I'm new in IOS programming. I have a question, how to get all value in array except x value. Let say i have array like below :
let array : [Any] = [1,2,3,4,5,6,7,8,9,0,11,22,33,44,55,66,77,200]
how to print all value except 1 and 2.
I have read this , its using filter and i try it with playground but i still not have the right value. Any answer will helpfull for me. Thanks in advance .
I don't know why you have defined the array as [Any] so I just removed that and the array is:-
let array = [1,2,3,4,5,6,7,8,9,0,11,22,33,44,55,66,77,200]
Next you can use filter as follows:-
let filtered = array.filter { (element) -> Bool in
return element != 1 && element != 2
}
You can test this out in the playground, it will print all values except 1 & 2
You can also use some syntactical sugar for filter as follows:-
array.filter({ return $0 != 1 && $0 != 2 })
And since the closure is a trailing argument, you can also separate it from the arguments as follows:-
array.filter { return $0 != 1 && $0 != 2 }
Another way to do this would be
let filterTheseOut = [1,2]
let anotherWay = array.filter { !filterTheseOut.contains($0) }
So here you can basically add all the elements to be filtered out in a separate array
for before and its work for me. but now i want to know if using filter. your answer is better than other$0 ?
You can do it like this
let array : [Int] = [1,2,3,4,5,6,7,8,9,0,11,22,33,44,55,66,77,200]
print(array.filter { $0 != 1 && $0 != 2 } )
or if you will have more than 1 or 2 values you can put them into array
let array : [Int] = [1,2,3,4,5,6,7,8,9,0,11,22,33,44,55,66,77,200]
let unwantedValues = [1,2]
print(array.filter { unwantedValues.contains($0) == false } )
Next time please paste your code, it will be easier to tell you what you're doing wrong, then giving you ready solution.
No need for filter use this:
for i in array {
if i != 1 && i != 2 {
print i
}
}
// This will print all values except 1 and 2
filterthat does not work[Any], it's[Int]. Do not annotate types the compiler can infer. In many cases like this one you make it worse.