At a glance
CHOCOPIEㅠㅠ ranks #6,009,383 globally for subscribers (19), #4,034,167 globally for views (31.1K), 26 videos published.
Subscribers
19
Views
31,154
Videos
26
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
1.19K
Total views ÷ total videos
Views per subscriber
1,640×
Total views ÷ subscribers
Subs per video
1
Subscribers ÷ total videos
Analytics tables
About
たくさんのKーPOPを聞いてきて、たくさんの大好きでお気に入りな曲達に出会う事が出来ました💓💭 KーPOPオタなら知ってるタイトル曲から、ファンの間で名曲だと言われるファンソング、アルバム曲、その魅力を他のグループのファンの方が知る機会になればと想い投稿しています🧸🌟 あくまで、韓国語勉強中の趣味として投稿しています! 誤訳等ございましたら、優しく教えて下さるとありがたいです! 特に英語は分野では無いので、、🙇♂️
- Handle
- @sui9795
- Joined
- May 2, 2022
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.