Suppose I have a list of IDs I want to filter a Collection by:
const ids = [1, 2, 3, 4];
How do I filter the collection to match just those IDs? Something like this doesn't work:
return Coll.fetch({_id: {$all: ids}});
This will work:
return Collection.find({_id: {$in: ids}}).fetch();
Here's how I might approach this:
function hasAllIds(collections, ids) {
for (let i = 0; i < collections.length; i++) {
let count = collections[i].find({
_id: {
$in: ids
}
}).count();
if (count === ids.length) {
return collections[i];
}
}
return null;
}
const colls = [Meteor.users, Games]; //etc.
const ids = [1, 2, 3];
const coll = hasAllIds(colls, ids);
if (coll) {
coll.find(); //or whatever
}
Related
I'm trying to create a copy/clone of a "List<List<Map'>>".
So far I tried:
dataFTY2 = dataFTY.map((element)=>element).toList();
dataFTY2 = json.decode(json.encode(dataFTY));
dataFTY2 = List.from(dataFTY);
Nothing seems to work. Whenever I change the copy "dataFTY2", dataFTY changes as well. I need this to be a completely independent copy. Please help. I cant seem to figure this out, its driving me crazy.
More code added for reference.
List failureDetails = [];
List trackIDs = [];
List dateTime = [];
var dataFTY2 = dataFTY.map((element) => element.map((ele) => Map.from(ele)).toList()).toList();
// get historyData for each one and sort through "F"s and put them into the table in a row?
for (var x in dataFTY2[4]) {
trackIDs.add(x["track_id"]);
dateTime.add(x["datetime"]);
}
List failuresOnly = List.filled(trackIDs.length, {}, growable: true);
for (var i = 0; i < trackIDs.length; i++) {
await fetchTrackIDTestDetails(context, trackIDs[i], dateTime[i], false);
failureDetails.add(MyGlobals().getTestCodeDetailsData());
}
//filter out only "F"s
for (var p = 0; p < failureDetails.length; p++) {
for (var t in failureDetails[p][0]) {
if (t["Status"] == "F") {
//add it to list, if pass do nothing
failuresOnly[p] = t;
}
}
}
//combine with FTY failure data, don't use new screen use old screen and toggle when pressed, add column on right side
//dataFTY2 = MyGlobals().getFTYFailureMoreDetails();
for (var i = 0; i < dataFTY2[4].length; i++) {
dataFTY2[4][i]["TestCode"] = failuresOnly[i]["TestCode"];
dataFTY2[4][i]["Status"] = failuresOnly[i]["Status"];
dataFTY2[4][i]["TestValue"] = failuresOnly[i]["TestValue"];
dataFTY2[4][i]["Lo_Limit"] = failuresOnly[i]["Lo_Limit"];
dataFTY2[4][i]["Up_Limit"] = failuresOnly[i]["Up_Limit"];
dataFTY2[4][i]["ProcTime"] = failuresOnly[i]["ProcTime"];
}
You can use Map.from named constructor to clone the Map like this,
dataFTY2 = dataFTY.map((element) => element.map((ele) => Map.from(ele)).toList()).toList();
I find it more straightforward to use collection-for and the spread (...) operator:
void main() {
var original = [
[
{'foo': 1, 'bar': 2},
{'foo': 3, 'bar': 4},
]
];
// Create a new List...
var copy = [
for (var sublist in original)
// ... where each element is a new List...
[
for (var map in sublist)
// ... where each element of the sublist is a new Map that
// copies all entries from `map`.
{...map},
],
];
original[0][0]['foo'] = -1;
print(original); // Prints: [[{foo: -1, bar: 2}, {foo: 3, bar: 4}]]
print(copy); // Prints: [[{foo: 1, bar: 2}, {foo: 3, bar: 4}]]
}
How I can get the most popular number from a list in dart without using any third party libraries?
var list = [0, 1, 1, 2, 2, 2, 3, 3, 4]; // most popular number is 2
If there are two or more popular numbers then the output should be a List with both values. Example:
One popular number:
var list = [0, 1, 1, 2, 2, 2, 3, 3, 4];
// Output should be [2]
Two or more popular numbers:
var list = [0, 1, 1, 2, 2, 2, 3, 3, 3];
// Output should be [2, 3]
Thank you in advance for your help!
This works...you can optimize it
var list = [1, 1, 2, 2, 3, 4, 5];
list.sort();
var popularNumbers = [];
List<Map<dynamic, dynamic>> data = [];
var maxOccurrence = 0;
var i = 0;
while (i < list.length) {
var number = list[i];
var occurrence = 1;
for (int j = 0; j < list.length; j++) {
if (j == i) {
continue;
}
else if (number == list[j]) {
occurrence++;
}
}
list.removeWhere((it) => it == number);
data.add({number: occurrence});
if (maxOccurrence < occurrence) {
maxOccurrence = occurrence;
}
}
data.forEach((map) {
if (map[map.keys.toList()[0]] == maxOccurrence) {
popularNumbers.add(map.keys.toList()[0]);
}
});
print(popularNumbers);
try this to count each element in list:
var list = [0, 1, 1, 2, 2, 2, 3, 3, 4];
var popular = Map();
list.forEach((l) {
if(!popular.containsKey(l)) {
popular[l] = 1;
} else {
popular[l] +=1;
}
});
I guess I found the solution.
Let me explain it to you:
I had queried through your list and checked whether the keys of the map contains the element or not. If the map does not contain the element as the key then, it will create a key from the element and pass 1 as the value. If the map does contain the element as a key then it will simply increment the value.
Once the map is ready, I had sorted the map values and stored them in a List. From the sorted map values I had taken the last element from the list of sorted values because we had sorted it in ascending order so the most popular value will be at last.
At last, I had queried through the map and check whether the value of the particular key is equal to the popularValue or not. If it is then we are adding the current key and value to the mostPopularValues list.
If I got something wrong please let me know.
void main() {
List list = [0, 1, 1, 1, 2, 2, 2, 3, 3, 4];
List mostPopularValues = [];
var map = Map();
list.forEach((element) {
if (!map.containsKey(element)) {
map[element] = 1;
} else {
map[element] += 1;
}
});
print(map);
// o/p : {0: 1, 1: 3, 2: 3, 3: 2, 4: 1}
List sortedValues = map.values.toList()..sort();
print(sortedValues);
// o/p : [1, 1, 2, 3, 3]
int popularValue = sortedValues.last;
print(popularValue);
// o/p : 3
map.forEach((k, v) {
if (v == popularValue) {
mostPopularValues.add("$k occurs $v time in the list");
}
});
print(mostPopularValues);
// o/p : [1 occurs 3 time in the list, 2 occurs 3 time in the list]
}
Not sure if that's the best solution, but it works pretty well. Let me know if there are any doubts.
final list = [0, 1, 1, 2, 2, 2, 3, 3, 4];
// Count occurrences of each item
final folded = list.fold({}, (acc, curr) {
acc[curr] = (acc[curr] ?? 0) + 1;
return acc;
}) as Map<dynamic, dynamic>;
// Sort the keys (your values) by its occurrences
final sortedKeys = folded.keys
.toList()
..sort((a, b) => folded[b].compareTo(folded[a]));
print('Most popular value: ${sortedKeys.first}'); // 1
print('Second most popular value: ${sortedKeys[1]}'); // 2
I have solved this problem by defining an extension on Iterable:
extension MostPopularItemsExtension<E> on Iterable<E> {
/// Returns the most popular items, where all items in the returned
/// list have the same number of occurances. If [this] is empty, returns an
/// empty list
///
/// Examples:
/// `[1,2,3,2].mostPopularItems() == [2]`
/// `[1,1,2,2].mostPopularItems() == [1,2]`
Iterable<E> mostPopularItems() {
if (isEmpty) return [];
final itemsCounted = <E, int>{};
for (final e in this) {
if (itemsCounted.containsKey(e)) {
itemsCounted[e] = itemsCounted[e]! + 1;
} else {
itemsCounted[e] = 1;
}
}
final highestCount = (itemsCounted.values.toList()..sort()).last;
return itemsCounted.entries
.where((e) => e.value == highestCount)
.map((e) => e.key);
}
}
The basic idea is to count all occurrences of each item in a Map object, get the highest count from this map and then return all items that have that specific number of occurrences.
I have some objects:
var array = [];
var obj1 = new Object();
obj1.category = 0;
obj1.sequence = 0;
obj1.otherAttr = 1;
array.push(obj1)
var obj2 = new Object();
obj2.category = 1;
obj2.sequence = 2;
obj1.otherAttr = 2;
array.push(obj2)
var obj3 = new Object();
obj3.category = 4;
obj4.sequence = 3;
obj1.otherAttr = 3;
array.push(obj3)
...
and I have collection in MongoDB:
{
category:0,
sequence:0,
anotherAttr:1
},
{
category:0,
sequence:1,
anotherAttr:2
},
{
category:0,
sequence:2,
anotherAttr:3
},
{
category:1,
sequence:0,
anotherAttr:4
},
{
category:1,
sequence:1,
anotherAttr:5
},
{
category:1,
sequence:2,
anotherAttr:6
},
...
