CODEWITHSUNDEEP
pythonpandasgroup-bypivot-tablemean
UserYmY
UserYmY

Reputation: 8554

group by in group by and average

I have a dataframe like this:

cluster  org      time
   1      a       8
   1      a       6
   2      h       34
   1      c       23
   2      d       74
   3      w       6

I would like to calculate the average of time per org per cluster.

Expected result:

cluster mean(time)
1       15 #=((8 + 6) / 2 + 23) / 2
2       54 #=(74 + 34) / 2
3       6

I do not know how to do it in Pandas, can anybody help?

Upvotes: 134

Views: 440628

Answers (3)

cottontail
cottontail

Reputation: 23171

Another possible solution is to reshape the dataframe using pivot_table() then take mean(). Note that it's necessary to pass aggfunc='mean' (this averages time by cluster and org).

df.pivot_table(index='org', columns='cluster', values='time', aggfunc='mean').mean()

Another possibility is to use level parameter of mean() after the first groupby() to aggregate:

df.groupby(['cluster', 'org']).mean().mean(level='cluster')

Upvotes: 1

Zero
Zero

Reputation: 76927

If you want to first take mean on the combination of ['cluster', 'org'] and then take mean on cluster groups, you can use:

In [59]: (df.groupby(['cluster', 'org'], as_index=False).mean()
            .groupby('cluster')['time'].mean())
Out[59]:
cluster
1          15
2          54
3           6
Name: time, dtype: int64

If you want the mean of cluster groups only, then you can use:

In [58]: df.groupby(['cluster']).mean()
Out[58]:
              time
cluster
1        12.333333
2        54.000000
3         6.000000

You can also use groupby on ['cluster', 'org'] and then use mean():

In [57]: df.groupby(['cluster', 'org']).mean()
Out[57]:
               time
cluster org
1       a    438886
        c        23
2       d      9874
        h        34
3       w         6

Upvotes: 191

Vince Payandeh
Vince Payandeh

Reputation: 383

I would simply do this, which literally follows what your desired logic was:

df.groupby(['org']).mean().groupby(['cluster']).mean()

Upvotes: 21

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