比较两个列表的元素

A = [a1, a2, a3...]  #a1<a2<a3...
B = [b1, b2...]  #b1<b2<b3...

A 和 B 不相交。我事先不知道A/B中元素的数量和它们的值。我想比较列表和删除元素中元素的值：

delete a[i+1] if there is no b[j] such that a[i]<b[j]<a[i+1]
delete b[i+1] if there is no a[j] such that b[i]<a[j]<b[i+1]

最后，我想分开列表，而不是A和B的组合。

例如，如果A[0] < B[0]，A = [1, 10, 40]，B = [15, 30]。比较A[1] and B[0]第一的。删除10因为 B 中没有元素在 1 到 15 之间。 然后删除15因为 15 和 30 之间不再存在任何元素。输出应该是：如果您尝试对新的 2 个列表的元素进行排序，它应该是A[0]<B[0]<A[1]<B[1]<...

If A[0] > B[0]， 反之亦然。

a = [1, 10, 40]
b = [15, 30]

srcs = [a, b]
dsts = [[], []]
prev_which = -1
while all(srcs):
    which = int(srcs[0][0] > srcs[1][0])
    elem = srcs[which].pop(0)
    if prev_which != which:
        dsts[which].append(elem)
    prev_which = which
for src, dst in zip(srcs,dsts):
    if src:
        dst.append(src.pop(0))
a, b = dsts

returns:

a = [1, 40]
b = [15]

and for

a = [3, 4, 6, 7, 8, 9]
b = [1, 2, 5, 10]

它返回[3, 6] and [1, 5, 10].

EDIT: 另一种可能：

import itertools as it
import operator as op

a = [3, 4, 6, 7, 8, 9]
b = [1, 2, 5, 10]
srcs = [a, b]
dsts = [[], []]

for which, elems in it.groupby(sorted((x, i) for i in (0,1) for x in srcs[i]), key=op.itemgetter(1)):
    dsts[which].append(next(elems)[0])
a, b = dsts