Frondicularia - genus of amoeboid protists. Article “Frondicularia” in Spanish Wikipedia has 20.9 points for quality (as of July 1, 2025). The article contains 6 references and 3 sections.
Since the creation of article “Frondicularia”, its content was written by 2 registered users of Spanish Wikipedia and edited by 2 registered Wikipedia users in all languages.
The article is cited 6 times in Spanish Wikipedia and cited 6 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 | Spanish: #423444 08.2008 |
Global: #4949379 08.2008 |
| Best Rank | Spanish: #48708 01.2016 |
Global: #895347 01.2016 |
On 28 January 2026 in multilingual Wikipedia, Internet users most often read articles on the following topics: 2025–26 UEFA Champions League, Nipah virus, Doomsday Clock, UEFA Champions League, Elena Rybakina, Donald Trump, 2026 European Men's Handball Championship, killing of Alex Pretti, Kristi Noem, deaths in 2026.
In Spanish Wikipedia the most popular articles on that day were: Liga de Campeones de la UEFA, Santiago Ascacíbar, Violencia vicaria, Kendry Páez, Ángel Romero Villamayor, Liga de Campeones de la UEFA 2025-26, Mariano Barbacid, José Mourinho, Glòria Serra, XXX.
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
$$Quality=1/c∑↙{i=1}↖c nm_i-RS=1/5(4.59+3.9+0+13.64+82.41)$$
where:
The results of the calculations may be slightly different because of rounded values of normalized metrics in formulas
