You almost had it. You just need to add an order property to your a1 terms aggregations, like this: GET myindex/_search { “size”:0, “aggs”: { “a1”: { “terms”: { “field”: “FIELD1”, “size”:0, “order”: {“a2”: “desc”} <— add this }, “aggs”:{ “a2”:{ “sum”:{ “field”:”FIELD2.SUBFIELD” } } } } } }
I’m going to ignore Aggregation. It is not a very clearly defined concept and in my opinion it causes more confusion than it is worth. Composition and Association are quite enough, Craig Larman told me so. It might not be the answer your instructor was looking for but it is unlikely to be implemented in … Read more
I see this post is pretty old and there are some useful answers but this is a pretty crazy straight forward method… Select SUM(DISTINCT(n & 0x01)) + SUM(DISTINCT(n & 0x02)) + SUM(DISTINCT(n & 0x04)) as OrN From BitValues
(WARNING – WM_CONCAT is an unsupported function that was removed in version 12c. Unless you’re using a very old database, you should avoid this function. You should probably use LISTAGG instead.) It depends on the version of Oracle you’re using. If it supports the wm_concat() function, then you can simply do something like this: SELECT … Read more
The agg method can accept a dict, in which case the keys indicate the column to which the function is applied: grouped.agg({‘numberA’:’sum’, ‘numberB’:’min’}) For example, import numpy as np import pandas as pd df = pd.DataFrame({‘A’: [‘foo’, ‘bar’, ‘foo’, ‘bar’, ‘foo’, ‘bar’, ‘foo’, ‘foo’], ‘B’: [‘one’, ‘one’, ‘two’, ‘three’, ‘two’, ‘two’, ‘one’, ‘three’], ‘number A’: … Read more
consider the dataframe df df = pd.DataFrame(dict(A=list(‘aabb’), B=[1, 2, 3, 4], C=[0, 9, 0, 9])) groupby is the standard use aggregater df.groupby(‘A’).mean() maybe you want these values broadcast across the whole group and return something with the same index as what you started with. use transform df.groupby(‘A’).transform(‘mean’) df.set_index(‘A’).groupby(level=”A”).transform(‘mean’) agg is used when you have specific … Read more
Yes, certainly you can! That’s what the constructor initializer list is for. This is an essential feature that you require to initialize members that don’t have default constructors, as well as constants and references: class Foo { Bar x; // requires Bar::Bar(char) constructor const int n; double & q; public: Foo(double & a, char b) … Read more
I don’t believe there is a way to get only the value. You could just do ${{ total_paid.amount__sum }} in your template. Or do total_paid = Payment.objects.all().aggregate(Sum(‘amount’)).get(‘amount__sum’, 0.00) in your view. EDIT As others have pointed out, .aggregate() will always return a dictionary with all of the keys from the aggregates present, so doing .get() … Read more
consider the dataframe df df = pd.DataFrame(dict(A=list(‘aabb’), B=[1, 2, 3, 4], C=[0, 9, 0, 9])) groupby is the standard use aggregater df.groupby(‘A’).mean() maybe you want these values broadcast across the whole group and return something with the same index as what you started with. use transform df.groupby(‘A’).transform(‘mean’) df.set_index(‘A’).groupby(level=”A”).transform(‘mean’) agg is used when you have specific … Read more
One approach is the difference of row numbers: select name, count(*) from (select t.*, (row_number() over (order by id) – row_number() over (partition by name order by id) ) as grp from t ) t group by grp, name; The logic is easiest to understand if you run the subquery and look at the values … Read more