Take the 2-minute tour ×
Academia Stack Exchange is a question and answer site for academics and those enrolled in higher education. It's 100% free, no registration required.

The eigenfactor is offered by ISI Web of Knowledge as an alternative journal score compared to the impact factor. It is explained on their site as follows:

The Eigenfactor Score calculation is based on the number of times articles from the journal published in the past five years have been cited in the JCR year, but it also considers which journals have contributed these citations so that highly cited journals will influence the network more than lesser cited journals. References from one article in a journal to another article from the same journal are removed, so that Eigenfactor Scores are not influenced by journal self-citation.

From this explanation, I am not sure whether the eigenfactor depends on the total number of articles that a journal publishes?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

Wikipedia’s article on the eigenfactor.org ranking of journals answers your question:

The Eigenfactor score is intended to measure the importance of a journal to the scientific community, by considering the origin of the incoming citations, and is thought to reflect how frequently an average researcher would access content from that journal. However, the Eigenfactor score is influenced by the size of the journal, so that the score doubles when the journal doubles in size (measured as published articles per year).

To give a simple example: the Journal of Physical Chemistry B is the highest-ranked journal in physical chemistry by this criterion… in good part because it carries the largest number of articles. It is a good journal, but under all other metrics it is not at the very top of its field.

share|improve this answer
    
Eingenfactor.org summarizes their methods here: eigenfactor.org/methods.php. A nice summary of this procedure would improve the answer. –  Ben Norris Nov 1 '12 at 22:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.