如何避免酸洗字典VS酸洗列表的處罰

Question

介紹

我有以下格式的詞典：

dict_list = {'S0':[[list of int],[list of int]], 'S1':[[list of int],[list of int]], ...} 

與S0的通過

dict_list['S0'][0] and dict_list['S0'][1] 

訪問的整數的列表

爲了提高代碼的可讀性，我將「列表清單」更改爲「列表字典」，如下所示：

dict_dict = {'S0': {'list0': [list of int], 'list1': [list of int]}, ...} 

訪問列表時，這導致更可讀代碼：

dict_dict['S0']['list0'] and dict_dict['S0']['list1'] 

味酸問題

然而，當我酸洗並保存到文件的dict_dict，原來的懲罰具有附加的字典鍵實際上與「S＃」條目的數量成比例地縮放。看起來，醃菜不是「巧妙地」存儲字典，因爲它分開存儲每個字母鍵。

現在，我們意識到，畢竟，這是醃菜應該如何工作，因爲每個'S＃'可能有不同的鑰匙組。鹹菜沒有辦法事先知道我們的dict_dict實際上只是一個有定期重複字段的表。

問題

我的問題是，是否有dict_list替代，其中整數的列表可以通過一個字符串鍵來訪問（如在dict_dict），但沒有上述鹹菜罰？

UPDATE：實驗基於評論鑑於

3,100 bytes - dict_list['S0'][0] (list.bin) 
3,314 bytes - dict_dict['S0']['list0'] (dict.bin) 
3,922 bytes - dict_class['S0'].list0 (class.bin) 
5,855 bytes - dict_namedtuple['S0'].list0 (namedtuple.bin) 

---

s_list = ['S0','S1','S2','S3','S4','S5','S6','S7','S8','S9','S10','S11','S12','S13','S14','S15','S0a','S1a','S2a','S3a','S4a','S5a','S6a','S7a','S8a','S9a','S10a','S11a','S12a','S13a','S14a','S15a','AA0','AA1','AA2','AA3','AA4','AA5','AA6','AA7','AA8','AA9','AA10','AA11','AA12','AA13','AA14','AA15','AA0a','AA1a','AA2a','AA3a','AA4a','AA5a','AA6a','AA7a','AA8a','AA9a','AA10a','AA11a','AA12a','AA13a','AA14a','AA15a','BB0','BB1','BB2','BB3','BB4','BB5','BB6','BB7','BB8','BB9','BB10','BB11','BB12','BB13','BB14','BB15','BB0a','BB1a','BB2a','BB3a','BB4a','BB5a','BB6a','BB7a','BB8a','BB9a','BB10a','BB11a','BB12a','BB13a','BB14a','BB15a'] 
num_of_s_entries = 32 
list_length = 5 

def pickle_n_save(dict_var, filename): 
    outfile = open(filename, "wb") 
    pickle.dump(dict_var, outfile) 
    outfile.close() 

# ------------------------------------------------------------dict_list['S0'][0] 
dict_list = {} 
for s in s_list[0:num_of_s_entries]: 
    dict_list[s] = [[],[]] 
    for pts in range(0,list_length): 
     dict_list[s][0].append(randrange(1,100)) 
     dict_list[s][1].append(randrange(1,100)*1000) 

pickle_n_save(dict_list, "list.bin") 

# -----------------------------------------------------dict_dict['S0']['list0'] 
dict_dict = {} 
for s in dict_list.keys(): 
    dict_dict[s] = {} 
    dict_dict[s]['list0'] = dict_list[s][0] 
    dict_dict[s]['list1'] = dict_list[s][1] 

pickle_n_save(dict_dict, "dict.bin") 

# -------------------------------------------------------dict_class['S0'].list0 
class S: 
    def __init__(self, list0, list1): 
     self.list0 = list0 
     self.list1 = list1 

dict_class = {} 
for s in dict_list.keys(): 
    dict_class[s] = S(dict_list[s][0],dict_list[s][1]) 

pickle_n_save(dict_class, "class.bin") 

# ---------------------------------------------------dict_namedtuple['S0'].list0 
S_namedtuple = namedtuple('S_namedtuple', ['list0','list1']) 
dict_namedtuple = {} 
for s in dict_list.keys(): 
    dict_namedtuple[s] = S_namedtuple(dict_list[s][0],dict_list[s][1]) 

pickle_n_save(dict_namedtuple, "namedtuple.bin") 

Answer 1

也許你想要一個namedtuple？