具有相同电话号码的文档分组

Question

我的数据库包含大量酒店的集合（大约121000家）。你知道吗

我的收藏是这样的：

{
    "_id" : ObjectId("57bd5108f4733211b61217fa"),
    "autoid" : 1,
    "parentid" : "P01982.01982.110601173548.N2C5",
    "companyname" : "Sheldan Holiday Home",
    "latitude" : 34.169552,
    "longitude" : 77.579315,
    "state" : "JAMMU AND KASHMIR",
    "city" : "LEH Ladakh",
    "pincode" : 194101,
    "phone_search" : "9419179870|253013",
    "address" : "Sheldan Holiday Home|Changspa|Leh Ladakh-194101|LEH Ladakh|JAMMU AND KASHMIR",
    "email" : "",
    "website" : "",
    "national_catidlineage_search" : "/10255012/|/10255031/|/10255037/|/10238369/|/10238380/|/10238373/",
    "area" : "Leh Ladakh",
    "data_city" : "Leh Ladakh"
}

每个文档可以有一个或多个电话号码，电话号码之间用“|”分隔符分隔。你知道吗

我得把电话号码相同的文件放在一起。你知道吗

我所说的实时，是指当用户打开一个特定的酒店在web界面上查看其详细信息时，我应该能够显示所有与之链接的酒店，这些酒店按普通电话号码分组。你知道吗

在分组时，如果一家酒店链接到另一家酒店，而该酒店链接到另一家酒店，则所有3家酒店都应分组在一起。你知道吗

Example : Hotel A has phone numbers 1|2, B has phone numbers 3|4 and C has phone numbers 2|3, then A, B and C should be grouped together.

from pymongo import MongoClient
from pprint import pprint #Pretty print 
import re #for regex
#import unicodedata

client = MongoClient()

cLen = 0
cLenAll = 0
flag = 0
countA = 0
countB = 0
list = []
allHotels = []
conContact = []
conId = []
hotelTotal = []
splitListAll = []
contactChk = []

#We'll be passing the value later as parameter via a function call 
#hId = 37443; 

regx = re.compile("^Vivanta", re.IGNORECASE)

#Connection
db = client.hotel
collection = db.hotelData

#Finding hotels wrt search input
for post in collection.find({"companyname":regx}):
    list.append(post)

#Copying all hotels in a list
for post1 in collection.find():
    allHotels.append(post1)

hotelIndex = 11 #Index of hotel selected from search result
conIndex = hotelIndex
x = list[hotelIndex]["companyname"] #Name of selected hotel
y = list[hotelIndex]["phone_search"] #Phone numbers of selected hotel

try:
    splitList = y.split("|") #Splitting of phone numbers and storing in a list 'splitList'
except:
    splitList = y


print "Contact details of",x,":"

#Printing all contacts...
for contact in splitList:   
    print contact 
    conContact.extend(contact)
    cLen = cLen+1

print "No. of contacts in",x,"=",cLen


for i in allHotels:
    yAll = allHotels[countA]["phone_search"]
    try:
        splitListAll.append(yAll.split("|"))
        countA = countA+1
    except:
        splitListAll.append(yAll)
        countA = countA + 1
#   print splitListAll

#count = 0 

#This block has errors
#Add code to stop when no new links occur and optimize the outer for loop
#for j in allHotels:
for contactAll in splitListAll: 
    if contactAll in conContact:
        conContact.extend(contactAll)
#       contactChk = contactAll
#       if (set(conContact) & set(contactChk)):
#           conContact = contactChk
#           contactChk[:] = [] #drop contactChk list
        conId = allHotels[countB]["autoid"]
    countB = countB+1

print "Printing the list of connected hotels..."
for final in collection.find({"autoid":conId}):
    print final

这是我用Python写的一段代码。在这个例子中，我尝试在for循环中执行线性搜索。我现在得到一些错误，但它应该工作时纠正。你知道吗

我需要一个优化的版本，因为线性搜索的时间复杂度很低。你知道吗

我是一个非常新的，所以任何其他建议，以改善代码是欢迎的。你知道吗

谢谢。你知道吗

Answer 1

对于任何Python内存搜索问题，最简单的答案是“使用dict”。Dicts给出O（ln）键访问速度，list给出O（N）。你知道吗

还要记住，您可以将一个Python对象放入尽可能多的dict（或list）中，也可以在一个dict或list中放入尽可能多的次数。它们不是复制的。只是个参考。你知道吗

所以必需品看起来像

for hotel in hotels:
   phones = hotel["phone_search"].split("|")
   for phone in phones:
       hotelsbyphone.setdefault(phone,[]).append(hotel)

在这个循环的末尾，hotelsbyphone["123456"]将是一个hotel对象的列表，这些对象的phone_search字符串之一是“123456”。键编码特性是.setdefault(key, [])方法，如果该键不在dict中，它将初始化一个空列表，这样您就可以附加到它。你知道吗

一旦你建立了这个索引，这将很快

try:
    hotels = hotelsbyphone[x]
    # and process a list of one or more hotels
except KeyError:
    # no hotels exist with that number

除了try ... except，测试if x in hotelsbyphone: