英文:
Given two lists, how to create a unique combinations and the elements in each combination are different? Python
问题
以下是已翻译的内容：
输入：list1 = ['Town1','Town2','Town3','Town4','Town5']list2 = ['Town1','Town2','Town3','Town4','Town5']输出：[('Town1', 'Town2'), ('Town3', 'Town4'), ('Town4', 'Town4'), ('Town4', 'Town5'), ('Town1', 'Town4'), ('Town3', 'Town5'), ('Town1', 'Town3'), ('Town2', 'Town4'), ('Town1', 'Town5'), ('Town2', 'Town3'), ('Town2', 'Town5')]我尝试了：comb = list(product(list1, list2))comb1 = [sorted(item) for item in comb]comb2 = list(set(map(tuple, comb1))）但结果中包括了('Town1', 'Town1')。
英文:
Inputs:
list1 = ['Town1','Town2','Town3','Town4','Town5']list2 = ['Town1','Town2','Town3','Town4','Town5']
Outputs:
[('Town1', 'Town2'), ('Town3', 'Town4'), ('Town4', 'Town4'), ('Town4', 'Town5'), ('Town1', 'Town4'), ('Town3', 'Town5'), ('Town1', 'Town3'), ('Town2', 'Town4'), ('Town1', 'Town5'), ('Town2', 'Town3'), ('Town2', 'Town5')]
I tried :
comb = list(product(list1,list2))comb1 = [sorted(item) for item in comb]comb2 = list(set(map(tuple,comb1)))
But ('Town1', 'Town1') is involved in the result.
答案1
得分: 1
你可以使用一个列表推导式来排除具有重复元素的配对：
from itertools import productlist1 = ['Town1', 'Town2', 'Town3', 'Town4', 'Town5']list2 = ['Town1', 'Town2', 'Town3', 'Town4', 'Town5']comb = [pair for pair in product(list1, list2) if pair[0] != pair[1]]print(comb)
输出结果是：
[('Town1', 'Town2'), ('Town1', 'Town3'), ('Town1', 'Town4'), ('Town1', 'Town5'), ('Town2', 'Town1'), ('Town2', 'Town3'), ('Town2', 'Town4'), ('Town2', 'Town5'), ('Town3', 'Town1'), ('Town3', 'Town2'), ('Town3', 'Town4'), ('Town3', 'Town5'), ('Town4', 'Town1'), ('Town4', 'Town2'), ('Town4', 'Town3'), ('Town4', 'Town5'), ('Town5', 'Town1'), ('Town5', 'Town2'), ('Town5', 'Town3'), ('Town5', 'Town4')]
英文:
You can exclude the pairs with repeated elements with a list comprehension:
from itertools import productlist1 = ['Town1','Town2','Town3','Town4','Town5']list2 = ['Town1','Town2','Town3','Town4','Town5']comb = [pair for pair in product(list1,list2) if pair[0] != pair[1]]print(comb)
The output is:
[('Town1', 'Town2'), ('Town1', 'Town3'), ('Town1', 'Town4'), ('Town1', 'Town5'), ('Town2', 'Town1'), ('Town2', 'Town3'), ('Town2', 'Town4'), ('Town2', 'Town5'), ('Town3', 'Town1'), ('Town3', 'Town2'), ('Town3', 'Town4'), ('Town3', 'Town5'), ('Town4', 'Town1'), ('Town4', 'Town2'), ('Town4', 'Town3'), ('Town4', 'Town5'), ('Town5', 'Town1'), ('Town5', 'Town2'), ('Town5', 'Town3'), ('Town5', 'Town4')]
答案2
得分: 0
list1 = ['Town1', 'Town2', 'Town3', 'Town4', 'Town5']list2 = ['Town1', 'Town2', 'Town3', 'Town4', 'Town5']result = []for i in list1:for j in list2:if i != j:result.append((i, j))print(result)
英文:
list1 = ['Town1','Town2','Town3','Town4','Town5']list2 = ['Town1','Town2','Town3','Town4','Town5']result = []for i in list1:for j in list2:if i != j:result.append((i, j))print(result)
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