Question

我是新手，现在已经工作了一个星期，但找不到任何解决方案。我希望有人可以帮我解决这个问题。

如何在列表中找到项目 - 列表项并单独输出项目？

listitems = ['Beer, Chicken', 'Cake, Chocolate', 'Lemon with ci, Chicken', 'Beer, Beer, Cake, Chocolate']

无论如何，我可以在收据清单中计算相关食物吗？

到目前为止，我只能找到一件食品。我的代码如下：

我还计算了每个项目的另一个列表供比较。

eachitems = ['Beer', 'Cake', 'Chocolate', 'Lemon with ci', 'Chicken']

Answer 1

我个人会使用基于每个项目的键的词典以及相关项目作为其值，也可以更容易地获得您想要的结果，而不是确切地知道如何从列表中完成它只做了。

从原始代码中，在末尾添加“print（combi）”，“print（checklist）”和“print（correlist）”，你会发现它并没有真正按照你想要的方式添加它。

Answer 2

在Python 3.5中：

import itertools

listOfStrings = ['Beer, Chicken', 'Cake, Chocolate', 'Lemon with ci, Chicken', 'Beer, Beer, Cake, Chocolate']
print ('Original list:', listOfStrings)
print ()

listOfLists = [group.replace (', ', ',') .split (',') for group in listOfStrings]
print ('Turned into list of lists:', listOfLists)
print ()

allFoods = set (itertools.chain (*listOfLists))
print ('All individual foods:', allFoods)
print ()

listOfSets = [set (group) for group in listOfLists]
print ('A list of sets is handy, since sets contain no duplicates:', listOfSets)
print ()

dictOfFoods = dict ([[food, set ()] for food in allFoods])
print ('Prepare a dictionary, where we can put the associated foods:', dictOfFoods)
print ()

for food in dictOfFoods:
    for foodSet in listOfSets:
        if food in foodSet:
            dictOfFoods [food] .update (foodSet)
    dictOfFoods [food] .remove (food)
print ('The dictionary is now filled:', dictOfFoods)
print ()

for food in dictOfFoods:
    print ('People who buy', food, 'also buy:')
    for otherFood in dictOfFoods [food]:
        print (otherFood)
    print ()

将打印：

Original list: ['Beer, Chicken', 'Cake, Chocolate', 'Lemon with ci, Chicken', 'Beer, Beer, Cake, Chocolate']

Turned into list of lists: [['Beer', 'Chicken'], ['Cake', 'Chocolate'], ['Lemon with ci', 'Chicken'], ['Beer', 'Beer', 'Cake', 'Chocolate']]

All individual foods: {'Chocolate', 'Lemon with ci', 'Chicken', 'Cake', 'Beer'}

A list of sets is handy, since sets contain no duplicates: [{'Chicken', 'Beer'}, {'Chocolate', 'Cake'}, {'Chicken', 'Lemon with ci'}, {'Chocolate', 'Cake', 'Beer'}]

Prepare a dictionary, where we can put the associated foods: {'Chocolate': set(), 'Lemon with ci': set(), 'Cake': set(), 'Beer': set(), 'Chicken': set()}

The dictionary is now filled: {'Chocolate': {'Cake', 'Beer'}, 'Lemon with ci': {'Chicken'}, 'Cake': {'Chocolate', 'Beer'}, 'Beer': {'Chocolate', 'Chicken', 'Cake'}, 'Chicken': {'Lemon with ci', 'Beer'}}

People who buy Chocolate also buy:
Cake
Beer

People who buy Lemon with ci also buy:
Chicken

People who buy Cake also buy:
Chocolate
Beer

People who buy Beer also buy:
Chocolate
Chicken
Cake

People who buy Chicken also buy:
Lemon with ci
Beer

如果你不想使用itertools和*，你也可以在循环中循环遍历listOfLists的所有元素，并将它们添加到最初为空的allFoods中。

Answer 3

我花了一些时间才准确理解你想要的东西，但这是一个有效的解决方案。

listitems = ['Beer, Chicken', 'Cake, Chocolate', 'Lemon with ci, Chicken', 'Beer, Beer, Cake, Chocolate']
eachitems = ['Beer', 'Cake', 'Chocolate', 'Lemon with ci', 'Chicken']

for item in eachitems:
    assoc = [associated for associated in listitems if item in associated]
    result = set()
    for itemlist in assoc:
        itemlist = itemlist.replace(', ', ',').split(',')
        itemlist = set(itemlist)

        itemlist.remove(item)

        result = result | itemlist
    print('People who buy {} also buy: '.format(item), ', '.join(sorted(result)))

输出

People who buy Beer also buy:  Cake, Chicken, Chocolate
People who buy Cake also buy:  Beer, Chocolate
People who buy Chocolate also buy:  Beer, Cake
People who buy Lemon with ci also buy:  Chicken
People who buy Chicken also buy:  Beer, Lemon with ci

这个解决方案的关键部分是使用Sets删除重复项和|（联合）运算符。

作为旁注，请不要像这样使用|

result = result | itemlist

您可以使用

修改设置

result.update(itemlist)

如何查找列表

3 个答案: