Python: classify text into the categories

Question

I have a part of training set

url  category
ebay.com/sch/Mens-Clothing-/1059/i.html?_from=R40&LH_BIN=1&Bottoms%2520Size%2520%2528Men%2527s%2529=33&Size%2520Type=Regular&_nkw=Джинсы&_dcat=11483&Inseam=33&rt=nc&_trksid=p2045573.m1684 Онлайн-магазин
google.ru/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=%D0%BA%D0%BA%D1%83%D0%BF%D0%BE%D0%BD%D1%8B%20aliexpress%202016  Search
google.ru/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#newwindow=1&q=%D0%BA%D1%83%D0%BF%D0%BE%D0%BD%D1%8B+aliexpress+2016    Search
google.ru/search?q=авито&oq=авито&aqs=chrome..69i57j0l5.1608j0j7&sourceid=chrome&es_sm=122&ie=UTF-8 Search
irecommend.ru/content/kogda-somnenii-byt-ne-mozhet-tolko-klear-blyu-pomozhet    Форумы и отзывы
ebay.com/sch/Mens-Clothing-/1059/i.html?_from=R40&LH_BIN=1&Bottoms%2520Size%2520%2528Men%2527s%2529=33&Size%2520Type=Regular&_dcat=11483&Inseam=33&_nkw=Джинсы&_sop=15  Онлайн-магазин
ebay.com/sch/Mens-Clothing-/1059/i.html?_from=R40&LH_BIN=1&Bottoms%2520Size%2520%2528Men%2527s%2529=33&Size%2520Type=Regular&_dcat=11483&Inseam=33&_nkw=Джинсы&_sop=15  Онлайн-магазин
irecommend.ru/content/gramotnyi-razvod-na-dengi-bolshe-ne-kuplyu-vret   Форумы и отзывы
google.ru/search?q=яндекс&oq=яндекс&aqs=chrome..69i57j69i61l3j69i59l2.1383j0j1&sourceid=chrome&es_sm=93&ie=UTF-8    Search
google.ru/search?q=авито&oq=авито&aqs=chrome..69i57j69i59j69i60.1095j0j1&sourceid=chrome&es_sm=93&ie=UTF-8  Search
otzovik.com/review_1399716.html#debug   Форумы и отзывы
svyaznoy.ru Онлайн-магазин
mvideo.ru/smartfony-sotovye-telefony/apple-iphone-2927  Онлайн-магазин
mvideo.ru/promo/rassrochka-0-0-12-mark24197850/f/category=iphone-914?sort=priceLow&_=1453896710474&categoryId=10    Онлайн-магазин
svyaznoy.ru/catalog/phone/224/tag/windows-phone Онлайн-магазин
google.it/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=%D0%B5%D0%B2%D1%80%D0%BE%D1%81%D0%B5%D1%82%D1%8C    Search
vk.com   Social network

it's a connection between url and category And also I have test set and I need to get category to every url.

url    
vk.com/topic-102849764_32295213
stats.stackexchange.com/questions/19048/what-is-the-difference-between-test-set-and-validation-set
google.ru/search?q=learning+sample&oq=learning+sample&aqs=chrome..69i57.4063j0j1&sourceid=chrome&ie=UTF-8#newwindow=1&q=machine+learning+test+and+learn
facebook.com
locals.ru
tvzvezda.ru/news/vstrane_i_mire/content/201609261038-k6n1.htm

I don't know, what algorithm should I use to solve this task. I need the best way to get the most accuracy. And I think it's a problem, that I have multiple categories.

I try first parse html tag title , because I think, that I can's determine category only with url .

Answer 1

Basically you will classify strings into categories. Therefore you will to use a classifier. But you will not just use one classifier but rather test several and chose the most accurate.

Yet firstly, you will have to think about features of each url. I expect that you will not achieve great accuracy if you are simply feeding the url as a string and as the only feature.

Rather you will preprocess each url to extract features. The choice of relevant/useful features strongly depends on the domain. A feature could be:

simple features

• the first word until the dot such as: facebook for "facebook.com"

• the length of the whole string

complex features

imagine you define keywords for each cluster such as for "online-shopping"-cluster you will define [promo, buy, shop, sell, price], then you can compute the number of keywords which occur in the string for each cluster as a feature

Therefore, you will have to continue firstly with feature-engineering and secondly with a comparing classifier performance.

Additional input:

Similiar question on SO (regarding URL features)

Text feature extraction

Fast Webpage Classification Using URL Features

EDIT: An example

url = "irecommend.ru/content/kogda-somnenii-byt-ne-mozhet-tolko-klear-blyu-pomozhet"    

f1  = len(url) = 76
f2 = base = str(url).split("/",1)[0] = "irecommend.ru"
f3 = segments = str(a).count("/") = 2

more solutions from here by Eiyrioü von Kauyf

import string
count = lambda l1,l2: sum([1 for x in l1 if x in l2])

f4 = count_punctuation = count(a,set(string.punctuation))
f5 = count_ascii = count(a,set(string.ascii_letters))

Yet all these examples are very simple features, which do not cover the semantic content of the URL. Depending on the depth/sophistication of your target variables (clusters), you might need to use features n-gram based features such as in here