pagerank

In SEO people talk a lot about Google PageRank . It's kind of a catch 22 because until your site is actually big and you don't really need search engines as much, it's unlikely that big sites will link to you and increase your PageRank! I've been told that it's easiest to simply get a couple high quality links to point to a site to raise it's PageRank. I've also been told that there are certain Open Directories like dmoz.org that Google pays special attention to (since they're human managed links). Can anyone speak to the validity of this or suggest another site/technique to increase a site's

I am working on building a transition matrix for implementing the PageRank algorithm. How could I use numpy to make sure that the columns add up to one. For example: 1 1 1 1 1 1 1 1 1 should be normalized to be .33 .33 .33 .33 .33 .33 .33 .33 .33 Divide the elements of each column by their column-summations - a/a.sum(axis=0,keepdims=1) # or simply : a/a.sum(0) For making the row-summations unity, change the axis input - a/a.sum(axis=1,keepdims=1) Sample run - In [78]: a = np.random.rand(4,5) In [79]: a Out[79]: array([[ 0.37, 0.74, 0.36, 0.41, 0.44], [ 0.51, 0.86, 0.91, 0.03, 0.76], [ 0.56, 0

I am trying to build a directed graph and compute personalized page rank over this graph. So suppose I have a graph with vertices {1,2,3,4} and edges going from 2, 3, and 4 to vertex 1, I would like to: (1) compute the personalized page rank of every vertex with respect to 1 (2) compute the personalized page rank of every vertex with respect to 2. The question is how I should pass this option in the personalized page rank function. The following code does not seem to do what I want: import networkx as nx G = nx.DiGraph() [G.add_node(k) for k in [1,2,3,4]] G.add_edge(2,1) G.add_edge(3,1) G.add

I'm playing around with networkx (graph library in python) and I found documentation saying the PageRank algorithm takes edge weights into account when scoring, but I was wondering if larger edge weights were better or lower weights better? Shortly, large weights are better for incoming nodes. PageRank works on a directed weighted graph. If page A has a link to page B, then the score for B goes up, i.e. the more input the page B (node) have, the higher is its score. Wikipedia article on PageRank for further details. Edit : let's make an experiment. Create a directed graph with 3 nodes and two

I have a list of domains and would like to get the: PageRank for all the domains. So just an integer, there must be an API that returns this. The number of results in Google The position of the word of the domainname. For example, "google.com" would be the position of the word "google". This could be position 20. denislexic This will have everything you need. https://github.com/eyecatchup/SEOstats Amazing. eyecatchUp I've published a Pagerank API that returns a json array, containing the checksum and the Pagerank of any URL. I also have a bunch of related tools and a Google code project that

In Wikidata ( Wikidata SPARQL endpoint ), is there a way to order the SPARQL query results with something like a PageRank? SELECT DISTINCT ?entity ?entityLabel WHERE { ?entity wdt:P31 wd:Q5. SERVICE wikibase:label { bd:serviceParam wikibase:language "en" . } } LIMIT 100 OFFSET 0 Can we specify a field to order the results by and that field expresses that the entity at the top is more notable/important/recognizable that the following one and so on? It seems that PageRank does not make much sense in relation to Wikidata. Obviously, large classes and large aggregates will be leaders. Also, unlike

Does google offer a way to programmatically see their search engine results for a certain query? I want to build a tracking application so that a user can see what rank on the google results their website is for certain keywords. EDIT: The behavior of the program would be: every day the program queries Google for the desired phrases, sees what position the user's websites are, and emails the users an update of their positions for their phrases. I want to be sure to comply with Google's terms of service too. John After finding this question I have been researching as the other answers seem out