Question

我有一个变量名列表和一个酸洗对象名称列表：

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)

如何将变量列表初始化为None以允许稍后分配？

Answer 1

列表按位置索引，而不是按名称索引，因此您可以执行此操作：

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)

如何将pickled对象加载到空变量列表中？

1 个答案: