排序字典 - 它被声明为python的顺序

Question

这里我添加了一个代码来生成xml_tags,这就是我的输出:

md5 value = "sdfhsdkjgfjw35378563"
repOperation value = "push"
size value = "toolarge"
images value = "/home/rakesh/from_t_jack/imag1.ipds"
status value = "completed"
replication value = "mode"
sdlist value = "f:"

我用过字典.我无法弄清楚sort我的字典.我想按照我声明的顺序对字典进行排序和打印.另一个问题是,如果在第6行观察到输出,则输出有点逆转,我期待:mode value = "replication".但输出是颠倒的.我不知道为什么它会以这种方式表现.我想知道如何解决这两个问题.这是我的代码.

  def generate_tag(name,val):

     full_tag = ''+name+' value = "'+val+'"'
     return full_tag



  def auto_xml(sdList,repOperation,images,status,md5):


            tags =          {'repOperationTag'  :{'repOperation_tag_name':'repOperation' ,
                                                  'repOperation_val'     : repOperation },

                            'modeTag'             :{'mode_tag_name':'mode' ,
                                                    'mode_val'     : 'replication' },


                             'imagesTag'        :{'images_tag_name': 'images',
                                                  'images_val'     : images  },

                             'statusTag'        :{'status_tag_name':'status' ,
                                                  'status_val'     : status},

                            'sizeTag'          :{'size_tag_name':'size' ,
                                                 'size_val'     : 'toolarge'},

                             'md5Tag'           :{'md5_tag_name'    : 'md5' ,
                                                  'md5_val'         : md5} ,

                             'sdListTag'        :{'sdList_tag_name': 'sdlist' , 
                                                  'sdList_val'     : sdList} }

            count = 0
            tag_len = len(tags)

            while count < tag_len :

                inner = tags[tags.keys()[count]]

                tag_name = inner[inner.keys()[0]]
                tag_val = inner[inner.keys()[1]]

                full_tag = generate_tag(tag_name,tag_val)

                print full_tag
                count = count + 1



  def myordered_dict(): 

      from collections import OrderedDict


      tags =         {'A_repOperationTag'  :{'tag_name':'repOperation' ,
                                      'tag_val'     : repOperation },

                'B_modeTag'             :{'tag_name':'mode' ,
                                        'tag_val'     : 'replication' },


                'C_imagesTag'        :{'tag_name': 'images',
                                      'tag_val'     : images  },

                'D_statusTag'        :{'tag_name':'status' ,
                                      'tag_val'     : status},

                'E_sizeTag'          :{'tag_name':'size' ,
                                     'tag_val'     : 'toolarge'},

                'F_md5Tag'           :{'tag_name'    : 'md5' ,
                                      'tag_val'         : md5} ,

                'G_sdListTag'        :{'tag_name': 'sdlist' , 
                                      'tag_val'     : sdList},

                'H_machinename'        :{'tag_name': 'hostname' , 
                                      'tag_val'     : 'xp_vm'} }

      tags = OrderedDict(sorted(tags.items(), key=lambda t: t[0]))

Answer 1

dict对象不是有序的 - 更准确地说,dict对象的顺序是依赖于实现,版本和插入顺序.这并不是说您插入密钥的顺序是您将其退回的顺序.它只暗示如果您将相同的键插入到两个不同的词典中,但顺序不同,那么您将退回的顺序并不保证是相同的.例如:

>>> print {1:1, 9:2}
{1: 1, 9: 2}
>>> print {9:1, 1:2}
{9: 1, 1: 2}

这并不意味着密钥保持其插入顺序.请注意如果我2在结尾处插入一个密钥会发生什么:

>>> print {9:1, 1:2, 2:4}
{9: 1, 2: 4, 1: 2}
>>> print {1:1, 9:2, 2:4}
{1: 1, 2: 4, 9: 2}

这是我回答的另一个答案,它讨论了如果你感兴趣的话,(C)python如何确定字典的顺序.

如果你需要一个dict维持秩序,你应该使用collections.OrderedDict.