Question

我有三个列表，想要排序并生成两个新列表。任何人都可以告诉我们该怎么做？

list1=[12,25,45], list2=[14,69], list3=[54,98,68,78,48]

我想打印输出

chosen1=[12,14,54], rest1=[25,45,69,98,68,78,48]
chosen2=[12,14,98], rest2=[25,45,69,54,68,78,48]

等等（所选列表的每种可能组合）

我试过写这个，但我不知道

list1=[12,25,45]
list2=[14,69]
list3=[54,98,68,78,48]
for i in xrange (list1[0],list1[2]):
for y in xrange(list2[0], list2[1]):
for z in xrange(list[0],list[4])
      for a in xrange(chosen[0],[2])
chosed1.append()
for a in xrange(chosen[0],[7])
rest1.append()
Print rest1
Print chosen1

Answer 1

itertools.product生成从不同的事物中选择一件事的所有排列：

import itertools

list1 = [12,25,45]
list2 = [14,69]
list3 = [54,98,68,78,48]

for i,(a,b,c) in enumerate(itertools.product(list1,list2,list3),1):
    # Note: Computing rest this way will *not* work if there are duplicates
    #       in any of the lists.
    rest1 = [n for n in list1 if n != a]
    rest2 = [n for n in list2 if n != b]
    rest3 = [n for n in list3 if n != c]
    rest = ','.join(str(n) for n in rest1+rest2+rest3)
    print('chosen{0}=[{1},{2},{3}], rest{0}=[{4}]'.format(i,a,b,c,rest))

输出：

chosen1=[12,14,54], rest1=[25,45,69,98,68,78,48]
chosen2=[12,14,98], rest2=[25,45,69,54,68,78,48]
chosen3=[12,14,68], rest3=[25,45,69,54,98,78,48]
chosen4=[12,14,78], rest4=[25,45,69,54,98,68,48]
chosen5=[12,14,48], rest5=[25,45,69,54,98,68,78]
chosen6=[12,69,54], rest6=[25,45,14,98,68,78,48]
chosen7=[12,69,98], rest7=[25,45,14,54,68,78,48]
chosen8=[12,69,68], rest8=[25,45,14,54,98,78,48]
chosen9=[12,69,78], rest9=[25,45,14,54,98,68,48]
chosen10=[12,69,48], rest10=[25,45,14,54,98,68,78]
chosen11=[25,14,54], rest11=[12,45,69,98,68,78,48]
chosen12=[25,14,98], rest12=[12,45,69,54,68,78,48]
chosen13=[25,14,68], rest13=[12,45,69,54,98,78,48]
chosen14=[25,14,78], rest14=[12,45,69,54,98,68,48]
chosen15=[25,14,48], rest15=[12,45,69,54,98,68,78]
chosen16=[25,69,54], rest16=[12,45,14,98,68,78,48]
chosen17=[25,69,98], rest17=[12,45,14,54,68,78,48]
chosen18=[25,69,68], rest18=[12,45,14,54,98,78,48]
chosen19=[25,69,78], rest19=[12,45,14,54,98,68,48]
chosen20=[25,69,48], rest20=[12,45,14,54,98,68,78]
chosen21=[45,14,54], rest21=[12,25,69,98,68,78,48]
chosen22=[45,14,98], rest22=[12,25,69,54,68,78,48]
chosen23=[45,14,68], rest23=[12,25,69,54,98,78,48]
chosen24=[45,14,78], rest24=[12,25,69,54,98,68,48]
chosen25=[45,14,48], rest25=[12,25,69,54,98,68,78]
chosen26=[45,69,54], rest26=[12,25,14,98,68,78,48]
chosen27=[45,69,98], rest27=[12,25,14,54,68,78,48]
chosen28=[45,69,68], rest28=[12,25,14,54,98,78,48]
chosen29=[45,69,78], rest29=[12,25,14,54,98,68,48]
chosen30=[45,69,48], rest30=[12,25,14,54,98,68,78]

Answer 2

如果您需要从两个列表中获得2位数组合，剩下的就是解决方案：

import itertools
list1 = [12,25,45]
list2 = [14,69]
list3 = [21,34,56,32]
chosen = []
leftover = []
mergedlist = list(set(list1 + list2 + list3)) 
mergedNewList = [x for x in itertools.permutations(mergedlist,3)]
for i,value in enumerate(mergedNewList):
    chosen.append(list(value))
    leftover.append([j for j in mergedlist if j not in chosen[i]])
    print chosen[i]
    print leftover[i]`

我已将chosen的单个变量中的值附加到leftover中的其余变量中，因为这是存储值的最pythonic方式。

划分列表并生成一系列新列表。每个列表中的一个并休息到其他列表中

2 个答案: