GROUPAVERAGE
- Juliane Wetzel
Syntax
GROUPAVERAGE(<number>)
Description
Returns the average value of a group by creating a sum for all values in the group and dividing by the number of values. A GROUPBY() function must first be used in order to then find the average arguments of another column in relation to it. The value of the GROUPAVERAGE() must be a number and is returned as a float.
Empty or <null> records of a group aren't calculated into GROUPAVERAGE().
Calculating a column with a (or multiple) <error> records always returns an error for its group.
This is an aggregate function.
Example
Given the following data set:
| user_screen_name | user_follower_count | user_friend_count | user_time_zone |
|---|---|---|---|
| ArteWorks_SEO | 2605 | 345 | Central Time (US & Canada) |
| evan_b | 460 | 316 | Pacific Time (US & Canada) |
| briancoffee | 185 | 321 | London |
| PhillisGene | 280 | <null> | London |
| SsReyes | 4 | 4 | Pacific Time (US & Canada) |
| addztickets | 291 | 0 | Central Time (US & Canada) |
| mynossseee | 190 | 94 | Melbourne |
| AprilBraswell | 14206 | 14082 | Pacific Time (US & Canada) |
| bebacklatersoon | 234 | 103 | Sydney |
| nicolebaker76 | 7 | 27 | Alaska |
| RNIHA | 9 | 3 | London |
| Object5 | 512 | <error> | Melbourne |
| tman2088 | 108 | 155 | Central Time (US & Canada) |
First create a group, e.g. GROUPBY(#RawData!user_time_zone)
| user_time_zone |
|---|
| Alaska |
| Central Time (US & Canada) |
| London |
| Melbourne |
| Pacific Time (US & Canada) |
| Sydney |
Then use the GROUPAVERAGE(#RawData!user_friends_count), and the average value of the arguments of that column appears in relation to the GROUPBY() column.
| user_time_zone | GROUPAVERAGE returns |
|---|---|
| Alaska | 27 |
| Central Time (US & Canada) | 166.6666667 |
| London | 162 |
| Melbourne | <error> |
| Pacific Time (US & Canada) | 4800.666667 |
| Sydney | 27 |