I have a dataset that looks very similar to the data below.
I would like to create two groups using values in the group column.
group1 - new &
group2 - old
Within the group1 - new, I would like to then groupby id and take the average of each grouped id and then sum
data:
id type backup free total Last refreshed
a old 1 4 5 11/1/2020
b old 1 4 5 11/1/2020
c old 1 4 5 11/1/2020
d new 2 1 3 11/1/2020
e new 2 1 3 11/1/2020
f old 1 1 2 11/1/2020
g old 1 1 2 11/1/2020
e new 2 2 4 11/1/2020
d new 2 2 4 11/1/2020
1st step - create group1 and group2
id group type used free total Last refreshed
a group2 old 1 4 5 11/1/2020
b group2 old 1 4 5 11/1/2020
c group2 old 1 4 5 11/1/2020
f group2 old 1 1 2 11/1/2020
g group2 old 1 1 2 11/1/2020
d group1 new 2 1 3 11/1/2020
e group1 new 2 1 3 11/1/2020
e group1 new 2 2 4 11/1/2020
d group1 new 2 2 4 11/1/2020
2nd step groupby id
id group type used free total Last refreshed
d new group1 2 1 3 11/1/2020
d new group1 2 2 4 11/1/2020
e new group1 2 1 3 11/1/2020
e new group1 2 2 4 11/1/2020
a old group2 1 4 5 11/1/2020
b old group2 1 4 5 11/1/2020
c old group2 1 4 5 11/1/2020
f old group2 1 1 2 11/1/2020
g old group2 1 1 2 11/1/2020
3rd step we groupby id and take the average of d (for each column) we groupby id and take the average of e (for each column) then we sum the values in the old group (for each column) then we sum these altogether
id used free total Last refreshed
d 2 1.5 3.5 11/1/2020
e 2 1.5 3.5 11/1/2020
old 5 14 19 11/1/2020
Desired Result
All summed together to get:
used free total Last refreshed
9 17 26 11/1/2020
graph:
This is what I am doing:
2.place the type in groups 2.create calculated field
zn(avg(if [Group]= 'Group1' end))
+
zn(sum(if [Group] = 'Group2' end))
any suggestion is appreciated



ids are different, no two are same. So if you take average of these single rows (in group2) the result will be same. Moreover, as suggested earlier you have to pivot the columns to show the measure values together. What I am trying to say is that (i) if the ids are group are distinct the problem can be solved much easily, (ii) With your desired result shown above, you cannot achieve the visual (simply because in your result there are three measures while in chart only one) - AnilGoyal