harrypotter

You can create list of lists for last 2 lists and then use zip and list_comprehension.

The problem with your code is that for every card aded in populateBook, you loop all the previous cards and add a click event listener, which means the second book gets 2 copies of this handler, the third 3 etc.

Instead of doing that, add a single event handler for clicking and handle appropriately:

The first part is quite simple to convert from Scala to Java, you only need to use map to split each line by comma to get a JavaRDD. Then using flatMap, for each row, split the last part of the array corresponding to Name, and using java streams, you can transform each element of the names list into a new list.

Here is a complete example:

You were not sufficiently clear about what was the result of your approaches. Do they give execution errors or don't they bring the results you expected?

For the first problem, it seems that alive and hogwartsStudents are of boolean type, so I recommend to use true instead of "true"

Your root is a dictionary not an array so

1. Mongodb creates an _id field any element that doesn't have it.
2. If _id already there, it won't overwrite it. (and throws an error instead).
3. If id is there, mongodb doesn't care. It won't modify, and will follow 1 and 2.

Let's run an example in the mongo shell:

The error clearly tells that the wand type should be Wand instead of String.

Also, for variables with multiple types, you can use enums.

Also, for options like gender, you should use enum instead.

So:

I don't think I fully understood your requirement especially about "similar name". So I assume that you want to group the records based on the "name" and all the records which have the same "name" will have the same "age".

The solution below is grouping the records using an object hash and keep concatenating the likings into the element. After finishing it, return the all elements of the object by calling Object.values() which should maintain the order of appearance of names.

Is this something you want, or at least give you some idea? Hope it helps.

You could use Array.filter(), then Array.sort() as follows: