使用字段名称访问字典

Question

我有一个释放函数，它返回一个dict：

def unpickle(file):
with open(file, 'rb') as fo:
    dict = pickle.load(fo, encoding='bytes')
return dict

和一个读取带有字段名的腌制对象的函数（不知道这是否是正确的定义）：

def do_sth():
    all_data = unpickle('mypickle.pickle')
    image_filenames = all_data["Filenames"]
    conditions = all_data["Labels"] 

为简便起见，我有两个列表，分别为Filenames = ['001.png','002.png']和Labels = ['0','1']，我需要在mypickle.pickle下腌制并保存，以便可以在do_sth函数下调用它们。到目前为止，我所做的是：

data = [Filenames,Labels]
with open("mypickle.pickle", "wb") as f:
pickle.dump(data, f)

和

data = dict(zip(file_paths, labels))
with open("mypickle.pickle", "wb") as f:
pickle.dump(data, f)

但是我得到KeyError :'Filenames'。我应使用哪种结构来保存这两个列表，以便它们可以正常工作。 谢谢。

Answer 1

将功能更改为此

def do_sth():
    all_data = unpickle('mypickle.pickle')     
    image_filenames = all_data[0]    
    conditions = all_data[1] 

说明

您将泡菜保存为列表。加载泡菜时，它仍然是列表。

或

实际上将其另存为字典

data = {"Filenames": Filenames, "Labels": Labels}
with open("mypickle.pickle", "wb") as f:
    pickle.dump(data,  f)