面向对象的条件更具特殊性-Python

这是我上一个问题的延续，使用第二个答案-

我在比较DNA中的核苷酸位置，这些位置以[染色体，起始，终止]的形式存在，我有一个现有位置的列表和一个新位置的列表。其目的是将新位置列表与现有位置进行比较，并添加一个新位置元素（如果它是唯一的或与现有位置重叠）。如果重叠，则应报告。只有当新的position元素完全包含在现有元素中时，才应将其丢弃

感谢@Hugh Bothwell-我目前拥有以下代码：

class ChromoSegments:
    def __init__(self, ChromoSegments_args=None):
        #Creates an empty default dict list which can be added to in the style "chromo[start,end]", and will stay ordered
        self.segments = defaultdict(list)
        #If a list is passed to the constructor you can add values to your list depending on conditions in 'add_seg'
        if ChromoSegments_args is not None:
            for chromo,start,end in ChromoSegments_args:
                try:
                    self.add_seg(chromo, start, end)
                except ValueError:
                    pass
    #function for adding positions to the expos list
    def add_seg(self, chromo, start, end):
        seg = self.segments[chromo]
        val = (start, end)
        ndx = bisect_left(seg, val)
        if (ndx == 0 or seg[ndx - 1][1] < start):           
            if (ndx == len(seg) or end < seg[ndx][0]):
                seg.insert(ndx, val)
            else:
                nstart, nend = seg[ndx]
                raise ValueError("Hit ({}, {}, {}) \t\t\t overlaps with ({}, {}, {})".format(chromo, start, end, chromo, nstart, nend))
        #collision with preceding element
        else:
             nstart, nend = seg[ndx - 1]
             raise ValueError("Hit ({}, {}, {}) \t\t\t overlaps with ({}, {}, {})".format(chromo, start, end, chromo, nstart, nend))

    def to_list(self):
        keys = sorted(self.segments.keys())
        return [(k, s, e) for k in keys for s,e in self.segments[k]]

def main():
    expos = ChromoSegments(expos_list)
    newpos = (newpos_list)

    error_file = open("discarded_hits.txt", "w")
    for seg in newpos:
        try:
        expos.add_seg(*seg)
        except ValueError, e:
                collision = str(e)
        error_file.write(collision + "\n")
    error_file.close()

    #convert results back into text files of positions
    updated_expos = expos.to_list()
    updated_expos_file = open(sys.argv[2], "w")
    for element in updated_expos:
        c1 = str(element[0])
        c2 = str(element[1])
        c3 = str(element[2])
        updated_expos_file.write(c1 + "\t" + c2 + "\t" + c3 + "\n")
    updated_expos_file.close()

if __name__ == "__main__":
    main()

因此，在一个成功的过程之后，我想要一个最终的曝光列表（记住第一个数字（目前在所有情况下为1）可以是任何数字：

expos_list = [[1, 12, 25], [1, 20, 40], [1, 60, 80], [1, 75, 90], [1, 100, 150]]

我试一下：

START = 1
END = 2
expos_list = [[1, 20, 40], [1, 60, 80]]
newpos_list = [[1, 12, 25], [1, 22, 38], [1, 75, 90], [1, 100, 150]]
# sort the new list
expos_list.extend(newpos_list)
expos_list.sort(lambda l1, l2: cmp(l1[0], l2[0]) or (l1[0] == l2[0] and (cmp(l1[1], l2[1]) or (l1[1] == l2[1] and (cmp(l1[2], l2[2]))))))

# remove and print the overlaps
i = 0
l = expos_list
while i < len(l) - 1:
    if l[i][END] > l[i + 1][START]:
        print 'overlap', l[i], l[i+1]
        if l[i][START] <= l[i+1][START] and l[i][END] >= l[i+1][END]:
            # i+1 is in i
            l.pop(i+1)
        elif l[i][START] >= l[i+1][START] and l[i][END] <= l[i+1][END]:
            # i is in i+1
            l.pop(i)
        else:
            # there is a partial overlap
            i += 1
    else:
        i += 1

# overlap [1, 12, 25] [1, 20, 40]
# overlap [1, 20, 40] [1, 22, 38]
# [1, 22, 38] # removed
# overlap [1, 60, 80] [1, 75, 90]

