Python实现嵌套列表及字典并按某一元素去重复功能示例

#! /usr/bin/env python
#coding=utf-8
class HostScheduler(object):
  def __init__(self, resource_list):
    self.resource_list = resource_list
  def MergeHost(self):
    allResource=[]
    allResource.append(self.resource_list[0])
    for dict in self.resource_list:
      #print len(l4)
      k=0
      for item in allResource:
        #print 'item'
        if dict['host'] != item['host']:
          k=k+1
          #continue
        else:
          break
        if k == len(allResource):
          allResource.append(dict)
    taskhost=[]
    for item in allResource:
      taskhost.append(item['host'])
    return taskhost
#该函数实现嵌套列表中，按某一元素去重复
def deleteRepeat():
  #1、列表中嵌套列表。按元素‘b'实现去重复
  l1=[['b',1],['b',2],['c',3],['a',1],['b',1],['b',1],]
  l2=[]
  l2.append(l1[0])
  for data in l1:
    #print len(l2)
    k=0
    for item in l2:
      #print 'item'
      if data[0] != item[0]:
        k=k+1
      else:
        break
      if k == len(l2):
        l2.append(data)
  print "l2: ",l2
  #2、列表中嵌套字典。按键值host实现去重复
  l3=[{'host':'compute21', 'cpu':2},{'host':'compute21', 'cpu':2},{'host':'compute22', 'cpu':2},
    {'host':'compute23', 'cpu':2},{'host':'compute22', 'cpu':2},{'host':'compute23', 'cpu':2},
    {'host':'compute24', 'cpu':2}]
  l4=[]
  l4.append(l3[0])
  for dict in l3:
    #print len(l4)
    k=0
    for item in l4:
      #print 'item'
      if dict['host'] != item['host']:
        k=k+1
        #continue
      else:
        break
      if k == len(l4):
        l4.append(dict)
  print "l4: ",l4
if __name__ == '__main__':
  #deleteRepeat()
  resource_list=[{'host':'compute21', 'cpu':2},{'host':'compute21', 'cpu':2},{'host':'compute22', 'cpu':2},
          {'host':'compute23', 'cpu':2},{'host':'compute22', 'cpu':2},{'host':'compute23', 'cpu':2},
          {'host':'compute24', 'cpu':2}]
  hostSchedule=HostScheduler(resource_list)
  taskhost=hostSchedule.MergeHost()
  print '测试结果: '
  print 'taskhost: '
  print taskhost