How to custom sort a field in MongoDB - mongodb

Question

I have a collection called INFODOCS which has a field called ML_PRIORITY(HIGH/MEDIUM/LOW) and STATUS(True/False/Null). I want to determine count of STATUS for each ML_PRIORITY and then sort the ML_PRIORITY in order High, Medium and Low.
[
{
"_id": "1",
"ML_PRIORITY" : "HIGH",
"STATUS" : "True"
},
{
"_id": "2",
"ML_PRIORITY" : "HIGH",
"STATUS" : ""
},
{
"_id": "3",
"ML_PRIORITY" : "HIGH",
"STATUS" : "False"
},
{
"_id": "4",
"ML_PRIORITY" : "MEDIUM",
"STATUS" : ""
},
{
"_id": "5",
"ML_PRIORITY" : "Low",
"STATUS" : ""
}
]
I was able to determine the count of STATUS for each ML_PRIORITY using below aggregation pipeline
but I am not sure how can I custom sort the ML_PRIORITY as $sort has only two option (1 and -1)
db.collection.aggregate([
{
'$group': {
'_id': '$ML_PRIORITY',
'QUALITYCHECKDONE': {
'$sum': {
'$cond': [
{
'$eq': [
'$STATUS', 'TRUE'
]
}, 1, 0
]
}
},
'QUALITYCHECKNOTDONE': {
'$sum': {
'$cond': [
{
'$eq': [
'$STATUS', ''
]
}, 1, 0
]
}
},
'QUALITYCHECKNOTREQ': {
'$sum': {
'$cond': [
{
'$eq': [
'$STATUS', 'FALSE'
]
}, 1, 0
]
}
}
}
}, {
'$project': {
'_id': 0,
'ML_PRIORITY': '$_id',
'QUALITYCHECKDONE': 1,
'QUALITYCHECKNOTDONE': 1,
'QUALITYCHECKNOTREQ': 1
}
}
])
Example - https://mongoplayground.net/p/anAwoqZk2Ys

Answer 1

One option is to replace your last step with 3 steps, in order to $set an order field, $sort, and $unset it:
[
{$set: {
order: {$indexOfArray: [["HIGH", "MEDIUM", "Low"], "$_id"]},
"ML_PRIORITY": "$_id"
}},
{$sort: {order: 1}},
{$unset: ["_id", "order"]}
]
See how it works on the playground example