At a glance
Perle的魔女手札 ranks #1,347,293 globally for subscribers (14.6K), #1,976,090 globally for views (1.02M), 39 videos published from TW.
Subscribers
14,629
Views
1,028,057
Videos
39
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
26.3K
Total views ÷ total videos
Views per subscriber
70×
Total views ÷ subscribers
Subs per video
375
Subscribers ÷ total videos
Analytics tables
About
晚安親愛的!這裡是Perle的魔女手札。 除了星露谷漫畫系列,魔女我還會做點小動畫, 這裡用來分享自製的動畫、或有機會開趕稿直播跟大家聊聊天。 🚫DO NOT REPOST🚫禁止轉載🚫 動畫 | 漫畫 | 水彩 | 星露谷物語 | 迷宮飯 Facebook: perle.artwork twitter: @Perle_Arte Instagram: @perle_arte
- Handle
- @perle_arte
- Joined
- July 10, 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.