Article "พราว" in Thai Wikipedia has 13.8 points for quality (as of November 1, 2023). The article contains 1 references and 7 sections. The article also contains quality flaw template, which reduce quality score.
In October 2023 the article "พราว" was edited by 1 authors in Thai Wikipedia (5543rd place) and written by 1 authors in all languages.
Since the creation of article "พราว", its content was written by 21 registered users of Thai Wikipedia (9910th place) and edited by 21 registered Wikipedia users in all languages.
The article is cited 23 times in Thai Wikipedia and cited 23 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 November 1, 2023 (including revision history and pageviews for previous years).
| Best Rank | Thai: #2245 04.2018 |
Global: #580571 04.2018 |
On 28 April 2024 in multilingual Wikipedia, Internet users most often read articles on the following topics: YouTube, Fallout, Facebook, Baby Reindeer, Challengers, Premier League, Fallout, deaths in 2024, Queen of Tears, Arsenal F.C..
In Thai Wikipedia the most popular articles on that day were: พิเศษ:ค้นหา, ปานปรีย์ พหิทธานุกร, องค์การกระจายเสียงและแพร่ภาพสาธารณะแห่งประเทศไทย, สโมสรฟุตบอลแมนเชสเตอร์ซิตี, สมศักดิ์ เทพสุทิน, ดวงใจเทวพรหม (ละครโทรทัศน์), ภาวะโลกร้อน, สโมสรฟุตบอลอาร์เซนอล, ฟุตซอลชิงแชมป์เอเชีย 2024, อสมท.
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 November, 2023. When calculating current popularity and AI of articles data from October 2023 was taken into account. For historical values of popularity and AI WikiRank used data from 2001 to 2023... More information
$$Quality=1/c∑↙{i=1}↖c nm_i-RS=1/5(11.39+0.88+20+33.33+7.06)-0.73$$
where:
The results of the calculations may be slightly different because of rounded values of normalized metrics in formulas