# outcome
print expos_list
# [[1, 12, 25], [1, 20, 40], [1, 60, 80], [1, 75, 90], [1, 100, 150]]

START=1
结束=2
expos_list=[[1,20,40]，[1,60,80]]
newpos_list=[[1,12,25]、[1,22,38]、[1,75,90]、[1100,150]]
#对新列表排序
expos\u list.extend（newpos\u list）
expos_list.sort（lambda l1，l2:cmp（l1[0]，l2[0]）或（l1[0]==l2[0]和（cmp（l1[1]，l2[1]）或（l1[1]==l2[1]和（cmp（l1[2]，l2[2]））；）
#删除并打印重叠部分
i=0
l=曝光列表
而il[i+1][START]：
打印“重叠”，l[i]，l[i+1]
如果l[i][START]=l[i+1][END]：
#i+1在i中
l、 流行音乐（i+1）
elif l[i][START]>=l[i+1][START]和l[i][END]看起来像是一个一维几何相交问题。你可以在“网络”上找到解决方案。@asmason似乎你有混合的制表符/空格。我修复了代码格式，请检查正确性。@alko-是的，当我把它放在问题中时，它添加了我原始代码中没有的空格。如果我没有编辑，很抱歉哼哼，不正确！另外：Ber-由于细节原因，找不到任何真正有用的东西。尤其是它以这种方式使用字典。4个问题：庞大的代码示例是否有助于理解如何插入列表？你能缩短问题以便更多人阅读吗？如果我插入
[2,20,40]会发生什么
转换为
[[1，…]，…，[1，…]]
？expos_list的元素是否总是在结果中粘在一起？Hi@User-我知道，它很长，但因为函数是链接的，所以我不想被问到变量来自何处等。核心是
add_seg
函数，但我希望人们有上下文。我在上一个问题中因为没有上下文而受到批评ng足够特异…第三个-它被视为与第一个数字（染色体）一样唯一是不同的，所以会被添加到列表中。第4-粘在一起？不确定你的意思，但是-
expos\u list
始终作为参考，并且
平分左
保持列表的顺序以便不断比较感谢答案，但我不只是想将列表连接在一起，而是要有选择性地选择如何
ex位置列表继续。使用上面的示例-从碱基20到40的染色体区域已添加到列表中。因此，从碱基22到38的区域已添加到列表中，您不希望重复。但是，如果存在重叠（提供新序列）这需要记录-例如区域12到25
[1，12，25]
。这就是为什么我需要一个额外的条件声明，我想。谢谢！非常感谢这一点-我已经根据我的输入调整了它，它工作得很好。很好，很简单！为strand添加了额外的条件（末端可能比开始的数字低）染色体。
START = 1
END = 2
expos_list = [[1, 20, 40], [1, 60, 80]]
newpos_list = [[1, 12, 25], [1, 22, 38], [1, 75, 90], [1, 100, 150]]
# sort the new list
expos_list.extend(newpos_list)
expos_list.sort(lambda l1, l2: cmp(l1[0], l2[0]) or (l1[0] == l2[0] and (cmp(l1[1], l2[1]) or (l1[1] == l2[1] and (cmp(l1[2], l2[2]))))))

# remove and print the overlaps
i = 0
l = expos_list
while i < len(l) - 1:
    if l[i][END] > l[i + 1][START]:
        print 'overlap', l[i], l[i+1]
        if l[i][START] <= l[i+1][START] and l[i][END] >= l[i+1][END]:
            # i+1 is in i
            l.pop(i+1)
        elif l[i][START] >= l[i+1][START] and l[i][END] <= l[i+1][END]:
            # i is in i+1
            l.pop(i)
        else:
            # there is a partial overlap
            i += 1
    else:
        i += 1

# overlap [1, 12, 25] [1, 20, 40]
# overlap [1, 20, 40] [1, 22, 38]
# [1, 22, 38] # removed
# overlap [1, 60, 80] [1, 75, 90]

# outcome
print expos_list
# [[1, 12, 25], [1, 20, 40], [1, 60, 80], [1, 75, 90], [1, 100, 150]]

lambda l1, l2: l1[0] < l2[0] or (l1[0] == l2[0] and (l1[1] < l2[1] or (l1[1] == l2[1] and (l1[2] <= l2[2]))))