At a glance
Suit Cat西猫 ranks #1,386,315 globally for subscribers (13.7K), #576,501 globally for views (20.9M), 920 videos published.
Subscribers
13,732
Views
20,932,450
Videos
920
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
22.7K
Total views ÷ total videos
Views per subscriber
1,524×
Total views ÷ subscribers
Subs per video
15
Subscribers ÷ total videos
Analytics tables
About
感谢兄弟们的订阅🗣️🔥 本频道将更新关于异环的任何咨询🗣️🔥(日后可能开发其他游戏咨询🔥) 有任何不懂的难题可以询问在留言区🔥或者在本频道的DC群询问🗣️🔥 🗣️里程碑: 2025/6/22日 5000订阅 2025/7/6日 6000订阅 2025/7/17日 7000订阅 2025/8/18日 8000订阅 2025/8/21日 9000订阅 2025/8/25日 10000订阅
- Handle
- @suitcat西猫
- Joined
- July 27, 2023
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.