比较两个列表的元素 [英] compare elements of two lists

问题描述

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

A和B不相交.我不知道元素的数量和它们在A/B中的价值是否提前.我想比较列表和删除元素iff中的元素的值:

A and B are disjoint. are I do not know the number of elements and the value of them in A/B in advance. I want to compare the value of the elements in both list and delete elements iff:

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，我想分隔列表，而不是A和B的组合.

At the end, I want to separate list, not a combination of A and B.

例如，如果A[0] < B[0]，则A = [1，10，40]，B = [15，30].首先比较A[1]和B[0].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:如果尝试对新的2个列表的元素进行排序，则应该为A[0]<B[0]<A[1]<B[1]<...

For example, If A[0] < B[0], A = [1, 10, 40], B = [15, 30]. Compare A[1] and B[0] first. Delete 10 because no element in B are in between 1 and 15. Then delete 15 since no element exist anymore btw 15 and 30. The output should be: if you try to order the elements of the new 2 lists, it should be A[0]<B[0]<A[1]<B[1]<...

如果是A[0] > B[0]，反之亦然.

推荐答案

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

返回:

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

和

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

它返回[3, 6]和[1, 5, 10].

编辑:另一种可能性:

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

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