Vziav by ya banduru - Ukrainian folk song. Article "Взяв би я бандуру" in Ukrainian Wikipedia has 30.7 points for quality (as of November 1, 2023). The article contains 14 references and 5 sections.
Since the creation of article "Взяв би я бандуру", its content was written by 10 registered users of Ukrainian Wikipedia and edited by 10 registered Wikipedia users in all languages.
The article is cited 11 times in Ukrainian Wikipedia and cited 11 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 | Ukrainian: #41715 02.2016 |
Global: #2546530 11.2021 |
| Best Rank | Ukrainian: #1987 08.2013 |
Global: #443555 08.2013 |
On 25 June 2024 in multilingual Wikipedia, Internet users most often read articles on the following topics: UEFA Euro 2024, Julian Assange, UEFA European Championship, UEFA Euro 2020, Shifty Shellshock, 2024 Copa América, WikiLeaks, YouTube, Inside Out 2, House of the Dragon.
In Ukrainian Wikipedia the most popular articles on that day were: Тищенко Микола Миколайович, Чемпіонат Європи з футболу 2024, Содоль Юрій Іванович, Гнатов Андрій Вікторович, 169-й навчальний центр «Десна», Єрмак Андрій Борисович, Балога Андрій Вікторович, YouTube, Чемпіонат Європи з футболу, Джуліан Ассанж.
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
$$Quality=1/c∑↙{i=1}↖c nm_i-RS=1/5(15.9+15.91+0+21.74+100)$$
where:
The results of the calculations may be slightly different because of rounded values of normalized metrics in formulas
