如何根据某个键的值对字典列表进行排序，而不使用Python中的sort()

Question

我试图根据key: positive_ratings的值对一个大的json文件进行排序，并且不使用内置的sort()函数。

small_example = [
{'id':10,'game':'Counterstrike','positive_ratings':150},
{'id':20,'game':'Bioshock Infinite','positive_ratings':50},
{'id':30,'game':'Rust','positive_ratings':300},
{'id':40,'game':'Portal','positive_ratings':200}
]

按降序排列的输出将是以下列表：['Rust', 'Portal', 'Counterstrike', 'Bioshock Infinite']

对于学校，我们必须做一个快速排序功能，排序列表如下所示。现在，我想重写它，以便对字典列表进行排序。

def quick_sort(sequence):
    length = len(sequence)
    if length <= 1:
        return sequence
    else:
        centre = sequence.pop()


    items_bigger = []
    items_smaller = []

    for item in sequence:
        if item > centre:
            items_bigger.append(item)

        else: items_smaller.append(item)

    return quick_sort(items_smaller) + [centre] + quick_sort(items_bigger)


print(quick_sort([1,2,5,6,2,10,34,54,23,1]))

Answer 1

在代码中，根据元素的值对列表进行排序。但是您想要的是基于element['positive_ratings']的排序列表。您只需稍微修改一下代码：

def quick_sort(sequence):
    length = len(sequence)
    if length <= 1:
        return sequence
    else:
        centre = sequence.pop()


    items_bigger = []
    items_smaller = []

    for item in sequence:
        if item['positive_ratings'] > centre['positive_ratings']: # I changed only this line
            items_bigger.append(item) 

        else: items_smaller.append(item)

    return quick_sort(items_smaller) + [centre] + quick_sort(items_bigger)

sort函数也是这样工作的，您只需指定键：

some_list.sort(key= lambda x: x['positive_ratings'])

我们可以调整您的代码，使其看起来类似于sort函数：

def quick_sort(sequence, key = lambda x: x):
    length = len(sequence)
    if length <= 1:
        return sequence
    else:
        centre = sequence.pop()


    items_bigger = []
    items_smaller = []

    for item in sequence:
        if key(item> key(centre): # I changed only this line
            items_bigger.append(item) 

        else: items_smaller.append(item)

    return quick_sort(items_smaller, key) + [centre] + quick_sort(items_bigger, key)

你可以这样称呼它：

quick_sort(small_example, key = lambda x: x['positive_ratings'])

编辑:我忘了在最后一行中添加键。多亏了达林G，我修正了这个问题

Answer 2

通过基于positive_ratings键对数据进行排序，可以对示例进行排序(首先对postive_ratings值进行排序，然后根据此排序返回输出)

small_example = [
{'id':10,'game':'Counterstrike','positive_ratings':150},
{'id':20,'game':'Bioshock Infinite','positive_ratings':50},
{'id':30,'game':'Rust','positive_ratings':300},
{'id':40,'game':'Portal','positive_ratings':200}
]



def func(data, key: int):
    dic = {}
    for i in data:
        if i[key] not in dic:
            dic[i[key]] = [i]
        else:
            dic[i[key]].append(i)
    
    dic_key = list(dic.keys())
    
    # sorting the dic_key, sorting data based on postive_raing values, you can
    # use any sort algo here
    
    for i in range(len(dic_key)):
        for j in range(i+1, len(dic_key)):
            if dic_key[i]>dic_key[j]:
                dic_key[i], dic_key[j] = dic_key[j], dic_key[i]
    
    result = []
    for i in dic_key:
        result.extend(dic[i])
    return result

sol =  func(small_example, 'positive_ratings')
print(solution)

输出

[{'id': 20, 'game': 'Bioshock Infinite', 'positive_ratings': 50},
 {'id': 10, 'game': 'Counterstrike', 'positive_ratings': 150},
 {'id': 40, 'game': 'Portal', 'positive_ratings': 200},
 {'id': 30, 'game': 'Rust', 'positive_ratings': 300}]