将列表转换为多值字典答案

【问题标题】：Converting a list into a multi-valued dict将列表转换为多值字典
【发布时间】：2016-02-04 06:33:06
【问题描述】：

我有一个这样的列表：

pokemonList = ['Ivysaur', 'Grass', 'Poison', '', 'Venusaur', 'Grass', 'Poison', '', 'Charmander', 'Fire', ''...]

注意模式是'Pokemon name', 'its type', ''...next pokemon

口袋妖怪有单型和双型两种。我该如何编码，以便每个口袋妖怪（键）都将它们各自的类型应用为它的值？

到目前为止我得到了什么：

types = ("", "Grass", "Poison", "Fire", "Flying", "Water", "Bug","Dark","Fighting", "Normal","Ground","Ghost","Steel","Electric","Psychic","Ice","Dragon","Fairy")
pokeDict = {}
    for pokemon in pokemonList:
        if pokemon not in types:
            #the item is a pokemon, append it as a key
        else:
            for types in pokemonList:
                #add the type(s) as a value to the pokemon

正确的字典如下所示：

{Ivysaur: ['Grass', 'Poison'], Venusaur['Grass','Poison'], Charmander:['Fire']}

【问题讨论】：

你的列表中是否总是在每个 pokemon 之后有一个空字符串？
是的。（谢谢美丽的汤）。

标签： python list python-3.x dictionary dictionary-comprehension

【解决方案1】：

只需迭代列表并适当地为 dict 构造项目..

current_poke = None
for item in pokemonList:
    if not current_poke:
        current_poke = (item, [])
    elif item:
        current_poke[1].append(item)
    else:
        name, types = current_poke
        pokeDict[name] = types
        current_poke = None

【讨论】：

【解决方案2】：

用于分割原始列表的递归函数，以及用于创建字典的字典理解：

# Slice up into pokemon, subsequent types
def pokeSlice(pl):
    for i,p in enumerate(pl):
        if not p:
            return [pl[:i]] + pokeSlice(pl[i+1:])      
    return []

# Returns: [['Ivysaur', 'Grass', 'Poison'], ['Venusaur', 'Grass', 'Poison'], ['Charmander', 'Fire']]

# Build the dictionary of 
pokeDict = {x[0]: x[1:] for x in pokeSlice(pokemonList)}

# Returning: {'Charmander': ['Fire'], 'Ivysaur': ['Grass', 'Poison'], 'Venusaur': ['Grass', 'Poison']}

【讨论】：

【解决方案3】：

这是一种低技术含量的方法：遍历列表并随时收集记录。

key = ""
values = []
for elt in pokemonList:
    if not key:
        key = elt
    elif elt:
        values.append(elt)
    else:
        pokeDict[key] = values
        key = ""
        values = []

【讨论】：

这段代码读起来非常简洁明了。我希望我能接受这两个答案！
没问题，接受您更喜欢的答案是您的特权。（但是您可以根据需要投票给尽可能多的答案 ;-)

【解决方案4】：

一个班轮。不是因为它有用，而是因为我开始尝试并且必须完成。

>>> pokemon = ['Ivysaur', 'Grass', 'Poison', '', 'Venusaur', 'Grass', 'Poison', '', 'Charmander', 'Fire', '']
>>> { pokemon[i] : pokemon[i+1:j] for i,j in zip([0]+[k+1 for k in [ brk for brk in range(len(x)) if x[brk] == '' ]],[ brk for brk in range(len(x)) if x[brk] == '' ]) }
{'Venusaur': ['Grass', 'Poison'], 'Charmander': ['Fire'], 'Ivysaur': ['Grass', 'Poison']}

【讨论】：