I have a datatable that's being returned by google.visualization.data.group()
var aggData = google.visualization.data.group(
view,
[0],
aggColumns
);
I want to set several columns to be of type string with a tooltip role, and converting values in them to an html string
for (var col=2; col < aggData.getNumberOfColumns(); col = col + 2){
aggData.setColumnProperties(col,{'type':'string', 'role':'tooltip', 'p':{'html':true}});
//looking to see if the column type was actually changed
console.log('Column '+col+' type: ' + aggData.getColumnProperty(col, 'type'))
for (var row = 0; row < aggData.getNumberOfRows(); row = row + 1){
aggData.setValue(row, col, getHTML(aggData.getValue(row, col)))
}
}
function getHTML(count) {;
return 'Projects Completed: <b>' + count + '</b>';
}
I checked the column data type in the log and it does return a string but when i set the value to a string it throws a type mismatch error.
Column 2 type: string
Uncaught Error: Type mismatch. Value Projects Completed: <b>2</b> does not match type number in column index 2
I also tried setting the column type using setColumnProperty() method but it's the same result. What am I missing?
================================================================================================
Below is a snippet of the larger script if needed
Sample input data looks like
"Oct 1, 2019, 12:00:00 AM",Team C,68
"Sep 23, 2019, 12:00:00 AM",Team C,68
"Nov 29, 2019, 12:00:00 AM",Team C,87
"Dec 31, 2019, 12:00:00 AM",Team C,62
....................................
"Nov 21, 2018, 12:00:00 AM",Team A,79
"Dec 29, 2018, 12:00:00 AM",Team A,58
"Nov 15, 2018, 12:00:00 AM",Team B,96
"Dec 29, 2018, 12:00:00 AM",Team B,77
The data is being read into a data table
var data = new google.visualization.DataTable();
data.addColumn('datetime', 'Year');
data.addColumn('string', 'Team');
data.addColumn('number', 'Total Score');
var groupData = google.visualization.data.group(
data,
[
{
column: 0,
modifier: getYear,
type: 'number'
},
1
],
[
{
column: 2,
aggregation: google.visualization.data.sum,
type: 'number'
},
{
column: 2,
aggregation: google.visualization.data.count,
type: 'number',
role: 'tooltip'
}
]
);
// create data view from groupData
var view = new google.visualization.DataView(groupData);
// sum column array
var aggColumns = [];
// use year (column 0) as first view column
var viewColumns = [0];
// build calculated view & agg columns for each team
groupData.getDistinctValues(1).forEach(function (team, index) {
// add a column to the view for each team
viewColumns.push({
calc: function (dt, row) {
if (dt.getValue(row, 1) === team) {
return dt.getValue(row, 2);
}
return null;
},
label: team,
type: 'number'
});
viewColumns.push({
calc: function (dt, row) {
if (dt.getValue(row, 1) === team) {
return dt.getValue(row, 3);
}
return null;
},
label: 'Number of Projects',
type: 'number'
});
// add sum column for each team
aggColumns.push({
aggregation: google.visualization.data.sum,
column: index*2 + 1,
label: team,
type: 'number'
});
aggColumns.push({
aggregation: google.visualization.data.sum,
column: index*2 + 2,
type: 'number',
role: 'tooltip',
});
});
// set view columns
view.setColumns(viewColumns);
var aggData = google.visualization.data.group(
view,
[0],
aggColumns
);
/*
The aggData looks like
"2,018",137,2,173,2,0,0
"2,019",864,12,"1,028",12,610,12
*/
for (var col=2; col < aggData.getNumberOfColumns(); col = col + 2){
aggData.setColumnProperties(col,{'type':'string', 'role':'tooltip', 'p':{'html':true}});
console.log('Column '+col+' type: ' + aggData.getColumnProperty(col, 'type'))
for (var row = 0; row < aggData.getNumberOfRows(); row = row + 1){
aggData.setValue(row, col, getHTML(aggData.getValue(row, col)))
}
}
data table method setColumnProperties isn't doing what you expect.
it only sets the properties portion of the column --> 'p':{'html':true}
so after your code runs, you end up with the following in your column properties.
'p': {'type':'string', 'role':'tooltip', 'p':{'html':true}}
and in fact, it is not possible to change a column's type,
once it has been created.
instead, you'll need to either use the addColumn or insertColumn method.
another option would be to use a data view.
then you could use a calculated column for the tooltip,
and exclude the original column you are trying to change,
using the setColumns method on the data view.
