Question

假设我有四个间隔：

a = pd.Interval(0, 2)
b = pd.Interval(1, 2)
c = pd.Interval(0, 1.1)
d = pd.Interval(-1,-0.5)

最常见/最常见的重叠是1到1.1之间，因为所有a，b和c（上面四个中的三个）都覆盖了该重叠。

在给定N个间隔列表的情况下，如何找到它？理想情况下，我想要一个可以接受上述4个间隔的列表并返回pd.Interval(1, 1.1)作为输出的函数。

Answer 1

我以为您正在尝试获取大多数其他间隔中包含的间隔，因为具有最常见重叠的间隔是(-1,1.1]而不是(1,1.1]，我不知道是否我错了，但是使用pd.IntervalsIndex.overlaps我得到了：

import pandas as pd
from itertools import product

a = pd.Interval(0, 2)
b = pd.Interval(1, 2)
c = pd.Interval(0, 1.1)
d = pd.Interval(-1,-0.5)
ls=[a,b,c,d]
indexinter = pd.IntervalIndex(ls)
print(indexinter.overlaps(pd.Interval(1, 1.1)))
print('Interval (1,1.1] overlaps ',sum(indexinter.overlaps(pd.Interval(1, 1.1))),' of 4')
print('\n')
print(indexinter.overlaps(pd.Interval(-1, 1.1)))
print('Interval (-1,1.1] overlaps ',sum(indexinter.overlaps(pd.Interval(-1, 1.1))),' of 4')

输出：

[ True  True  True False]
Interval (1,1.1] overlaps  3  of 4


[ True  True  True  True]
Interval (-1,1.1] overlaps  4  of 4

没什么，要获取大多数其他间隔中包含的间隔，您可以尝试以下方法：

第一个选项

import pandas as pd
import portion as P

d = pd.Interval(0, 2)
b = pd.Interval(1, 2)
c = pd.Interval(0, 1.1)
a = pd.Interval(-1,-0.5)

def getmaxoverlapwrap(ls,inter):
    intervals=[P.openclosed(i.left, i.right) for i in ls]
    for i in intervals[1:]:
        newinter=inter&i
        if newinter.empty==False:
            inter=newinter     
    return inter
def getmaxoverlap(ls):
    intervals=[P.openclosed(i.left, i.right) for i in ls]
    ls1=[getmaxoverlapwrap(ls,i) for i in intervals]
    index = list(map(lambda x: sum([(x in s) for s in ls]), ls1)).index(max(list(map(lambda x: sum([(x in s) for s in ls]), ls1))))
    return ls1[index]
ls=[a,b,c,d]
indexinter = pd.IntervalIndex(ls)
print(getmaxoverlap(ls))

您必须安装库portion（以前以python-intervals的形式分发），该库为Python 3.5+中的间隔提供了数据结构和操作，例如：-支持任意（可比）间隔）对象，-封闭或开放，有限或（半）无限间隔，-间隔集（原子间隔的联合），-间隔的自动简化等。

第二个选项

import pandas as pd
from itertools import product

a = pd.Interval(0, 2)
b = pd.Interval(1, 2)
c = pd.Interval(0, 1.1)
d = pd.Interval(-1,-0.5)

def getmostcontained(ls):
    indexinter = pd.IntervalIndex([pd.Interval(inter.left, inter.right, closed='both') for inter in ls])
    rights = [i.right for i in ls]
    lefts=[i.left for i in ls]
    allcomb = [pd.Interval(i[0],i[1]) for i in list(set(list(product(lefts, rights)))) if i[0]<=i[1]]
    count=0
    maxi=sum(indexinter.contains(allcomb[0].left))+sum(indexinter.contains(allcomb[0].right))
    for i in allcomb:
        newmaxi=sum(indexinter.contains(i.left))+sum(indexinter.contains(i.right))

        if newmaxi>maxi:
            count=allcomb.index(i)

    return allcomb[count]

ls=[a,b,c,d]
print(getmostcontained(ls))

第三种选择

import pandas as pd

a = pd.Interval(0, 2)
b = pd.Interval(1, 2)
c = pd.Interval(0, 1.1)
d = pd.Interval(-1,-0.5)

def getmaxoverlap(ls):
    indexinter = pd.IntervalIndex(ls)
    start=ls[0].left
    end=ls[0].right
    checkstart=sum(indexinter.contains(start+0.00000000099))
    checkend=sum(indexinter.contains(end))


    for i in ls:
        if sum(indexinter.contains(i.left+0.000000000099))>checkstart:
            checkstart=sum(indexinter.contains(i.left+0.000000000099))
            start=i.left
        if sum(indexinter.contains(i.right))>checkend:
            checkend=sum(indexinter.contains(i.right))
            end=i.right

    return pd.Interval(start, end)

ls=[a,b,c,d]
print(getmostcontained(ls))

所有选项的输出：

(1,1.1]

第一个选项花费0.025967300000047544 seconds运行脚本的100倍，并花费27.19270740000138 seconds运行脚本的100000倍，第二个选项花费0.15769210000144085 seconds运行的脚本100倍，并且花费115.13360900000043 seconds运行脚本的100000倍，第三种选择花费0.08970480000061798 seconds运行脚本的100倍，并花费107.52801619999991 seconds运行脚本的100000倍。至少在第二个和第三个选项上，这种差异可能几乎不明显，但是这里是我得到的时间戳，以便您知道并选择一个。

如何找到大熊猫重叠最多的区间

1 个答案:

第一个选项

第二个选项

第三种选择