Question

我有一个列表：

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

请注意，模式为'Pokemon name', 'its type', ''...next pokemon

口袋妖怪有单一和双重形式。我该如何对其进行编码，以便每个口袋妖怪（密钥）都将其各自的类型应用为其值？

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

正确的字典将如下所示：

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

Answer 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

Answer 2

用于切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']}

Answer 3

这是执行此操作的低技术方法：迭代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 = []

Answer 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。

let myRouteRequest = MKDirectionsRequest()
myrouteRequest.transportType = .Automobile
myRouteRequest.setSource(MKMapItem.mapItemForCurrentLocation())
myRouteRequest.setDestination(MKMapItem(myPlacemark))

将列表转换为多值dict

4 个答案: