At a glance
小方看反差 ranks #2,410,763 globally for subscribers (2.58K), #1,060,633 globally for views (5.79M), 51 videos published from HK.
Subscribers
2,580
Views
5,796,597
Videos
51
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
113K
Total views ÷ total videos
Views per subscriber
2,247×
Total views ÷ subscribers
Subs per video
51
Subscribers ÷ total videos
Analytics tables
About
大家好,我是小方。 想了解一个国家,不能只听别人怎么说,更要看看普通人的真实生活。 这个频道会聊中国的发展,也会对比中国与世界各国在科技、制造业、基建、医疗、住房和生活方式上的差异。 我会结合外国博主的真实体验,用简单、客观的话,把现象背后的原因讲清楚。 不盲目夸赞,也不刻意贬低,只希望大家少一些偏见,多看一些事实。 欢迎订阅小方,一起看中国,也一起看懂这个变化中的世界。
- Handle
- @алланетипан-л4е
- Joined
- June 10, 2026
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.