Federico Scribani - Italian actor. Article “Federico Scribani” in Italian Wikipedia has 15.4 points for quality (as of July 1, 2025). The article contains 3 references and 10 sections.
In June 2025 the article “Federico Scribani” was edited by 1 authors in Italian Wikipedia and written by 1 authors in all languages.
Since the creation of article “Federico Scribani”, its content was written by 6 registered users of Italian Wikipedia and edited by 6 registered Wikipedia users in all languages.
The article is cited 41 times in Italian Wikipedia and cited 41 times in all languages.
The highest Authors Interest rank from 2001:
The highest popularity rank from 2008:
There is 1 language version for this article in the WikiRank database (of the considered 55 Wikipedia language editions).
The quality and popularity assessment was based on Wikipédia dumps from July 1, 2025 (including revision history and pageviews for previous years).
| Best Rank | Italian: #51786 03.2025 |
Global: #1013376 03.2025 |
| Best Rank | Italian: #11010 12.2024 |
Global: #469206 12.2024 |
On 16 March 2026 in multilingual Wikipedia, Internet users most often read articles on the following topics: 98th Academy Awards, Michael B. Jordan, Sinners, Jessie Buckley, Paul Thomas Anderson, Sean Penn, One Battle After Another, Amy Madigan, Hamnet, Timothée Chalamet.
In Italian Wikipedia the most popular articles on that day were: Premi Oscar 2026, Una battaglia dopo l'altra, Michael B. Jordan, I peccatori, Andrea Kimi Antonelli, Jessie Buckley, Sean Penn, Paul Thomas Anderson, Amy Madigan, Hamnet - Nel nome del figlio.
The WikiRank project is intended for automatic relative evaluation of the articles in the various language versions of Wikipedia. At the moment the service allows to compare over 44 million Wikipedia articles in 55 languages. Quality scores of articles are based on Wikipedia dumps from July, 2025. When calculating current popularity and AI of articles data from June 2025 was taken into account. For historical values of popularity and AI WikiRank used data from 2001 to 2025... More information
$$Quality=1/c∑↙{i=1}↖c nm_i-RS=1/5(14.63+1.78+5.26+43.48+11.93)$$
where:
The results of the calculations may be slightly different because of rounded values of normalized metrics in formulas
