토머스 브로디생스터

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토머스 브로디생스터

Quality:

Thomas Brodie-Sangster - English actor. This person is the 1317th most popular in the global Wikipedia ranking of people. Article "토머스 브로디생스터" in Korean Wikipedia has 3.8 points for quality (as of July 1, 2025). The article contains 0 references and 2 sections.

This article has the best quality in English Wikipedia. Also, this article is the most popular in that language version.

Achievements in all the time:
Global Wikipedia:
The 4407th most popular in all topics.
The 1317th most popular in people.
Achievements in the last month:
Global Wikipedia:
The 3320th most popular in all topics in the last month.

Since the creation of article "토머스 브로디생스터", its content was written by 9 registered users of Korean Wikipedia and edited by 2047 registered Wikipedia users in all languages.

Thomas Brodie-Sangster is on the 1317th place in global ranking of people on Wikipedia in all the time.

The article is cited 38 times in Korean Wikipedia and cited 1257 times in all languages.

The highest Authors Interest rank from 2001:

  • Local (Korean): #1209 in January 2018
  • Global: #402 in February 2018

The highest popularity rank from 2008:

  • Local (Korean): #2068 in January 2018
  • Global: #104 in December 2020

There are 35 language versions 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).

The table below shows the language versions of the article with the highest quality.

Languages with the highest quality

#LanguageQuality gradeQuality score
1English (en)
Thomas Brodie-Sangster
54.4675
2Russian (ru)
Броди-Сангстер, Томас
43.3907
3Bulgarian (bg)
Томас Броуди-Сангстър
40.6549
4Indonesian (id)
Thomas Brodie-Sangster
38.2984
5Uzbek (uz)
Thomas Brodie-Sangster
36.7225
6Spanish (es)
Thomas Brodie-Sangster
36.6
7Chinese (zh)
托马斯·桑斯特
34.1136
8Thai (th)
โทมัส แซงสเตอร์
32.324
9Armenian (hy)
Թոմաս Սանգստեր
31.4384
10Turkish (tr)
Thomas Brodie-Sangster
29.63
More...

The following table shows the most popular language versions of the article.

Most popular in all the time

The most popular language versions of the article "토머스 브로디생스터" in all the time
#LanguagePopularity awardRelative popularity
1English (en)
Thomas Brodie-Sangster
24 504 950
2French (fr)
Thomas Brodie-Sangster
3 404 672
3German (de)
Thomas Brodie-Sangster
3 320 809
4Russian (ru)
Броди-Сангстер, Томас
3 225 803
5Spanish (es)
Thomas Brodie-Sangster
2 671 847
6Italian (it)
Thomas Brodie-Sangster
1 600 231
7Portuguese (pt)
Thomas Brodie-Sangster
937 844
8Chinese (zh)
托马斯·桑斯特
731 226
9Japanese (ja)
トーマス・ブロディ=サングスター
652 240
10Dutch (nl)
Thomas Brodie-Sangster
519 694
More...

The following table shows the language versions of the article with the highest popularity in the last month.

Most popular in June 2025

The most popular language versions of the article "토머스 브로디생스터" in June 2025
#LanguagePopularity awardRelative popularity
1English (en)
Thomas Brodie-Sangster
157 888
2German (de)
Thomas Brodie-Sangster
16 784
3Spanish (es)
Thomas Brodie-Sangster
11 117
4French (fr)
Thomas Brodie-Sangster
9 865
5Russian (ru)
Броди-Сангстер, Томас
9 433
6Italian (it)
Thomas Brodie-Sangster
3 920
7Japanese (ja)
トーマス・ブロディ=サングスター
2 566
8Chinese (zh)
托马斯·桑斯特
2 430
9Persian (fa)
توماس سنگستر
1 996
10Portuguese (pt)
Thomas Brodie-Sangster
1 913
More...

The following table shows the language versions of the article with the highest Authors’ Interest.

The highest AI

Language versions of the article "토머스 브로디생스터" with the highest Authors Interest (number of authors). Only registered Wikipedia users were taken into account.
#LanguageAI awardRelative AI
1English (en)
Thomas Brodie-Sangster
770
2French (fr)
Thomas Brodie-Sangster
186
3Spanish (es)
Thomas Brodie-Sangster
148
4Italian (it)
Thomas Brodie-Sangster
120
5German (de)
Thomas Brodie-Sangster
110
6Russian (ru)
Броди-Сангстер, Томас
98
7Dutch (nl)
Thomas Brodie-Sangster
70
8Portuguese (pt)
Thomas Brodie-Sangster
67
9Hebrew (he)
תומאס ברודי-סאנגסטר
49
10Czech (cs)
Thomas Brodie-Sangster
40
More...

The following table shows the language versions of the article with the highest Authors’ Interest in the last month.

The highest AI in June 2025

Language versions of the article "토머스 브로디생스터" with the highest AI in June 2025
#LanguageAI awardRelative AI
1English (en)
Thomas Brodie-Sangster
3
2Spanish (es)
Thomas Brodie-Sangster
3
3Czech (cs)
Thomas Brodie-Sangster
2
4Finnish (fi)
Thomas Brodie-Sangster
1
5Italian (it)
Thomas Brodie-Sangster
1
6Arabic (ar)
توماس سانجستر
0
7Bulgarian (bg)
Томас Броуди-Сангстър
0
8Danish (da)
Thomas Brodie-Sangster
0
9German (de)
Thomas Brodie-Sangster
0
10Greek (el)
Τόμας Μπρόντι-Σάνγκστερ
0
More...

The following table shows the language versions of the article with the highest number of citations.

The highest CI

Language versions of the article "토머스 브로디생스터" with the highest Citation Index (CI)
#LanguageCI awardRelative CI
1English (en)
Thomas Brodie-Sangster
179
2Russian (ru)
Броди-Сангстер, Томас
97
3French (fr)
Thomas Brodie-Sangster
86
4German (de)
Thomas Brodie-Sangster
65
5Spanish (es)
Thomas Brodie-Sangster
60
6Italian (it)
Thomas Brodie-Sangster
57
7Portuguese (pt)
Thomas Brodie-Sangster
52
8Japanese (ja)
トーマス・ブロディ=サングスター
48
9Indonesian (id)
Thomas Brodie-Sangster
47
10Polish (pl)
Thomas Brodie-Sangster
45
More...

Scores

Estimated value for Wikipedia:
Korean:
Global:
Popularity in June 2025:
Korean:
Global:
Popularity in all years:
Korean:
Global:
Authors in June 2025:
Korean:
Global:
Registered authors in all years:
Korean:
Global:
Citations:
Korean:
Global:

Quality measures

Interwikis

#LanguageValue
arArabic
توماس سانجستر
bgBulgarian
Томас Броуди-Сангстър
csCzech
Thomas Brodie-Sangster
daDanish
Thomas Brodie-Sangster
deGerman
Thomas Brodie-Sangster
elGreek
Τόμας Μπρόντι-Σάνγκστερ
enEnglish
Thomas Brodie-Sangster
esSpanish
Thomas Brodie-Sangster
etEstonian
Thomas Brodie-Sangster
faPersian
توماس سنگستر
fiFinnish
Thomas Brodie-Sangster
frFrench
Thomas Brodie-Sangster
heHebrew
תומאס ברודי-סאנגסטר
huHungarian
Thomas Brodie-Sangster
hyArmenian
Թոմաս Սանգստեր
idIndonesian
Thomas Brodie-Sangster
itItalian
Thomas Brodie-Sangster
jaJapanese
トーマス・ブロディ=サングスター
kkKazakh
Томас Саңстер
koKorean
토머스 브로디생스터
msMalay
Thomas Brodie-Sangster
nlDutch
Thomas Brodie-Sangster
noNorwegian
Thomas Brodie-Sangster
plPolish
Thomas Brodie-Sangster
ptPortuguese
Thomas Brodie-Sangster
roRomanian
Thomas Brodie-Sangster
ruRussian
Броди-Сангстер, Томас
simpleSimple English
Thomas Brodie-Sangster
svSwedish
Thomas Brodie-Sangster
thThai
โทมัส แซงสเตอร์
trTurkish
Thomas Brodie-Sangster
ukUkrainian
Томас Сангстер
uzUzbek
Thomas Brodie-Sangster
viVietnamese
Thomas Brodie-Sangster
zhChinese
托马斯·桑斯特

Popularity rank trends

Best Rank Korean:
#2068
01.2018
Global:
#104
12.2020

AI rank trends

Best Rank Korean:
#1209
01.2018
Global:
#402
02.2018

Local popularity rank history

Global popularity rank history

Local AI rank history

Global AI rank history

Languages comparison

Important global interconnections (July 2024 – June 2025)

Wikipedia readers most often find their way to information on Thomas Brodie-Sangster from Wikipedia articles about Talulah Riley, Love Actually, The Queens Gambit, The Maze Runner and Nanny McPhee. Whereas reading the article about Thomas Brodie-Sangster people most often go to Wikipedia articles on Talulah Riley, List of A Song of Ice and Fire characters, Love Actually, The Artful Dodger and Wolf Hall: The Mirror and the Light.

Cumulative results of quality and popularity of the Wikipedia article

List of Wikipedia articles in different languages (starting with the most popular):

News from 12 August 2025

On 12 August 2025 in multilingual Wikipedia, Internet users most often read articles on the following topics: Cristiano Ronaldo, Wednesday, Georgina Rodríguez, ChatGPT, Weapons, Jenna Ortega, deaths in 2025, Taylor Swift, Miguel Uribe Turbay, 2025–26 UEFA Champions League.

In Korean Wikipedia the most popular articles on that day were: 문화방송, 김건희, 한국방송공사, 한국교육방송공사, 케이팝 데몬 헌터스, 윤미향, 유경촌, 노란봉투법, 윤석열, 강미정.

About WikiRank

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