问题
Similar questions asked here before:
Count items for a single key: jq count the number of items in json by a specific key
Calculate the sum of object values: How do I sum the values in an array of maps in jq?
Question
How to emulate the COUNT aggregate function which should behave similarly to its SQL original? Let's extend this question even more to include other regular SQL functions:
- COUNT
- SUM / MAX/ MIN / AVG
- ARRAY_AGG
The last one is not a standard SQL function - it's from PostgreSQL but is quite useful.
At input comes a stream of valid JSON objects. For demonstration let's pick a simple story of owners and their pets.
Model and data
Base relation: Owner
id name age
1 Adams 25
2 Baker 55
3 Clark 40
4 Davis 31
Base relation: Pet
id name litter owner_id
10 Bella 4 1
20 Lucy 2 1
30 Daisy 3 2
40 Molly 4 3
50 Lola 2 4
60 Sadie 4 4
70 Luna 3 4
Source
From above we get a derivative relation Owner_Pet (a result of SQL JOIN of the above relations) presented in JSON format for our jq queries (the source data):
{ "owner_id": 1, "owner": "Adams", "age": 25, "pet_id": 10, "pet": "Bella", "litter": 4 }
{ "owner_id": 1, "owner": "Adams", "age": 25, "pet_id": 20, "pet": "Lucy", "litter": 2 }
{ "owner_id": 2, "owner": "Baker", "age": 55, "pet_id": 30, "pet": "Daisy", "litter": 3 }
{ "owner_id": 3, "owner": "Clark", "age": 40, "pet_id": 40, "pet": "Molly", "litter": 4 }
{ "owner_id": 4, "owner": "Davis", "age": 31, "pet_id": 50, "pet": "Lola", "litter": 2 }
{ "owner_id": 4, "owner": "Davis", "age": 31, "pet_id": 60, "pet": "Sadie", "litter": 4 }
{ "owner_id": 4, "owner": "Davis", "age": 31, "pet_id": 70, "pet": "Luna", "litter": 3 }
Requests
Here are sample requests and their expected output:
- COUNT the number of pets per owner:
{ "owner_id": 1, "owner": "Adams", "age": 25, "pets_count": 2 }
{ "owner_id": 2, "owner": "Baker", "age": 55, "pets_count": 1 }
{ "owner_id": 3, "owner": "Clark", "age": 40, "pets_count": 1 }
{ "owner_id": 4, "owner": "Davis", "age": 31, "pets_count": 3 }
- SUM up the number of whelps per owner and get their MAX (MIN/AVG):
{ "owner_id": 1, "owner": "Adams", "age": 25, "litter_total": 6, "litter_max": 4 }
{ "owner_id": 2, "owner": "Baker", "age": 55, "litter_total": 3, "litter_max": 3 }
{ "owner_id": 3, "owner": "Clark", "age": 40, "litter_total": 4, "litter_max": 4 }
{ "owner_id": 4, "owner": "Davis", "age": 31, "litter_total": 9, "litter_max": 4 }
- ARRAY_AGG pets per owner:
{ "owner_id": 1, "owner": "Adams", "age": 25, "pets": [ "Bella", "Lucy" ] }
{ "owner_id": 2, "owner": "Baker", "age": 55, "pets": [ "Daisy" ] }
{ "owner_id": 3, "owner": "Clark", "age": 40, "pets": [ "Molly" ] }
{ "owner_id": 4, "owner": "Davis", "age": 31, "pets": [ "Lola", "Sadie", "Luna" ] }
回答1:
Extended jq solution:
Custom count() function:
jq -sc 'def count($k): group_by(.[$k])[] | length as $l | .[0]
| .pets_count = $l
| del(.pet_id, .pet, .litter);
count("owner_id")' source.data
The output:
{"owner_id":1,"owner":"Adams","age":25,"pets_count":2}
{"owner_id":2,"owner":"Baker","age":55,"pets_count":1}
{"owner_id":3,"owner":"Clark","age":40,"pets_count":1}
{"owner_id":4,"owner":"Davis","age":31,"pets_count":3}
Custom sum() function:
jq -sc 'def sum($k): group_by(.[$k])[] | map(.litter) as $litters | .[0]
| . + {litter_total: $litters | add, litter_max: $litters | max}
| del(.pet_id, .pet, .litter);
sum("owner_id")' source.data
The output:
{"owner_id":1,"owner":"Adams","age":25,"litter_total":6,"litter_max":4}
{"owner_id":2,"owner":"Baker","age":55,"litter_total":3,"litter_max":3}
{"owner_id":3,"owner":"Clark","age":40,"litter_total":4,"litter_max":4}
{"owner_id":4,"owner":"Davis","age":31,"litter_total":9,"litter_max":4}
Custom array_agg() function:
jq -sc 'def array_agg($k): group_by(.[$k])[] | map(.pet) as $pets | .[0]
| .pets = $pets | del(.pet_id, .pet, .litter);
array_agg("owner_id")' source.data
The output:
{"owner_id":1,"owner":"Adams","age":25,"pets":["Bella","Lucy"]}
{"owner_id":2,"owner":"Baker","age":55,"pets":["Daisy"]}
{"owner_id":3,"owner":"Clark","age":40,"pets":["Molly"]}
{"owner_id":4,"owner":"Davis","age":31,"pets":["Lola","Sadie","Luna"]}
回答2:
This is a nice exercise, but SO is not a programming service, so I will focus here on some key concepts for generic solutions in jq that are efficient, even for very large collections.
GROUPS_BY
The key to efficiency here is avoiding the built-in group_by, as it requires sorting. Since jq is fundamentally stream-oriented, the following definition of GROUPS_BY is likewise stream-oriented. It takes advantage of the efficiency of key-based lookups, while avoiding calling tojson on strings:
# emit a stream of the groups defined by f
def GROUPS_BY(stream; f):
reduce stream as $x ({};
($x|f) as $s
| ($s|type) as $t
| (if $t == "string" then $s else ($s|tojson) end) as $y
| .[$t][$y] += [$x] )
| .[][] ;
distinct and count_distinct
# Emit an array of the distinct entities in `stream`, without sorting
def distinct(stream):
reduce stream as $x ({};
($x|type) as $t
| (if $t == "string" then $x else ($x|tojson) end) as $y
| if (.[$t] | has($y)) then . else .[$t][$y] += [$x] end )
| [.[][]] | add ;
# Emit the number of distinct items in the given stream
def count_distinct(stream):
def sum(s): reduce s as $x (0;.+$x);
reduce stream as $x ({};
($x|type) as $t
| (if $t == "string" then $x else ($x|tojson) end) as $y
| .[$t][$y] = 1 )
| sum( .[][] ) ;
Convenience function
def owner: {owner_id,owner,age};
Example: "COUNT the number of pets per owner"
GROUPS_BY(inputs; .owner_id)
| (.[0] | owner) + {pets_count: count_distinct(.[]|.pet_id)}
Invocation: jq -nc -f program1.jq input.json
Output:
{"owner_id":1,"owner":"Adams","age":25,"pets_count":2}
{"owner_id":2,"owner":"Baker","age":55,"pets_count":1}
{"owner_id":3,"owner":"Clark","age":40,"pets_count":1}
{"owner_id":4,"owner":"Davis","age":31,"pets_count":3}
Example: "SUM up the number of whelps per owner and get their MAX"
GROUPS_BY(inputs; .owner_id)
| (.[0] | owner)
+ {litter_total: (map(.litter) | add)}
+ {litter_max: (map(.litter) | max)}
Invocation: jq -nc -f program2.jq input.json
Output: as given.
Example: "ARRAY_AGG pets per owner"
GROUPS_BY(inputs; .owner_id)
| (.[0] | owner) + {pets: distinct(.[]|.pet)}
Invocation: jq -nc -f program3.jq input.json
Output:
{"owner_id":1,"owner":"Adams","age":25,"pets":["Bella","Lucy"]}
{"owner_id":2,"owner":"Baker","age":55,"pets":["Daisy"]}
{"owner_id":3,"owner":"Clark","age":40,"pets":["Molly"]}
{"owner_id":4,"owner":"Davis","age":31,"pets":["Lola","Sadie","Luna"]}
来源：https://stackoverflow.com/questions/48321235/sql-style-group-by-aggregate-functions-in-jq-count-sum-and-etc