I wanna insert these objects. But Before insert these, I want check overlapping category and sequence between these and document in MongoDB unrelated to otherAttr or anotherAttr. For example, I have objects like top, the result I need is :
{
category:0,
sequence:0,
anotherAttr:1
},
{
category:1,
sequence:1,
anotherAttr:5
}
I tried to $match. But can't check some conditions in many attributes.
let's say I have this array:
array = [1,2,3,4,5,6,7,8,9,10];
Then I want to divide this array for each 2 batch with 5 array value:
a[0] = [1,2,3,4,5];
a[1] = [6,7,8,9,10];
This is my code until now keep trying.It should insert to db when batch has been slice for each job for 100 array value.
BatchList = new Meteor.Collection('batch_list');
var arrayAssetList = Asset.find({}, { _id: 1});
var data = [];
arrayAssetList.forEach(function(key){
data.push(key._id)
});
var arraySize = arrayAssetList.count() / 100;
var slice, sliceTo;
for(var i=0; i<arraySize; i++){
if(i == 0){
slice = 0
} {
slice = 100*i
}
if(i>1){
sliceTo = 100*i;
}
BatchList.insert({ data: data.slice(slice,sliceTo)});
}
it should be like this:
[{
_id: '()#()##)#*()#*#',
size: 100,
data: [1,2,3,4,100]
},{
_id: 'sdfsdfnnn3',
size: 100,
data: [100,101,102,103,200]
}]
I have a collection users in Mongo and I execute this map reduce which I believe is the equivalent of a COUNT(*) GROUP BY origin:
> m = function() { for (i in this.membership) {
... emit( this.membership[i].platform_profile.origin, 1 );
... } }
function () {
for (i in this.membership) {
emit(this.membership[i].platform_profile.origin, 1);
}
}
> r = function( id, values ) { var result = 0;
... for ( var i = 0; i < values.length; i ++ ) { result += values[i]; }
... return result; }
function (id, values) {
var result = 0;
for (var i = 0; i < values.length; i++) {
result += values[i];
}
return result;
}
> db.users.mapReduce(m, r, {out : { inline: 1}});
{
"results" : [
{
"_id" : 0,
"value" : 15
},
{
"_id" : 1,
"value" : 449
},
...
}
But if I try to count how many documents have this field set to a specific value like 1, I get fewer results:
db.users.count({"membership.platform_profile.origin": 1});
424
What am I missing?
Are your count queries using a sparse index by any chance? My only guess there would be if some other query criteria resulted in documents absent from from index to be ignored from the count.
I recreated your schema with some fixture data and the results between map/reduce and simple count queries are in agreement:
db.users.drop();
var map = function() {
for (i in this.membership) {
emit(this.membership[i].platform_profile.origin, 1);
}
};
var reduce = function(id, values ) {
var result = 0;
for (var i = 0; i < values.length; i++) {
result += values[i];
}
return result;
}
var origins = {1: "a", 2: "b", 3: "c", 4: "d"};
for (var i = 0; i < 1000; ++i) {
var membership = [];
for (var o in origins) {
if (0 == i % o) {
membership.push({ platform_profile: { origin: origins[o] }});
}
}
db.users.save({ membership: membership });
}
db.users.mapReduce(map, reduce, {out: {inline: 1}}).results.forEach(function(result){
print(result["_id"] + ": " + result["value"]);
});
for (var o in origins) {
print(origins[o] + ": " + db.users.count({"membership.platform_profile.origin": origins[o]}));
}
Here's the output:
$ mongo --quiet mr_count.js
a: 1000
b: 500
c: 334
d: 250
a: 1000
b: 500
c: 334
d: 250
You can use the following map/reduce for the equivalent of COUNT(*) GROUP BY origin
Map/Reduce Functions :
map = function() {
if(!this.membership) return;
for (i in this.membership) {
if(!this.membership[i].platform_profile || !this.membership[i].platform_profile.origin) return;
emit(this.membership[i].platform_profile.origin, 1);
}
}
reduce = function(key, values) {
var count = 0;
for (v in values) {
count += values[v];
}
return count;
}
result = db.runCommand({
"mapreduce" : "users",
"map" : map,
"reduce" : reduce,
"out" : "users_count"
});
I had the same issue. I replaced x.length by Array.sum(x) in the reduce function (assuming you emit 1 in the map function) and it works. I agree x.length should work too, but I cannot explain why it does not.