Creating columns by subaggregating by condition in Snowflake SQL
I have following table:
id1 | id2 | n_products | daydiff
a | 1 | 12 | 12
a | 1 | 11 | 13
a | 1 | 90 | 46
a | 2 | 5 | 5
b | 2 | 15 | 15
b | 2 | 15 | 21
c | 3 | 90 | 7
I need to aggregate this table by id and using daydiff in the following manner:
- if
daydiffis less than 14 - if
daydiffis between 14 and 28 - if
daydiffis more than 28.
this should be aggregated using mean.
The result should be:
id1 | id2 | sub 14 | 14_28 | 28+
a | 1 | 11.5 | 0 | 46
a | 2 | 5 | 0 | 0
b | 2 | 0 | 15 | 0
a | 3 | 7 | 0 | 0
How can I achieve this? I guess this would involve some group by statements, but I am not sure how should they be applied
sql group-by snowflake-datawarehouse
edited Oct 17 at 6:45
Marcin Zukowski
1,8251819
asked Oct 12 at 12:07
johnnyheineken
156114
add a comment |
I have following table:
id1 | id2 | n_products | daydiff
a | 1 | 12 | 12
a | 1 | 11 | 13
a | 1 | 90 | 46
a | 2 | 5 | 5
b | 2 | 15 | 15
b | 2 | 15 | 21
c | 3 | 90 | 7
I need to aggregate this table by id and using daydiff in the following manner:
- if
daydiffis less than 14 - if
daydiffis between 14 and 28 - if
daydiffis more than 28.
this should be aggregated using mean.
The result should be:
id1 | id2 | sub 14 | 14_28 | 28+
a | 1 | 11.5 | 0 | 46
a | 2 | 5 | 0 | 0
b | 2 | 0 | 15 | 0
a | 3 | 7 | 0 | 0
How can I achieve this? I guess this would involve some group by statements, but I am not sure how should they be applied
sql group-by snowflake-datawarehouse
edited Oct 17 at 6:45
Marcin Zukowski
1,8251819
asked Oct 12 at 12:07
johnnyheineken
156114
add a comment |
I have following table:
id1 | id2 | n_products | daydiff
a | 1 | 12 | 12
a | 1 | 11 | 13
a | 1 | 90 | 46
a | 2 | 5 | 5
b | 2 | 15 | 15
b | 2 | 15 | 21
c | 3 | 90 | 7
I need to aggregate this table by id and using daydiff in the following manner:
- if
daydiffis less than 14 - if
daydiffis between 14 and 28 - if
daydiffis more than 28.
this should be aggregated using mean.
The result should be:
id1 | id2 | sub 14 | 14_28 | 28+
a | 1 | 11.5 | 0 | 46
a | 2 | 5 | 0 | 0
b | 2 | 0 | 15 | 0
a | 3 | 7 | 0 | 0
How can I achieve this? I guess this would involve some group by statements, but I am not sure how should they be applied
sql group-by snowflake-datawarehouse
edited Oct 17 at 6:45
Marcin Zukowski
1,8251819
asked Oct 12 at 12:07
johnnyheineken
156114
I have following table:
id1 | id2 | n_products | daydiff
a | 1 | 12 | 12
a | 1 | 11 | 13
a | 1 | 90 | 46
a | 2 | 5 | 5
b | 2 | 15 | 15
b | 2 | 15 | 21
c | 3 | 90 | 7
I need to aggregate this table by id and using daydiff in the following manner:
- if
daydiffis less than 14 - if
daydiffis between 14 and 28 - if
daydiffis more than 28.
this should be aggregated using mean.
The result should be:
id1 | id2 | sub 14 | 14_28 | 28+
a | 1 | 11.5 | 0 | 46
a | 2 | 5 | 0 | 0
b | 2 | 0 | 15 | 0
a | 3 | 7 | 0 | 0
How can I achieve this? I guess this would involve some group by statements, but I am not sure how should they be applied
sql group-by snowflake-datawarehouse
sql group-by snowflake-datawarehouse
edited Oct 17 at 6:45
Marcin Zukowski
1,8251819
asked Oct 12 at 12:07
johnnyheineken
156114
edited Oct 17 at 6:45
Marcin Zukowski
1,8251819
asked Oct 12 at 12:07
johnnyheineken
156114
edited Oct 17 at 6:45
Marcin Zukowski
1,8251819
edited Oct 17 at 6:45
Marcin Zukowski
1,8251819
edited Oct 17 at 6:45
Marcin Zukowski
1,8251819
1,8251819
asked Oct 12 at 12:07
johnnyheineken
156114
asked Oct 12 at 12:07
johnnyheineken
156114
asked Oct 12 at 12:07
johnnyheineken
156114
156114
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Gordon's answer is cross platform correct, but for myself I prefer the snowflake IFF syntax
SELECT id1, id2,
AVG(IFF(datediff < 14, n_products, NULL)) as avg_lt14,
AVG(IFF(datediff >= 14 and datediff <= 28, n_products, NULL)) as avg_14_28,
AVG(IFF(datediff > 29, n_products, NULL)) as avg_29pl
FROM t
GROUP BY id1, id2;
answered Nov 12 at 3:56
Simeon Pilgrim
5,1111929
@johnnyheinekin . . . Do you really prefer non-standard functions such asIIF()to standard SQL constructs such asCASE?
– Gordon Linoff
Nov 12 at 11:28
add a comment |
Use conditional aggregation:
select id1, id2,
avg(case when datediff < 14 then n_products end) as avg_lt14,
avg(case when datediff >= 14 and datediff <= 28 then n_products end) as avg_14_28,
avg(case when datediff > 29 then n_products end) as avg_29pl
from t
group by id1, id2;
Some databases calculate the averages of integers as an integer. I don't know if Snowflake does this. If so, then change n_products to n_products * 1.0.
answered Oct 12 at 12:10
Gordon Linoff
755k35290397
snowflake returns 3dp floats
– Simeon Pilgrim
Nov 12 at 3:53
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Gordon's answer is cross platform correct, but for myself I prefer the snowflake IFF syntax
SELECT id1, id2,
AVG(IFF(datediff < 14, n_products, NULL)) as avg_lt14,
AVG(IFF(datediff >= 14 and datediff <= 28, n_products, NULL)) as avg_14_28,
AVG(IFF(datediff > 29, n_products, NULL)) as avg_29pl
FROM t
GROUP BY id1, id2;
answered Nov 12 at 3:56
Simeon Pilgrim
5,1111929
@johnnyheinekin . . . Do you really prefer non-standard functions such asIIF()to standard SQL constructs such asCASE?
– Gordon Linoff
Nov 12 at 11:28
add a comment |
Gordon's answer is cross platform correct, but for myself I prefer the snowflake IFF syntax
SELECT id1, id2,
AVG(IFF(datediff < 14, n_products, NULL)) as avg_lt14,
AVG(IFF(datediff >= 14 and datediff <= 28, n_products, NULL)) as avg_14_28,
AVG(IFF(datediff > 29, n_products, NULL)) as avg_29pl
FROM t
GROUP BY id1, id2;
answered Nov 12 at 3:56
Simeon Pilgrim
5,1111929
@johnnyheinekin . . . Do you really prefer non-standard functions such asIIF()to standard SQL constructs such asCASE?
– Gordon Linoff
Nov 12 at 11:28
add a comment |
Gordon's answer is cross platform correct, but for myself I prefer the snowflake IFF syntax
SELECT id1, id2,
AVG(IFF(datediff < 14, n_products, NULL)) as avg_lt14,
AVG(IFF(datediff >= 14 and datediff <= 28, n_products, NULL)) as avg_14_28,
AVG(IFF(datediff > 29, n_products, NULL)) as avg_29pl
FROM t
GROUP BY id1, id2;
answered Nov 12 at 3:56
Simeon Pilgrim
5,1111929
Gordon's answer is cross platform correct, but for myself I prefer the snowflake IFF syntax
SELECT id1, id2,
AVG(IFF(datediff < 14, n_products, NULL)) as avg_lt14,
AVG(IFF(datediff >= 14 and datediff <= 28, n_products, NULL)) as avg_14_28,
AVG(IFF(datediff > 29, n_products, NULL)) as avg_29pl
FROM t
GROUP BY id1, id2;
answered Nov 12 at 3:56
Simeon Pilgrim
5,1111929
answered Nov 12 at 3:56
Simeon Pilgrim
5,1111929
answered Nov 12 at 3:56
Simeon Pilgrim
5,1111929
answered Nov 12 at 3:56
Simeon Pilgrim
5,1111929
5,1111929
@johnnyheinekin . . . Do you really prefer non-standard functions such asIIF()to standard SQL constructs such asCASE?
– Gordon Linoff
Nov 12 at 11:28
add a comment |
@johnnyheinekin . . . Do you really prefer non-standard functions such asIIF()to standard SQL constructs such asCASE?
– Gordon Linoff
Nov 12 at 11:28
@johnnyheinekin . . . Do you really prefer non-standard functions such as
IIF() to standard SQL constructs such as CASE?– Gordon Linoff
Nov 12 at 11:28
@johnnyheinekin . . . Do you really prefer non-standard functions such as
IIF() to standard SQL constructs such as CASE?– Gordon Linoff
Nov 12 at 11:28
add a comment |
Use conditional aggregation:
select id1, id2,
avg(case when datediff < 14 then n_products end) as avg_lt14,
avg(case when datediff >= 14 and datediff <= 28 then n_products end) as avg_14_28,
avg(case when datediff > 29 then n_products end) as avg_29pl
from t
group by id1, id2;
Some databases calculate the averages of integers as an integer. I don't know if Snowflake does this. If so, then change n_products to n_products * 1.0.
answered Oct 12 at 12:10
Gordon Linoff
755k35290397
snowflake returns 3dp floats
– Simeon Pilgrim
Nov 12 at 3:53
add a comment |
Use conditional aggregation:
select id1, id2,
avg(case when datediff < 14 then n_products end) as avg_lt14,
avg(case when datediff >= 14 and datediff <= 28 then n_products end) as avg_14_28,
avg(case when datediff > 29 then n_products end) as avg_29pl
from t
group by id1, id2;
Some databases calculate the averages of integers as an integer. I don't know if Snowflake does this. If so, then change n_products to n_products * 1.0.
answered Oct 12 at 12:10
Gordon Linoff
755k35290397
snowflake returns 3dp floats
– Simeon Pilgrim
Nov 12 at 3:53
add a comment |
Use conditional aggregation:
select id1, id2,
avg(case when datediff < 14 then n_products end) as avg_lt14,
avg(case when datediff >= 14 and datediff <= 28 then n_products end) as avg_14_28,
avg(case when datediff > 29 then n_products end) as avg_29pl
from t
group by id1, id2;
Some databases calculate the averages of integers as an integer. I don't know if Snowflake does this. If so, then change n_products to n_products * 1.0.
answered Oct 12 at 12:10
Gordon Linoff
755k35290397
Use conditional aggregation:
select id1, id2,
avg(case when datediff < 14 then n_products end) as avg_lt14,
avg(case when datediff >= 14 and datediff <= 28 then n_products end) as avg_14_28,
avg(case when datediff > 29 then n_products end) as avg_29pl
from t
group by id1, id2;
Some databases calculate the averages of integers as an integer. I don't know if Snowflake does this. If so, then change n_products to n_products * 1.0.
answered Oct 12 at 12:10
Gordon Linoff
755k35290397
answered Oct 12 at 12:10
Gordon Linoff
755k35290397
answered Oct 12 at 12:10
Gordon Linoff
755k35290397
answered Oct 12 at 12:10
Gordon Linoff
755k35290397
755k35290397
snowflake returns 3dp floats
– Simeon Pilgrim
Nov 12 at 3:53
add a comment |
snowflake returns 3dp floats
– Simeon Pilgrim
Nov 12 at 3:53
snowflake returns 3dp floats
– Simeon Pilgrim
Nov 12 at 3:53
snowflake returns 3dp floats
– Simeon Pilgrim
Nov 12 at 3:53
add a comment |
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