Python3 reduce函数

Question

我在JS中使用了以下包含重复项的dict/JSON：

{
  0:{
    "id" : 1,
    "name": "test"
  },
  1:{
    "id" : 2,
    "name": "other name"
  },
  2:{
    "id" : 1,
    "name": "does not have to be the same name"
  },{
    "id" : 3,
    "name": "but they could be the same"
  },
  3:{
    "id" : 2,
    "name": "other name"
  }
}

我想把它简化为：

{
  0:{
    "id" : 1,
    "name": "test"
  },
  1:{
    "id" : 2,
    "name": "other name"
  },
  2:{
    "id" : 3,
    "name": "but they could be the same"
  }
}

标准是id必须是唯一的，而不管项的其余部分是否不同。

对于JS，我使用了以下代码：

const data = [ { "id" : 1, "name": "test" }, { "id" : 2, "name": "other name" }, { "id" : 1, "name": "does not have to be the same name" },{ "id" : 3, "name": "but they could be the same" },{ "id" : 2, "name": "other name" }, ],
      unique = Object.values(data.reduce((r, o) => {
        r[o.id] = r[o.id] || o;
        return r;
      },{}));
console.log(unique);

然而，我似乎不能用python3.9做同样的事情。我找到了functools.reduce()函数，并尝试将其与lambas一起使用，但dics与JSON对象不同。

Answer 1

我使用以下代码解决了这个问题：

data = [i for n, i in enumerate(data) if i['id'] not in {d['id'] for j,d in enumerate(data[n+1:])}] 

Answer 2

提醒一下，您提供的输入是无效的JSON。根据您的JS，结构应该是这样的。

foo = [
    {
        "id" : 1,
        "name": "test"
    },
    {
        "id" : 2,
        "name": "other name"
    },
    {
        "id" : 1,
        "name": "does not have to be the same name"
    },
    {
         "id" : 3,
         "name": "but they could be the same"
    },
    {
         "id" : 2,
         "name": "other name"
    }
]

也就是说，如果你想要一个使用reduce的解决方案，像这样的东西是可行的。

from functools import reduce
foo = [
    {
        "id" : 1,
        "name": "test"
    },
    {
        "id" : 2,
        "name": "other name"
    },
    {
        "id" : 1,
        "name": "does not have to be the same name"
    },
    {
         "id" : 3,
         "name": "but they could be the same"
    },
    {
         "id" : 2,
         "name": "other name"
    }
]
# in my opinion, the readability suffers here
# sometimes you need to be ok with using more than one line
result = reduce(lambda x, y: x + [y] if not any(i['id'] == y['id'] for i in x) else x, foo, [])
print(result)

一个更具可读性的例子是使用一个实际的函数而不是lambda。

from functools import reduce
foo = [
    {
        "id" : 1,
        "name": "test"
    },
    {
        "id" : 2,
        "name": "other name"
    },
    {
        "id" : 1,
        "name": "does not have to be the same name"
    },
    {
         "id" : 3,
         "name": "but they could be the same"
    },
    {
         "id" : 2,
         "name": "other name"
    }
]
def build_list(result_list, list_element):
    if not any(list_element['id'] == i['id'] for i in result_list):
        return result_list + [list_element]
    return result_list

result = reduce(build_list, foo, [])
print(result)

但这仍然不是我写它的风格。就我个人而言，我会使用set跟踪重复的内容，并完全避免使用comprehension / reduce。

foo = [
    {
        "id" : 1,
        "name": "test"
    },
    {
        "id" : 2,
        "name": "other name"
    },
    {
        "id" : 1,
        "name": "does not have to be the same name"
    },
    {
         "id" : 3,
         "name": "but they could be the same"
    },
    {
         "id" : 2,
         "name": "other name"
    }
]
result = []
ids = set()
for dct in foo:
    if not dct['id'] in ids:
        ids.add(dct['id'])
        result.append(dct)
print(result)