at `stop` subtract that value again
        heappush(self.hq, (stop, -value))

    def intervals(self):
        hq = self.hq
        start, val = heappop(hq)
        while hq:
            point, value = heappop(hq)
            yield start, point, val
            # just maintain the current value and where the interval started
            val += value
            start = point
        assert val == 0

summers = defaultdict(Summer)
for d in all.values():
    for (name, start, stop), value in d.iteritems():
        summers[name].additem(start, stop, value)

for name,s in summers.iteritems():
    print name, list(s.intervals())

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