At a glance
ItsLadynoir_ ranks #319,557 globally for subscribers (181K), #388,009 globally for views (40.2M), 158 videos published from US.
Subscribers
181,000
Views
40,277,394
Videos
158
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
254K
Total views ÷ total videos
Views per subscriber
223×
Total views ÷ subscribers
Subs per video
1.14K
Subscribers ÷ total videos
Analytics tables
About
🐞𝐇𝐞𝐥𝐥𝐨!🐞 ✨𝑰 𝒉𝒐𝒑𝒆 𝒚𝒐𝒖 𝒂𝒍𝒍 𝒆𝒏𝒋𝒐𝒚 𝒘𝒂𝒕𝒄𝒉𝒊𝒏𝒈 𝒎𝒚 𝒗𝒊𝒅𝒆𝒐𝒔. 𝑰𝒇 𝒚𝒐𝒖 𝒅𝒐 𝒃𝒆 𝒔𝒖𝒓𝒆 𝒕𝒐 𝒍𝒊𝒌𝒆 𝒂𝒏𝒅 𝒔𝒖𝒃𝒔𝒄𝒓𝒊𝒃𝒆 𝒇𝒐𝒓 𝒎𝒐𝒓𝒆✨ ⚠️𝑷𝒍𝒆𝒂𝒔𝒆 𝒅𝒐𝒏'𝒕 𝒓𝒆 𝒖𝒑𝒍𝒐𝒂𝒅 𝒎𝒚 𝒗𝒊𝒅𝒆𝒐𝒔 𝒐𝒓 𝒖𝒑𝒍𝒐𝒂𝒅 𝒂𝒏𝒚 𝒐𝒇 𝒕𝒉𝒆𝒎 𝒘𝒊𝒕𝒉𝒐𝒖𝒕 𝒎𝒚 𝒑𝒆𝒓𝒎𝒊𝒔𝒔𝒊𝒐𝒏…
- Handle
- @itsladynoir_
- Joined
- August 31, 2017
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.