How to flatten values in a row in Google Big Query
guys
I've a problem with my data in big query. I have a dataset like pic #1 and I need to rank my users and group their ranking notes in a unique row (like pic #2). To be clear, my ranking column type is int, not an array.
Is it possible to handle this w/bigquery or do I need to transport this dataset to python and do this transformation there ?
PIC #1 ORIGINAL DATASET
PIC #2 GROUPED AND FORMATED DATASET
google-bigquery flatten edited Nov 15 '18 at 8:01
Sagar Zala
2,33441236
asked Nov 14 '18 at 20:09
João
51
add a comment |
guys
I've a problem with my data in big query. I have a dataset like pic #1 and I need to rank my users and group their ranking notes in a unique row (like pic #2). To be clear, my ranking column type is int, not an array.
Is it possible to handle this w/bigquery or do I need to transport this dataset to python and do this transformation there ?
PIC #1 ORIGINAL DATASET
PIC #2 GROUPED AND FORMATED DATASET
google-bigquery flatten edited Nov 15 '18 at 8:01
Sagar Zala
2,33441236
asked Nov 14 '18 at 20:09
João
51
add a comment |
guys
I've a problem with my data in big query. I have a dataset like pic #1 and I need to rank my users and group their ranking notes in a unique row (like pic #2). To be clear, my ranking column type is int, not an array.
Is it possible to handle this w/bigquery or do I need to transport this dataset to python and do this transformation there ?
PIC #1 ORIGINAL DATASET
PIC #2 GROUPED AND FORMATED DATASET
google-bigquery flatten edited Nov 15 '18 at 8:01
Sagar Zala
2,33441236
asked Nov 14 '18 at 20:09
João
51
guys
I've a problem with my data in big query. I have a dataset like pic #1 and I need to rank my users and group their ranking notes in a unique row (like pic #2). To be clear, my ranking column type is int, not an array.
Is it possible to handle this w/bigquery or do I need to transport this dataset to python and do this transformation there ?
PIC #1 ORIGINAL DATASET
PIC #2 GROUPED AND FORMATED DATASET
google-bigquery flatten google-bigquery flatten edited Nov 15 '18 at 8:01
Sagar Zala
2,33441236
asked Nov 14 '18 at 20:09
João
51
edited Nov 15 '18 at 8:01
Sagar Zala
2,33441236
asked Nov 14 '18 at 20:09
João
51
edited Nov 15 '18 at 8:01
Sagar Zala
2,33441236
edited Nov 15 '18 at 8:01
Sagar Zala
2,33441236
edited Nov 15 '18 at 8:01
Sagar Zala
2,33441236
2,33441236
asked Nov 14 '18 at 20:09
João
51
asked Nov 14 '18 at 20:09
João
51
asked Nov 14 '18 at 20:09
João
51
51
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Below example for BigQuery Standard SQL
#standardSQL
WITH `project.dataset.table` AS (
SELECT 'a' user, 10 ranking UNION ALL
SELECT 'b', 2 UNION ALL
SELECT 'a', 12 UNION ALL
SELECT 'a', 14 UNION ALL
SELECT 'c', 22 UNION ALL
SELECT 'd', 21
)
SELECT
user,
MAX(ranking) AS ranking_max,
STRING_AGG(CAST(ranking AS STRING)) ranking_list
FROM `project.dataset.table`
GROUP BY user
with result
Row user ranking_max ranking_list
1 a 14 10,12,14
2 b 2 2
3 c 22 22
4 d 21 21
Note: if you need ranking_list to be ordered - you can use ORDER BY in STRING_AGG as in below
STRING_AGG(CAST(ranking AS STRING) ORDER BY ranking) ranking_list
answered Nov 14 '18 at 20:15
Mikhail Berlyant
55.9k43368
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Below example for BigQuery Standard SQL
#standardSQL
WITH `project.dataset.table` AS (
SELECT 'a' user, 10 ranking UNION ALL
SELECT 'b', 2 UNION ALL
SELECT 'a', 12 UNION ALL
SELECT 'a', 14 UNION ALL
SELECT 'c', 22 UNION ALL
SELECT 'd', 21
)
SELECT
user,
MAX(ranking) AS ranking_max,
STRING_AGG(CAST(ranking AS STRING)) ranking_list
FROM `project.dataset.table`
GROUP BY user
with result
Row user ranking_max ranking_list
1 a 14 10,12,14
2 b 2 2
3 c 22 22
4 d 21 21
Note: if you need ranking_list to be ordered - you can use ORDER BY in STRING_AGG as in below
STRING_AGG(CAST(ranking AS STRING) ORDER BY ranking) ranking_list
answered Nov 14 '18 at 20:15
Mikhail Berlyant
55.9k43368
add a comment |
Below example for BigQuery Standard SQL
#standardSQL
WITH `project.dataset.table` AS (
SELECT 'a' user, 10 ranking UNION ALL
SELECT 'b', 2 UNION ALL
SELECT 'a', 12 UNION ALL
SELECT 'a', 14 UNION ALL
SELECT 'c', 22 UNION ALL
SELECT 'd', 21
)
SELECT
user,
MAX(ranking) AS ranking_max,
STRING_AGG(CAST(ranking AS STRING)) ranking_list
FROM `project.dataset.table`
GROUP BY user
with result
Row user ranking_max ranking_list
1 a 14 10,12,14
2 b 2 2
3 c 22 22
4 d 21 21
Note: if you need ranking_list to be ordered - you can use ORDER BY in STRING_AGG as in below
STRING_AGG(CAST(ranking AS STRING) ORDER BY ranking) ranking_list
answered Nov 14 '18 at 20:15
Mikhail Berlyant
55.9k43368
add a comment |
Below example for BigQuery Standard SQL
#standardSQL
WITH `project.dataset.table` AS (
SELECT 'a' user, 10 ranking UNION ALL
SELECT 'b', 2 UNION ALL
SELECT 'a', 12 UNION ALL
SELECT 'a', 14 UNION ALL
SELECT 'c', 22 UNION ALL
SELECT 'd', 21
)
SELECT
user,
MAX(ranking) AS ranking_max,
STRING_AGG(CAST(ranking AS STRING)) ranking_list
FROM `project.dataset.table`
GROUP BY user
with result
Row user ranking_max ranking_list
1 a 14 10,12,14
2 b 2 2
3 c 22 22
4 d 21 21
Note: if you need ranking_list to be ordered - you can use ORDER BY in STRING_AGG as in below
STRING_AGG(CAST(ranking AS STRING) ORDER BY ranking) ranking_list
answered Nov 14 '18 at 20:15
Mikhail Berlyant
55.9k43368
Below example for BigQuery Standard SQL
#standardSQL
WITH `project.dataset.table` AS (
SELECT 'a' user, 10 ranking UNION ALL
SELECT 'b', 2 UNION ALL
SELECT 'a', 12 UNION ALL
SELECT 'a', 14 UNION ALL
SELECT 'c', 22 UNION ALL
SELECT 'd', 21
)
SELECT
user,
MAX(ranking) AS ranking_max,
STRING_AGG(CAST(ranking AS STRING)) ranking_list
FROM `project.dataset.table`
GROUP BY user
with result
Row user ranking_max ranking_list
1 a 14 10,12,14
2 b 2 2
3 c 22 22
4 d 21 21
Note: if you need ranking_list to be ordered - you can use ORDER BY in STRING_AGG as in below
STRING_AGG(CAST(ranking AS STRING) ORDER BY ranking) ranking_list
answered Nov 14 '18 at 20:15
Mikhail Berlyant
55.9k43368
answered Nov 14 '18 at 20:15
Mikhail Berlyant
55.9k43368
answered Nov 14 '18 at 20:15
Mikhail Berlyant
55.9k43368
answered Nov 14 '18 at 20:15
Mikhail Berlyant
55.9k43368
55.9k43368
add a comment |
add a comment |
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