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-8class 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: ",l4if __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 'VEVB武林网测试结果: '  print 'taskhost: '  print taskhost