Im trying to filter column 'time' in visualization data table using getFilteredRows(filters) method.I provided column value with minimum and maximum values as,
var timesheet_dataTable = new google.visualization.DataTable(data, 0.6);
var time_filter = timesheet_dataTable.getFilteredRows([{column: 3, minValue: '2:28 PM', maxValue: '3:01 PM'}]);
and then created data view with setRows method to display the data but the table displayed without filtering the data.I checked with other column values and received proper output.So whether 'timeofday' data type is supported in this type of filters?
Is there any other method to filter column based on time?
Update:
This is the code for formatting and passing value to the visualization table.Value of variable startTime will be like '14:28:12'.
val datetimeStart: String = "Date(0,0,0,"
val datetimeEnd: String = ")"
val simpleDateTimeFormat = new SimpleDateFormat("HH,mm,ss")
Json.obj("v" -> JsString(datetimeStart + (simpleDateTimeFormat.format(tsl.startTime)).toString() + datetimeEnd))
before displaying in visualization table i have used formatter as:
var formatter_short1 = new google.visualization.DateFormat({pattern:'h:mm aa'});
formatter_short1.format(timesheet_dataTable,3);
The "timeofday" data type is supported by the filter method, you just need to use it correctly:
// filter column 3 from 2:28PM to 3:01PM
var time_filter = timesheet_dataTable.getFilteredRows([{
column: 3,
minValue: [14, 28, 0, 0],
maxValue: [15, 1, 0, 0]
}]);
var view = new google.visualization.DataView(timesheet_dataTable);
view.setRows(time_filter);
Make sure you are using the view you create to draw your chart, instead of the DataTable:
chart.draw(view, options);
[edit - example for filtering "datetime" type column]
// filter column 3 from 2:28PM to 3:01PM
var time_filter = timesheet_dataTable.getFilteredRows([{
column: 3,
minValue: new Date(0, 0, 0, 14, 28, 0, 0),
maxValue: new Date(0, 0, 0, 15, 1, 0, 0)
}]);
var view = new google.visualization.DataView(timesheet_dataTable);
view.setRows(time_filter);
How do I delete duplicates from a list without fooling around with a set? Is there something like list.distinct()? or list.unique()?
void main() {
print("Hello, World!");
List<String> list = ['abc',"abc",'def'];
list.forEach((f) => print("this is list $f"));
Set<String> set = new Set<String>.from(list);
print("this is #0 ${list[0]}");
set.forEach((f) => print("set: $f"));
List<String> l2= new List<String>.from(set);
l2.forEach((f) => print("This is new $f"));
}
Hello, World!
this is list abc
this is list abc
this is list def
this is #0 abc
set: abc
set: def
This is new abc
This is new def
Set seems to be way faster!! But it loses the order of the items :/
Use toSet and then toList
var ids = [1, 4, 4, 4, 5, 6, 6];
var distinctIds = ids.toSet().toList();
Result: [1, 4, 5, 6]
Or with spread operators:
var distinctIds = [...{...ids}];
I didn't find any of the provided answers very helpful.
Here is what I generally do:
final ids = Set();
myList.retainWhere((x) => ids.add(x.id));
Of course you can use any attribute which uniquely identifies your objects. It doesn't have to be an id field.
Benefits over other approaches:
Preserves the original order of the list
Works for rich objects not just primitives/hashable types
Doesn't have to copy the entire list to a set and back to a list
Update 09/12/21
You can also declare an extension method once for lists:
extension Unique<E, Id> on List<E> {
List<E> unique([Id Function(E element)? id, bool inplace = true]) {
final ids = Set();
var list = inplace ? this : List<E>.from(this);
list.retainWhere((x) => ids.add(id != null ? id(x) : x as Id));
return list;
}
}
This extension method does the same as my original answer. Usage:
// Use a lambda to map an object to its unique identifier.
myRichObjectList.unique((x) => x.id);
// Don't use a lambda for primitive/hashable types.
hashableValueList.unique();
Set works okay, but it doesn't preserve the order. Here's another way using LinkedHashSet:
import "dart:collection";
void main() {
List<String> arr = ["a", "a", "b", "c", "b", "d"];
List<String> result = LinkedHashSet<String>.from(arr).toList();
print(result); // => ["a", "b", "c", "d"]
}
https://api.dart.dev/stable/2.4.0/dart-collection/LinkedHashSet/LinkedHashSet.from.html
Try the following:
List<String> duplicates = ["a", "c", "a"];
duplicates = duplicates.toSet().toList();
Check this code on Dartpad.
