第2集 最佳拍档
Published May 29, 2013
At a glance
Views
842,014
Likes
710
Comments
0
Performance vs. channel
All-time outlier score
Engagement rate
More features coming
We're working on new analytics and tools. Stay tuned.
About
뽀로로의 새로운 영상을 가장 빠르게 만나는 방법! 지금 바로 뽀롱뽀롱 뽀로로 공식 유튜브 채널 [구독]버튼을 눌러주세요. ★뽀로로 유튜브 구독하기: https://www.youtube.com/channel/UCPUeGC_AL-OnDQORKhRm6iA?sub_confirmation=1 小企鹅 啵乐乐
- Published
- May 29, 2013
- Made for kids
- Yes
More features coming
We're working on new analytics and tools. Stay tuned.
How 第2集 最佳拍档 is doing
It has 842K views in total. It ranks #5,395,989 among tracked YouTube videos by views.
Key facts about 第2集 最佳拍档
- Total views
- 842K
- Total likes
- 710
- Total comments
- 0
- Published
- May 29, 2013
- Global rank by views
- #5,395,989
Frequently asked questions
How many views does 第2集 最佳拍档 have?
第2集 最佳拍档 has 842,014 views, 710 likes, ranking #5,395,989 among tracked YouTube videos by views.
How accurate are these YouTube video analytics?
View, like, and comment 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 the creator sees privately in YouTube Studio.
What is the outlier score?
How this video's views compare to the channel's usual Shorts or long-form video. A 2x score means double the usual views for that channel, 0.5x means half.
What is engagement rate and how is it calculated?
Likes and comments per view, compared to this channel's typical engagement rate — shown as above or below what's usual for the channel.
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 count — not actual creator earnings, which depend on ad rates, audience geography, and monetization settings YouTube doesn't expose publicly.
Is this YouTube video analytics checker free?
Yes. You get 10 free channel and video analytics lookups per 24 hours without an account — logging in, which is also free, removes that limit entirely.