At a glance
판슥 ranks #143,294 globally for subscribers (478K), #78,837 globally for views (350M), 444 videos published from KR.
Subscribers
478,554
Views
350,671,551
Videos
444
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
789K
Total views ÷ total videos
Views per subscriber
733×
Total views ÷ subscribers
Subs per video
1.07K
Subscribers ÷ total videos
Analytics tables
About
MC사회문의, 행사문의, 업체광고문의, 프랜차이즈 브랜딩 S.V 라이브 커머스 (Live Commerce), 라이브 쇼핑 (Live Shopping) Corporate Franchise Branding, Franchise Brand Development ※전략 비즈니스 상세문의 💌Email: [email protected] 🟡KakaoTalk: kms1084
- Handle
- @판슥
- Joined
- August 13, 2018
More features coming
We're working on new analytics and tools. Stay tuned.
More features coming
We're working on new analytics and tools. Stay tuned.
Frequently asked questions
How accurate are these YouTube channel analytics?
Subscriber, view, and video counts come from YouTube's public API and refresh regularly. Growth and revenue figures are estimates built from tracked historical data, so they can differ slightly from what a channel owner sees privately in YouTube Studio.
How is a channel's global and country rank calculated?
Rank compares this channel's subscriber count and view count against every other channel Socialcounts.org tracks, both worldwide and within its home country.
How is the revenue estimate calculated?
It's a rough public estimate based on a $0.7–$2.0 CPM (revenue per 1,000 views) applied to view counts — not actual creator earnings, which depend on ad rates, audience geography, and monetization settings YouTube doesn't expose publicly.
Can I check analytics for any YouTube channel, not just my own?
Yes — search any public YouTube channel by name, handle, or channel URL. No login is required to see the basics.
Is this YouTube channel analytics checker free?
Yes. You get 10 free channel and video analytics lookups per rolling hour without an account — logging in, which is also free, removes that limit entirely.