If you want to keep ordering or are dealing with more complex objects than primitive types, store seen ids to the Set and filter away those ones that are already in the set.
final list = ['a', 'a', 'b'];
final seen = Set<String>();
final unique = list.where((str) => seen.add(str)).toList();
print(unique); // => ['a', 'b']
//This easy way works fine
List<String> myArray = [];
myArray = ['x', 'w', 'x', 'y', 'o', 'x', 'y', 'y', 'r', 'a'];
myArray = myArray.toSet().toList();
print(myArray);
// result => myArray =['x','w','y','o','r', 'a']
I am adding this to atreeon's answer. For anyone that want use this with Object:
class MyObject{
int id;
MyObject(this.id);
#override
bool operator ==(Object other) {
return other != null && other is MyObject && hashCode == other.hashCode;
}
#override
int get hashCode => id;
}
main(){
List<MyObject> list = [MyObject(1),MyObject(2),MyObject(1)];
// The new list will be [MyObject(1),MyObject(2)]
List<MyObject> newList = list.toSet().toList();
}
Remove duplicates from a list of objects:
class Stock {
String? documentID; //key
Make? make;
Model? model;
String? year;
Stock({
this.documentID,
this.make,
this.model,
this.year,
});
}
List of stock, from where we want to remove duplicate stocks
List<Stock> stockList = [stock1, stock2, stock3];
Remove duplicates
final ids = stockList.map((e) => e.documentID).toSet();
stockList.retainWhere((x) => ids.remove(x.documentID));
Using Dart 2.3+, you can use the spread operators to do this:
final ids = [1, 4, 4, 4, 5, 6, 6];
final distinctIds = [...{...ids}];
Whether this is more or less readable than ids.toSet().toList() I'll let the reader decide :)
For distinct list of objects you can use Equatable package.
Example:
// ignore: must_be_immutable
class User extends Equatable {
int id;
String name;
User({this.id, this.name});
#override
List<Object> get props => [id];
}
List<User> items = [
User(
id: 1,
name: "Omid",
),
User(
id: 2,
name: "Raha",
),
User(
id: 1,
name: "Omid",
),
User(
id: 2,
name: "Raha",
),
];
print(items.toSet().toList());
Output:
[User(1), User(2)]
Here it is, a working solution:
var sampleList = ['1', '2', '3', '3', '4', '4'];
//print('original: $sampleList');
sampleList = Set.of(sampleList).toList();
//print('processed: $sampleList');
Output:
original: [1, 2, 3, 3, 4, 4]
processed: [1, 2, 3, 4]
Using the fast_immutable_collections package:
[1, 2, 3, 2].distinct();
Or
[1, 2, 3, 2].removeDuplicates().toList();
Note: While distinct() returns a new list, removeDuplicates() does it lazily by returning an Iterable. This means it is much more efficient when you are doing some extra processing. For example, suppose you have a list with a million items, and you want to remove duplicates and get the first five:
// This will process five items:
List<String> newList = list.removeDuplicates().take(5).toList();
// This will process a million items:
List<String> newList = list.distinct().sublist(0, 5);
// This will also process a million items:
List<String> newList = [...{...list}].sublist(0, 5);
Both methods also accept a by parameter. For example:
// Returns ["a", "yk", "xyz"]
["a", "yk", "xyz", "b", "xm"].removeDuplicates(by: (item) => item.length);
If you don't want to include a package into your project but needs the lazy code, here it is a simplified removeDuplicates():
Iterable<T> removeDuplicates<T>(Iterable<T> iterable) sync* {
Set<T> items = {};
for (T item in iterable) {
if (!items.contains(item)) yield item;
items.add(item);
}
}
Note: I am one of the authors of the fast_immutable_collections package.
void uniqifyList(List<Dynamic> list) {
for (int i = 0; i < list.length; i++) {
Dynamic o = list[i];
int index;
// Remove duplicates
do {
index = list.indexOf(o, i+1);
if (index != -1) {
list.removeRange(index, 1);
}
} while (index != -1);
}
}
void main() {
List<String> list = ['abc', "abc", 'def'];
print('$list');
uniqifyList(list);
print('$list');
}
Gives output:
[abc, abc, def]
[abc, def]
As for me, one of the best practices is sort the array, and then deduplicate it. The idea is stolen from low-level languages. So, first make the sort by your own, and then deduplicate equal values that are going after each other.
