Question

我有以下课程：

class Document:
  _queryID  = ""
  _name = []
  _docId = ""

在代码中我创建了一个名为listDocuments的列表。我用列表类型的对象填充列表。

import pickle 
if __name__ == "__main__": 
   listDocuments = []
   for i in range(0,10):
      e = Document()
      e._queryID = str(i)
      e._docId   = str(i+1)
      e._name.append("h"+str(i))
      listDocuments.append(e)
   save_data(listDocuments, "Variations.dat")
   list = load_data("Variations.dat")

   for obj  in listEntities:
     print obj._queryID  + "==>"+ obj._docId
     for var in obj._name:
       print var +"  "
     print "----\n"

我使用了Python: how to save a list with objects in a file中的以下函数：

#save list of variations
def save_data(data, path):
    with open(path, "wb") as f:
        pickle.dump(data, f)

#load list of variations (Correctd after comments thank you)
def load_data(var_file):
    try:
        with open(var_file) as f:
            listDocument = pickle.load(f)
    except:
        listDocument = []

    return listDocument

作为输出我只能打印_queryID。我错过了什么？我不能打印列表_name和_docID

Answer 1

好的，有几个问题。

首先：

def load_data(var_file):
    try:
        with open(var_file) as f:
            listDocument = pickle.load(f)
    except:
        listDocuments = []

    return listDocuments

您同时使用listDocument和listDocuments。（注意一个有一个尾随s）。此外，您正在外部程序中使用这些变量listDocuments来隐藏您的错误。让我们将函数内的变量重命名为data，就像save_data函数一样。出于同样的一致性原因，我们还会将var_file重命名为path。

考虑：

def load_data(path):
    data = []
    try:
        with open(path) as f:
            data = pickle.load(f)
    except: pass

    return data

接下来，您使用的是list这是一个保留的内置功能 - 让我们将其更改为lst。所以现在你有了：

class Document:
    _queryID  = ""
    _name = []
    _docId = ""

def save_data(data, path):
    with open(path, "wb") as f:
        pickle.dump(data, f)

#load list of variations
def load_data(path):
    data = []
    try:
        with open(path) as f:
            data = pickle.load(f)
    except: pass

    return data

import pickle
if __name__ == "__main__":
    listDocuments = []
    for i in range(0,10):
        e = Document()
        e._queryID = str(i)
        e._docId   = str(i+1)
        e._name.append("h"+str(i))
        listDocuments.append(e)
    save_data(listDocuments, "Variations.dat")
    lst = load_data("Variations.dat")

    for elem in lst:
        print elem._queryID
        print elem._docId
        print elem._name
        print "---"

哪个有效，并打印：

0
1
['h0', 'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8', 'h9']
---
1
2
['h0', 'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8', 'h9']
---
2
3
['h0', 'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8', 'h9']
---
3
4
['h0', 'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8', 'h9']
---
4
5
['h0', 'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8', 'h9']
---
5
6
['h0', 'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8', 'h9']
---
6
7
['h0', 'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8', 'h9']
---
7
8
['h0', 'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8', 'h9']
---
8
9
['h0', 'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8', 'h9']
---
9
10
['h0', 'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8', 'h9']
---

修改

根据您的修改，您没有listEntities变量 - 我认为您的意思是（最近重命名的）lst：

for obj in lst: print obj._queryID + "==>"+ obj._docId for var in obj._name: print var +" " print "----\n"

最后，如果您想知道_name类的所有实例的Document变量为何相同，那是因为您将_name声明为类属性（因此它对所有实例都是一样的），而不是实例属性。

您可以将Document类定义替换为：

class Document: def __init__(self): self._name = []

你可能对结果更满意：

0==>1 h0 ---- 1==>2 h1 ---- 2==>3 h2 ---- 3==>4 h3 ---- 4==>5 h4 ---- 5==>6 h5 ---- 6==>7 h6 ---- 7==>8 h7 ---- 8==>9 h8 ---- 9==>10 h9 ----

如何使用pickle python将包含列表作为成员的对象列表保存到文件中？

1 个答案: