I want to know how this work it should remove the item in list number 3 which is 'four' and print 'one','two','Three' why it is print three, four
here is the code:
final numbers = <String>['one', 'two', 'three', 'four'];
numbers.removeWhere((item) => item.length == 3);
print(numbers); // [three, four]`
removeWhere :Removes all objects from this list that satisfy test.
An object o satisfies test if test(o) is true.
removeWhere will go through the list and find the matched item, means where we do our logic and return true.
You can see item.length==3 is true for 'one' and 'two' because its string length is 3, that's why these elements are removed. You can find more on List
You can expand the method like
numbers.removeWhere((item) {
bool isItemContains3Char = item.length == 3;
print("${isItemContains3Char} : $item length : ${item.length}");
return isItemContains3Char;
});
All you need to return true based on your logic from removeWhere to remove elements.
final numbers = <String>['one', 'two', 'three', 'four'];
numbers.removeAt(2);
print(numbers);
output
[one, two, four]
Related
Assuming I have a List as below:
final numbers = [1, 0, -1, 1, -3, 0];
I'd like to generate another list where 1 results in true and 0 results in false and skip the others. So, in other words, I'd like a method like map where I can also skip some elements. In terms of test, it would assert to:
expect(newList, equals([true, false, true, false]));
In this result, I'd like to skip -1 and -3 in the list.
How can I achieve this?
Environment
Dart 2.18.5
Use the .where() method to get a new iterable with only the elements satisfying a predicate, then use .map() to apply the transformation on the result.
final numbers = [1, 0, -1, 1, -3, 0];
final result = numbers.where((n) => n == 1 || n == 0).map((n) => n == 1);
print(result);
\\ (true, false, true, false)
The most direct version would be:
var result = numbers.expand<bool>((n) =>
n == 0
? const <bool>[false]
: n == 1
? const <bool>[true]
: const <bool>[]);
The expand method can do everything map and where can, plus more,
Not particularly efficient, but not all bad either.
Another approach is to use a sync* function:
var result = () sync* {
for (var number in numbers) {
if (number == 0) {
yield false;
} else if (number == 1) {
yield true;
}
}();
If you don't care about creating a list eagerly, that's a also the approach for a list literal:
var result = [for (var number in numbers)
if (number == 0) false else if (number == 1) true
];
Or:
const _map = {0: [false], 1: [true]};
var result = [for (var number in numbers} ...?_map[number]];
The options are endless. In practice, doing where and map is probably more readable.
Ariel's answer gave me a better idea. In the example of the question, what we'd like to filter out are two values, which are 1 and 0. The real problem I'm dealing with right now is more complicated than this one, it has a wide range of lookup list. I haven't specifically asked "What is a more programmatic way to do this?" or "What if the lookup list has more values?".
So, if the values I'm looking for are more than these two values, a List of lookup values and contains method would be better for my case.
final numbers = [1, 0, -1, 1, -3, 0];
final lookup = {1, 0};
final result = numbers.where((n) => lookup.contains(n)).map((n) => n == 1 ? true : false).toList();
print(result);
This has some drawbacks though, some of which that come to my mind are:
Performance is definitely worse than Ariel's answer. My solution uses contains, which will perform look-ups in lookup rather than a simple byte-comparison with n == 1. So, it isn't great with large lists.
hashCode must be overriden for non-primitives. It is easy to compare int with an int in this simple example, but if you have an instance, you need to override hashCode, otherwise Dart will compare the memory addresses rather than values.
Using another collection type for lookup other than Set might also have performance impact because Set is inherently O(1) while the others might depend on their standard library implementation, but it's safe to assume they're going to be worse than Set. (See Ariel's comment).
BTW, lookup does not have to be a Set. I've done it Set because lookup contains distinct values and the order is not important.
Iterable.map is a 1:1 mapping; if your input has n elements, then the output must have n elements too.
Using numbers.where(...).map(...) works, but alternatively:
Use Iterable.map first to map either to desired values or to an invalid sentinel value. If your sentinel value is null, then you can use Iterable.whereType to filter out them out:
var transformed = numbers.map((n) {
switch (n) {
case 0:
return false;
case 1:
return true;
default:
return null;
}
}).whereType<bool>();
You might prefer this if you want all of your logic in a single callback.
Use collection-for instead of Iterable.map and collection-if instead of Iterable.where:
var transformed = [
for (var n in numbers)
if (isValidValue(n))
transform(n),
];
where you define isValidValue and transform functions with your filtering and transformation logic respectively.
We have two lists
List<String> list1 = ['foo', 'bar', 'blah', 'bee', 'fog'];
List<String> list2 = ['bee', 'bar'];
and we'd like to remove from the first the items in the second.
We can iterate, removing those in the second one by one
for (var v in list2) {
list1.removeWhere((item) => item == v);
}
but that's silly. It'll take time proportional to the product of the length of the two lists.
We can convert to sets and use Dart's Set.difference(), then return a list.
n = (n.toSet().difference(m.toSet())).toList();
but we lose the order of the items in list1.
What's a good way for determining stable list set-difference in Dart?
A solution would be to convert list2 into a Set to make it more efficient to ask if an element from list1 is part of list2. We can then use removeWhere to remove elements from list1:
void main() {
List<String> list1 = ['foo', 'bar', 'blah', 'bee', 'fog'];
List<String> list2 = ['bee', 'bar'];
list1.removeWhere(list2.toSet().contains);
print(list1); // [foo, blah, fog]
}
im trying to compare 2 arrays but i dont know how
for example:
local array1 = { 'friend', 'work', 'privat' }
local array2 = { 'apple', 'juice', 'privat' }
if both arrays have the same value it should do a print.
i know i need to work with something like this
for x in ipairs(array1) do
if x == array2 then
print ("Hi")
end
end
but ofcourse it didnt work.
so how can i check if the array1 value contains a values from array2?
Think of it this way: You have to check each element in the first array to its counterpart in the second. If any element is not equal, you know right away that the arrays aren't equal. If every element checks out as equal, the arrays are equal.
local function arrayEqual(a1, a2)
-- Check length, or else the loop isn't valid.
if #a1 ~= #a2 then
return false
end
-- Check each element.
for i, v in ipairs(a1) do
if v ~= a2[i] then
return false
end
end
-- We've checked everything.
return true
end
how can i check if the array1 value contains a values from array2?
#luther's answer will not always work for your question..
If the arrays are different sizes, it completely fails.
If you have an array where similar element are not in the exact same index, it can return a false negative.
for example a = {'one', 'two'}; b = {'two', 'one'} will return false
Using table.sort to solve this would be a band-aid solution without fixing the real problem.
The function below will work with arrays of different sizes containing elements in any order
function array_compare(a, b)
for ia, va in ipairs(a) do
for ib, vb in ipairs(b) do
if va == vb then
print("matching:",va)
end
end
end
end
In array_compare we go through all the combinations of elements in table a and table b, compare them, and print if they are equal.
ipairs(table) uses index, value (instead of just value)
For example
local array1 = { 'friend', 'work', 'privat' }
local array2 = { 'apple', 'juice', 'privat' }
array_compare(array1, array2)
will print
matching: privat
(I'm writing a second answer to account for another possible interpretation of the question.)
If you want to see if array1 contains any value that's also in array2, you can do the following:
Convert array1 to a set. A set is a new table where the array's values become keys whose values are true.
Iterate through array2 to see if any of its values are a key in the set.
local set = {}
for _, v in ipairs(array1) do
set[v] = true
end
for _, v in ipairs(array2) do
if set[v] then
print'Hi'
-- Use a break statement if you only want to say hi once.
end
end
If the arrays are large, this algorithm should be faster than a nested loop that compares every value in array1 to every value in array2.
