如何將pickled對象加載到一個空變量列表中？

Question

我有變量名的列表，以及醃製對象的名稱列表：

pickle_var_names = object1, object2 # both empty, which throws an error 
pickle_names = ['object1', 'object2'] 

我想叫pickle_names泡菜裝入pickle_var_names，分別爲：

for pickle_var_name,pickle_name in zip(pickle_var_names,pickle_names): 
    with open(pickle_name,'r') as pickle_fp: 
     pickle_var_name = cPickle.load(pickle_fp) 

我如何請初始化一個變量列表None，以便稍後分配它們？

Answer 1

列表是按位置索引，叫不上名字，所以你可以這樣做：

picklevars = [] 
picklenames = ['object1', 'object2'] 
for picklename in picklenames: 
    with open(pickle_name, 'r') as pickle_fp: 
     picklevars.append(cPickle.load(pickle_fp)) 

或者，如果你能夠使用字典而不是列表：

picklevars = {} 
picklenames = ['object1', 'object2'] 
for picklename in picklenames: 
    with open(pickle_name, 'r') as pickle_fp: 
     picklevars[picklename] = cPickle.load(pickle_fp)