正在创建JSON对象的置换？

3条回答

网友

1楼 · 编辑于 2024-09-28 21:22:03

我想你要找的是笛卡尔积，而不是组合，你能为每个人得到的每个可能的名字-城市对。也就是这些列表的笛卡尔积：

[('Robert', 'New York'), ('Bob', 'New York')]
[('Allison', 'Los Angeles'), ('Ally', 'Los Angeles')]
[('James', 'Chicago'), ('Jim', 'Chicago')]
[('Hannah', 'Toronto')]
[('Elizabeth', 'London')]

首先需要一个函数，将人名city short dict转换为姓名city/short city对的列表：

def pairs(person):
    names = [(person['name'], person['city'])]
    try:
        names.append((person['short'], person['city']))
    except KeyError:
        pass
    return names

现在，您需要将此应用于具有map或理解力的所有人：

allpairs = map(pairs, people)

现在您有了一些可以直接调用产品的功能：

results = itertools.product(*allpairs)

现在你只需要把每一对的列表变成一个dict：

output = [dict(result) for result in results]

然后你必须编写代码，将它们按你想要的奇怪顺序排列（也许你只是想把它们反向生成，然后再反向生成？），并以看起来有点像Python dict repr或JSON对象但实际上两者都不是的格式输出它。（另外，如果每个dict的顺序很重要，并且您没有使用python3.7或更高版本，则需要使用collections.OrderedDict而不是dict。）

网友

2楼 · 编辑于 2024-09-28 21:22:03

这里有一个唯一的jq解决方案，它确实使用了内置函数combinations。但是，它确实假设输入是有效的JSON。你知道吗

[ map([ {(.name):.city},
        (if .short then {(.short):.city} else empty end) ] )
  | combinations
  | add ]

输出

[
  {
    "Robert": "New York",
    "Allison": "Los Angeles",
    "James": "Chicago",
    "Hannah": "Toronto",
    "Elizabeth": "London"
  },
  {
    "Robert": "New York",
    "Allison": "Los Angeles",
    "Jim": "Chicago",
    "Hannah": "Toronto",
    "Elizabeth": "London"
  },
  {
    "Robert": "New York",
    "Ally": "Los Angeles",
    "James": "Chicago",
    "Hannah": "Toronto",
    "Elizabeth": "London"
  },
  {
    "Robert": "New York",
    "Ally": "Los Angeles",
    "Jim": "Chicago",
    "Hannah": "Toronto",
    "Elizabeth": "London"
  },
  {
    "Bob": "New York",
    "Allison": "Los Angeles",
    "James": "Chicago",
    "Hannah": "Toronto",
    "Elizabeth": "London"
  },
  {
    "Bob": "New York",
    "Allison": "Los Angeles",
    "Jim": "Chicago",
    "Hannah": "Toronto",
    "Elizabeth": "London"
  },
  {
    "Bob": "New York",
    "Ally": "Los Angeles",
    "James": "Chicago",
    "Hannah": "Toronto",
    "Elizabeth": "London"
  },
  {
    "Bob": "New York",
    "Ally": "Los Angeles",
    "Jim": "Chicago",
    "Hannah": "Toronto",
    "Elizabeth": "London"
  }
]

网友

3楼 · 编辑于 2024-09-28 21:22:03

如果您正在寻找一个Javascript解决方案，您需要自己的笛卡尔积函数，但其余的与上面的Python解决方案非常相似。这个函数使用了一个可变的笛卡尔函数，所以需要额外的步骤cartesian.apply(null, arr)。但除此之外，还应该直截了当：

；

const cartesian = (xs, ys, ...zs) => ys 
  ? cartesian([].concat(...xs.map(x => ys.map(y => [].concat(x, y)))), ...zs) 
  : xs;

const toObj = xs => xs.reduce((a, b) => Object.assign(a, b), {})

const toNames = people => people.map(person => Object.keys(person)
  .filter(name => ['name', 'short'].includes(name))
  .map(name => ({[person[name]]: person.city}))
)

const nameCombos = people => cartesian.apply(null, toNames(people)).map(toObj)

const people = [
    { name: "Robert",    city: "New York",    short: "Bob"  },
    { name: "Allison",   city: "Los Angeles", short: "Ally" },
    { name: "James",     city: "Chicago",     short: "Jim"  },
    { name: "Hannah",    city:"Toronto"                     },
    { name: "Elizabeth", city: "London"                     }
]

console.log(nameCombos(people))

；

输出

相关问题更多 >

编程相关推荐

热门问题

热门文章