Hello I am new to dart and trying to find an item by property name in a list of list.
class Product{
String id;
String title;
Product(this.id,this.title);
}
void main(){
List<List<Product>> allProdcuts=[
//New Prodcuts
[
Product("1","Hammer"),
Product("3","Nails"),
Product("2","Screws"),
],
futureItems,
//Old Prodcuts
[
Product("4","Rock"),
Product("5","Paper"),
Product("6","Scissor"),
],
//Rare Items
[
Product("7","Plank"),
Product("8","Wires"),
Product("9","Box"),
],
];
print(allProdcuts.where((itemsList)=>itemsList.contains((product)=>product.title='Wires')));
//Returns ()
}
I have tried using for a single List:
List<Product> futureItems= [
Product("101","Galactic Hammer"),
Product("301","Galactic Nails"),
Product("201","Galactic Screws"),
];
print(newProduct.firstWhere((p)=>p.title=='Hammer'));
//Instance of 'Product'
Also tried this:
print(allProdcuts.map((itemList)=>itemList.firstWhere((p)=>p.title=='Nails')));
// Bad state: No elementError: Bad state: No element.
But there is an element with the title='Nails'.I don't understand what I am doing wrong.
You are calling itemList.firstWhere((p)=>p.title=='Nails') on each list, also the ones with no element with title "Nails". Since firstWhere throws if there is no matching value, it does that for two of your three lists. Also, in the example, itemsList.contains(...) does not take a callback, so you are just checking whether a function is in the list, which it isn't. You might want to use any for that, but it won't solve the problem here.
To do this efficiently, I'd probably create helper function:
Product findByTitle(List<List<Product>> allProducts, String title) {
for (var products in allProducts) {
for (var product in products) {
if (product.title == title) return product;
}
}
// Or return `null`.
throw ArgumentError.value(title, "title", "No element with that title");
}
The return in the middle allows you to skip out of the double iteration the moment you have a match, something which is harder to do with firstWhere/map/forEach etc.
One alternative solutions would be:
var product = allProducts.expand((l) => l.where((p) => p.title == title)).first;
which finds all the products with the given title and flattens them into a single iterable, then picks the first one (if there are any). Because iterables are lazy, it will actually stop at the first match.
There are many ways to solve this.
One example is to use the forEach() method:
allProdcuts.forEach(
(List<Product> l)=>l.forEach(
(Product p){
if (p.title=="Nails")
print(p.id);
}
)
);
The for each method receives a function and applies this function to every element on the list. If you have a lists of lists, you can do this twice to get a function applied to each element of the sub lists.
The above code prints 3, which is the desired result.
Another solution would be to flatten the list first, so you can have an easier search later.
print(allProdcuts.any((innerListOfProducts) =>
innerListOfProducts.any((product) => product.title == 'Wires')));
This code will return true if 'Wires' is in the inner list, and false otherwise.
I have a list and I want to put a condition on it. for example, I want to have items from list lst that value greater than 10:
var lst = [{"value":5 , "name":"test1"},
{"value":12 , "name":"test2"},
{"value":8 , "name":"test3"},
{"value":23 , "name":"test4"}];
/*
output: value greater than 10 =>
[{"value":12 , "name":"test2"},
{"value":23 , "name":"test4"}]
*/
You can either use the where function on iterables to filter its elements, then convert the resulting iterable back to a list, or use the list literal syntax, or a combination of the two where you "spread" the result of where:
var list = [ ... ];
var filtered1 = list.where((e) => e["value"] > 10).toList();
var filtered2 = [for (var e in list) if (e["value"] > 10) e];
var filtered3 = [... list.where((e) => e["value"] > 10)];
To filter a list base on a condition you can use List.where which takes a test function and returns a new Iterable that contains the elements that match the test.
To get a list with only the values greater than 10 you can filter you list of maps as such:
lst.where((e) => e['value'] > 10); //Returns a lazy Iterable
if you need to modify your list later you can append a .toList(), to get a new list.
try to use this code:
List lst = [{"value":5 , "name":"test1"} ,{"value":12 , "name":"test2"} , {"value":8 , "name":"test3"} , {"value":23 , "name":"test4"} ];
List newLst = lst.where( (o) => o['value'] > 5).toList();
print(newLst);
> Just try this Function, catogory_id == 1 is condition here
List<dynamic> chooseCatogory(List<dynamic> list) {
List newlist = list.where((o) => o['category_id'] == '1').toList();
return newlist;
}