@parsonline vs @kimeechan: Instagram Profiles Compared
A side-by-side comparison of the Instagram profiles of Parsonline | پارس آنلاين (@parsonline) and 김희찬 (@kimeechan) on 6 figures. @parsonline has 49.7K followers and @kimeechan has 100.1K, so @kimeechan has 2.0x as many. Over the 12 and 12 recent posts we analysed, @parsonline averages 90 likes and 12 comments a post, @kimeechan 1.2K and 21. Measured against followers, that is an engagement rate of 0.20% against 1.19%. How we calculate these numbers.
Parsonline | پارس آنلاين
@parsonline
پارسآنلایناولينارائهدهندهاينترنتپرسرعتدرايران ۱۵۸۵:شمارهسراسری☎️
View full stats page →| @parsonline | Metric | @kimeechan |
|---|---|---|
| 49.7K | Followers | 100.1K |
| not available | Total Posts | 230 |
| 12 | Recent Posts Analyzed | 12 |
| 90 | Avg Likes / Post | 1.2K |
| 12 | Avg Comments / Post | 21 |
| 102 | Avg Engagement / Post | 1.2K |
| 171 | Best Post Likes | 5.2K |
| 0.20% | Engagement Rate (%) | 1.19% |
| not available | Followers per Post | 435 |
Get this comparison as data
@parsonline and @kimeechan side by side as a CSV — followers, posts, engagement and 30-day growth. Add up to 25 accounts.
Download @parsonline in full
49.7K followers, 0 posts — followers, following and post history as CSV.
See export options →Download @kimeechan in full
100.1K followers, 230 posts — followers, following and post history as CSV.
See export options →Frequently Asked Questions
Who has more Instagram followers, @parsonline or @kimeechan?
@kimeechan has more Instagram followers than @parsonline: 100,058 vs 49,690.
Which account has higher engagement, @parsonline or @kimeechan?
@kimeechan has a higher Instagram engagement rate (1.19%) than @parsonline (0.2%).
How is this comparison calculated?
Follower count, post count, and verification status come directly from each Instagram profile. Engagement metrics (likes, comments, engagement rate) are computed from a sample of each account's most recent posts captured by IGDataHub. Engagement rate = (likes + comments) / followers per post × 100.