如何使用权重将字典合并在一起？

Question

d1 = {'weight':1, 'data': { 'apples': 8, 'oranges': 7 } }
d2 = {'weight':3, 'data': { 'apples': 4, 'bananas': 3 } }
all_dictionaries = [d1, d2, ... ]

def mergeDictionariesWithWeight(all_dictionaries)

如何将这些字典合并在一起（如果重叠，则多个值与权重）

该函数将返回：

{ 'apples': 4, 'oranges': 7, 'bananas': 3 }

Apples is 4 因为 8 * .25 + 4 * .75

编辑：我刚刚写了一篇取平均值的文章，类似这样。但当然，这与我想要做的确实不同，因为我将所有内容都放在列表中，然后除以长度。

result = {}
keymap = {}
for the_dict in dlist:
    for (k, v) in the_dict.items():
        if not keymap.has_key(k):
            keymap[k] = []
        keymap[k].append(v)
for (k, v) in keymap.items():
    average = sum(int(x) for x in keymap[k]) / float(len(keymap[k]))
    result[k] = float(average)
return result

Answer 1

>>> from collections import defaultdict
>>> d=defaultdict(lambda:(0,0))
>>> for D in all_dictionaries:
...   weight = D['weight']
...   for k,v in D['data'].items():
...     d[k]=d[k][0]+weight*v,d[k][1]+weight
... 
>>> dict((k,v[0]/v[1]) for k,v in d.items())
{'apples': 5, 'oranges': 7, 'bananas': 3}

如果需要浮点结果

>>> dict((k,1.*v[0]/v[1]) for k,v in d.items())
{'apples': 5.0, 'oranges': 7.0, 'bananas': 3.0}

关于defaultdict的注释

通常您会看到 defaultdict(int) 或 defaultdict(list) 甚至可能是 defaultdict(set)。 defaultdict 的参数必须可以不带参数地调用。每当发现密钥丢失时，都会使用调用此参数的结果。 的默认值，

即 - 调用此函数会返回字典

>>> d=defaultdict(int)
>>> d[1]
0
>>> d['foo']
0

这通常用于计数，因为 int() 返回 0。如果您希望默认值为1 而不是 0，它更棘手，因为您无法将参数传递给 int，但您所需要的只是一个返回 1 的可调用函数。通过使用 lambda 函数。

>>> d=defaultdict(lambda:1)
>>> d[1]
1
>>> d['foo']
1

在这个答案中，我想跟踪加权总和以及权重总和。我可以通过使用二元组作为默认值来做到这一点。

>>> d=defaultdict(lambda:(0,0))
>>> d[1]
(0, 0)
>>> d['foo']
(0, 0)

Answer 2

这是一个首先使用临时字典将项目收集到列表中的解决方案，然后计算最终的加权字典。可能不需要临时就可以完成，但这很容易理解。

from collections import defaultdict

def mergeDictionariesWithWeight(dlist):
    tmp = defaultdict(list)
    for d in dlist:
        weight = d['weight']
        for k, v in d['data'].items():
            tmp[k].append((weight, v))
    r = {}
    for k, v in tmp.items():
        # If there's just one item, ignore the weight
        if len(v) == 1:
            r[k] = v[0][1]
        else:
            total_weight = sum((x[0] for x in v), 0.0)
            r[k] = sum(x[1] * x[0]/total_weight for x in v)
    return r

返回：{'苹果'：5.0，'橙子'：7，'香蕉'：3}（因为 8 * .25 + 4 * .75 = 5.0）

Answer 3

试试这个：

def mergeDictionariesWithWeight(all_dictionaries):
    weightSum = 0
    weightDictionary ={}    

    for dictionary in all_dictionaries: 

        weight = dictionary['weight']
        data = dictionary['data']

        #find the total weight of the elements in data
        for (k,v) in data.items(): 
            if k in weightDictionary:
                weightDictionary[k] += weight*v
        weightSum += weight 
        #normalize the results by deviding by the weight sum
        for (key, value) in weightDictionary:
            weightDictionary[key] = value / float(weightSum)
    return weightDictionary 

d1 = {'weight':1, 'data': { 'apples': 8, 'oranges': 7 } }
d2 = {'weight':3, 'data': { 'apples': 4, 'bananas': 3 } }
all_dictionaries = [d1, d2]

mergeDictionariesWithWeight(all_dictionaries)

Answer 4

from collections import defaultdict

def merge_dictionaries_with_weight(all_dictionaries):
    totals = defaultdict(int)
    result = defaultdict(int)
    for each in all_dictionaries:
        weight = float(each['weight'])
        for key, value in each['data'].items():
            totals[key] += weight
            result[key] += weight * value
    for key, total in totals.items():
       result[key] /= total
    return result

Answer 5

在算法上与gnibbler的无法区分，但不知何故生成器表达式让我很高兴。

>>> from collections import defaultdict
>>> weights, values = defaultdict(int), defaultdict(int)
>>> key_weight_value = ((key, d['weight'], value) 
                        for d in all_dictionaries 
                        for key, value in d['data'].iteritems())
>>> for k, w, v in key_weight_value:
...     weights[k], values[k] = weights[k] + w, values[k] + w * v
... 

>>> dict((k, values[k] * 1.0 / weights[k]) for k in weights)
{'apples': 5.0, 'oranges': 7.0, 'bananas': 3.0}