如何将腌制对象加载到空变量列表中？答案

【问题标题】：how to load pickled objects into a list of empty variables?如何将腌制对象加载到空变量列表中？
【发布时间】：2015-11-03 18:48:51
【问题描述】：

我有一个变量名列表和一个腌制对象名称列表：

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

我希望将pickle_names中调用的泡菜分别加载到pickle_var_names中：

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)

我如何将变量列表初始化为None，以便以后分配它们？

【问题讨论】：

标签： python python-2.7 serialization pickle

【解决方案1】：

列表是按位置而不是名称来索引的，所以你可以这样做：

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

或者，如果您能够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)

【讨论】：