At a glance
파가시-아파트 가격시황 ranks #454,674 globally for subscribers (113K), #954,764 globally for views (7.53M), 316 videos published.
Subscribers
113,000
Views
7,531,124
Videos
316
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
23.8K
Total views ÷ total videos
Views per subscriber
67×
Total views ÷ subscribers
Subs per video
358
Subscribers ÷ total videos
Analytics tables
About
부동산 시장 분석 전문가 유재성입니다. 대학원에서 부동산학을 전공했습니다. 겸임교수로 재직하면서 대원과학대 부동산과에서 부동산 시장분석, 대원대학교 철도경영과에서 지하철 역세권 분석을 강의했습니다. 현재는 아파트 가격 시황과 이를 둘러싼 경제 시황을 실시간으로 분석해 영상을 제작하는 유튜브 채널 '파가시 - 아파트 가격 시황'을 운영하고 있습니다.
- Handle
- @pagasii
- Joined
- October 19, 2020
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.