// Easy example
void dedup<T>(List<T> list, {removeLast: true}) {
int shift = removeLast ? 1 : 0;
T compareItem;
for (int i = list.length - 1; i >= 0; i--) {
if (compareItem == (compareItem = list[i])) {
list.removeAt(i + shift);
}
}
}
// Harder example
void dedupBy<T, I>(List<T> list, I Function(T) compare, {removeLast: true}) {
int shift = removeLast ? 1 : 0;
I compareItem;
for (int i = list.length - 1; i >= 0; i--) {
if (compareItem == (compareItem = compare(list[i]))) {
list.removeAt(i + shift);
}
}
}
void main() {
List<List<int>> list = [[1], [1], [2, 1], [2, 2]];
print('$list');
dedupBy(list, (innerList) => innerList[0]);
print('$list');
print('\n removeLast: false');
List<List<int>> list2 = [[1], [1], [2, 1], [2, 2]];
print('$list2');
dedupBy(list2, (innerList) => innerList[0], removeLast: false);
print('$list2');
}
Output:
[[1], [1], [2, 1], [2, 2]]
[[1], [2, 1]]
removeLast: false
[[1], [1], [2, 1], [2, 2]]
[[1], [2, 2]]
This is another way...
final reducedList = [];
list.reduce((value, element) {
if (value != element)
reducedList.add(value);
return element;
});
reducedList.add(list.last);
print(reducedList);
It works for me.
var list = [
{"id": 1, "name": "Joshua"},
{"id": 2, "name": "Joshua"},
{"id": 3, "name": "Shinta"},
{"id": 4, "name": "Shinta"},
{"id": 5, "name": "Zaidan"}
];
list.removeWhere((element) => element.name == element.name.codeUnitAt(1));
list.sort((a, b) => a.name.compareTo(b.name));
Output:
[{"id": 1, "name": "Joshua"},
{"id": 3, "name": "Shinta"},
{"id": 5, "name": "Zaidan"}]
List<Model> bigList = [];
List<ModelNew> newList = [];
for (var element in bigList) {
var list = newList.where((i) => i.type == element.type).toList();
if(list.isEmpty){
newList.add(element);
}
}
Create method to remove duplicates from Array and return Array of unique elements.
class Utilities {
static List<String> uniqueArray(List<String> arr) {
List<String> newArr = [];
for (var obj in arr) {
if (newArr.contains(obj)) {
continue;
}
newArr.add(obj);
}
return newArr;
}
}
You can use the following way:
void main(List <String> args){
List<int> nums = [1, 2, 2, 2, 3, 4, 5, 5];
List<int> nums2 = nums.toSet().toList();
}
NOTE: This will not work if the items in the list are objects of class and have the same attributes. So, to solve this, you can use the following way:
void main() {
List<Medicine> objets = [Medicine("Paracetamol"),Medicine("Paracetamol"), Medicine("Benylin")];
List <String> atributs = [];
objets.forEach((element){
atributs.add(element.name);
});
List<String> noDuplicates = atributs.toSet().toList();
print(noDuplicates);
}
class Medicine{
final String name;
Medicine(this.name);
}
This is my solution
List<T> removeDuplicates<T>(List<T> list, IsEqual isEqual) {
List<T> output = [];
for(var i = 0; i < list.length; i++) {
bool found = false;
for(var j = 0; j < output.length; j++) {
if (isEqual(list[i], output[j])) {
found = true;
}
}
if (found) {
output.add(list[i]);
}
}
return output;
}
Use it like this:
var theList = removeDuplicates(myOriginalList, (item1, item2) => item1.documentID == item2.documentID);
or...
var theList = removeDuplicates(myOriginalList, (item1, item2) => item1.equals(item2));
or...
I have a library called Reactive-Dart that contains many composable operators for terminating and non-terminating sequences. For your scenario it would look something like this:
final newList = [];
Observable
.fromList(['abc', 'abc', 'def'])
.distinct()
.observe((next) => newList.add(next), () => print(newList));
Yielding:
[abc, def]
I should add that there are other libraries out there with similar features. Check around on GitHub and I'm sure you'll find something suitable.