细分嵌套字典，而在Python

Question

施加从值百分比欲细分一个字典，同时施加的值（其是百分比，到另一个的值

我有那些数据集：

{'C-STD-B&M-SUM': {datetime.date(2015, 5, 20): 0.21484699999999998, 
       datetime.date(2015, 5, 21): 0.245074, 
       datetime.date(2015, 5, 22): 0.27874} 

{'G-CAM-BAC-SUM': {datetime.date(2015, 5, 20): 0.13294399999999998, 
       datetime.date(2015, 5, 21): 0.151648, 
       datetime.date(2015, 5, 22): 0.17248, 
       datetime.date(2015, 5, 23): 0.195664} 

{'G-CAM-BAC-XS': 0.06, 'G-CAM-BAC-XXS': 0.01, 'G-CAM-BAC-XL': 0.11, 'G-CAM-BAC-S': 0.19, 'G-CAM-BAC-L': 0.26, 'G-CAM-BAC-XXL': 0.03, 'G-CAM-BAC-M': 0.35} 

{'C-STD-B&M-XL': 0.3, 'C-STD-B&M-XXL': 0.11, 'C-STD-B&M-S': 0.06, 'C-STD-B&M-M': 0.2, 'C-STD-B&M-XS': 0, 'C-STD-B&M-L': 0.32} 

预期输出：

{'C-STD-B&M-XL': {datetime.date(2015, 5, 20): 0.21484699999999998*0.3, 
      datetime.date(2015, 5, 21): 0.245074*0.3, 
      datetime.date(2015, 5, 22): 0.27874*0.3} 

{'C-STD-B&M-XXS': {datetime.date(2015, 5, 20): 0.21484699999999998*0.1, 
      datetime.date(2015, 5, 21): 0.245074*0.1, 
      datetime.date(2015, 5, 22): 0.27874*0.1} 

{'C-STD-B&M-XXL': {datetime.date(2015, 5, 20): 0.21484699999999998*0.11, 
      datetime.date(2015, 5, 21): 0.245074*0.11, 
      datetime.date(2015, 5, 22): 0.27874*0.11} 

等等，所有的词典请注意，我需要的值相乘的结果，而不是语句，刚刚离开。让他们更清楚。

到目前为止我的代码（部分）：

def apply_size_distribution(dictionary_with_temporal_distribution): 
    gown_cap_size = get_size_distribution('G2:G7', 'H2:H7') 
    cap_medium_demand = gown_cap_size['C-STD-B&M-M'] 
    for k, v in dictionary_with_temporal_distribution.items(): 
     if k == "C-STD-B&M-SUM": 
      dictionary_with_temporal_distribution['C-STD-B&M-M'] = dictionary_with_temporal_distribution.pop('C-STD-B&M-SUM') 
      for k, v in dictionary_with_temporal_distribution['C-STD-B&M-M'].items(): 
       dictionary_with_temporal_distribution["{}".format(k)] = v * cap_medium_demand 

但是我却越来越好老字典迭代过程中发生变化。此外，使用我的代码，我将不得不复制并粘贴每个大小的代码，只将Key的名称更改为适当的大小。我想知道是否有更强大的方法。

Answer 1

你所得到的dictionary_with_temporal_distribution在你的代码，因为改变你改变它 -

dictionary_with_temporal_distribution["{}".format(k)] = v * cap_medium_demand 

相反，你应该考虑在开始创建一个新的字典，也许和元素添加到它，你继续处理的最终回报它。

此外，而不是硬编码值，像你这样 - gown_cap_size['C-STD-B&M-M'] - k == "C-STD-B&M-SUM"，你应该使用string.rsplit('-', 1)[0]找出共同部分 - C-STD-B&M，然后使用该平等。

你的代码看起来像 -

def apply_size_distribution(dictionary_with_temporal_distribution): 
    gown_cap_size = get_size_distribution('G2:G7', 'H2:H7') 
    retlist = {} 
    for cap_demand in grown_cap_size: 
     for k, v in dictionary_with_temporal_distribution.items(): 
      if k.rsplit('-', 1)[0] == cap_demand.rsplit('-', 1)[0]: 
       if cap_demand not in retlist: 
        retlist[cap_demand] = {} 
       temp = retlist[cap_demand] 
       for k1, v1 in dictionary_with_temporal_distribution[k].items(): 
        temp[k1] = v1 * grown_cap_size[cap_demand] 
    return retlist