The intersection 'xxx & xxx' was reduced to 'never' because property 'xxx' has conflicting types in some constituents

Question

I got an error while using ts generics, here is the simple code:

Typescript Playground

On the last line, ts reports the following error:

error TS2345: Argument of type 'Task<"build"> | Task<"repair">' is not assignable to parameter of type 'never'.
  The intersection 'Task<"build"> & Task<"repair">' was reduced to 'never' because property 'type' has conflicting types in some constituents.
   
44 action_map[task.type].execute(task);

I try to use switch to avoid errors:

function execute<T extends TaskType>(task: Task<T>) {
    switch (task.type) {
        case "build":
            // now type of `task` should be `Task<"build">`
            BuildAction.execute(task);
            break;
        case "repair":
            // now type of `task` should be `Task<"repair">`
            RepairAction.execute(task);
            break;
        default:
            // now type of `task` should be `Task<never>`
            console.log("Error");
    }
}

But it even worse:

error TS2345: Argument of type 'Task<T>' is not assignable to parameter of type 'Task<"build">'.
  Type 'T' is not assignable to type '"build"'.
    Type 'keyof TMap' is not assignable to type '"build"'.
      Type '"repair"' is not assignable to type '"build"'.

50    BuildAction.execute(task);

I noticed that vscode's type hint for task is always Task<T> instead of what I expected.

So, what should I do?

Answer 1

The fastest way is to transform both in sets and print the difference:

>>> print(set(x).difference(set(y)))
{6}

This code print numbers present in x but not in y

Answer 2

x =[1,2,3,4,5,6]
y = [1,2,3,4,5]

for i in x:
    if i in y:
        print(f"{i} found")
    else:
        print(f"{i} not found")

Answer 3

This is the best option in my opinion.

x =[1,2,3,4,5,6]
y = [1,2,3,4,5]

for number in x:
    if number not in y:
        print(f"{number} not found")

Answer 4

to get not matches:

def returnNotMatches(a, b):
    return [[x for x in a if x not in b], [x for x in b if x not in a]]

or

new_list = list(set(list1).difference(list2))

to get the intersection:

list1 =[1,2,3,4,5,6]
list2 = [1,2,3,4,5]
list1_as_set = set(list1)
intersection = list1_as_set.intersection(list2)

output:

{1, 2, 3, 4, 5}

you can also transfer it to a list:

intersection_as_list = list(intersection)

or:

new_list = list(set(list1).intersection(list2))