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I have a 3-tuple list like the following [I added line breaks for readability]:
(2, 127, 3)
(12156, 127, 3)
(4409, 127, 2) <-- 4409 occurs 2x
(1312, 127, 12) <-- 1312 occurs 3x
(4409, 128, 1) <--
(12864, 128, 1)
(1312, 128, 1) <--
(2664, 128, 2)
(12865, 129, 1)
(183, 129, 1)
(12866, 129, 2)
(1312, 129, 10) <--
I want to sum up based on the first entry. The first entry should be unique.
The result should look like this:
(2, 127, 3)
(12156, 127, 3)
(4409, 127, 3) <- new sum = 3
(1312, 127, 23) <- new sum = 23
(12864, 128, 1)
(2664, 128, 2)
(12865, 129, 1)
(183, 129, 1)
(12866, 129, 2)
How can I achieve this in Scala?
777
Use indexing to get the first element of each tuple Use a for-loop to iterate through a list of tuples. Within the for-loop, use the indexing tuple[0] to access the first element of each tuple, and append it.
How to find a sum of a tuple in Python. To find a sum of the tuple in Python, use the sum() method. Define a tuple with number values and pass the tuple as a parameter to the sum() function; in return, you will get the sum of tuple items.
Using sum() to get the total in a tuple The built-in sum() function in Python is used to return the sum of an iterable. To get the sum total of a tuple of numbers, you can pass the tuple as an argument to the sum() function.
A list of tuple basically contains tuples enclosed in a list. The map function applies a given function/operation to every item in an iterable (such as list, tuple). It returns a list as the result. The 'sum' method can be used to add the elements in an iterable.
Try this:
list groupBy {_._1} mapValues {v => (v.head._1, v.head._2, v map {_._3} sum)}
The middle entry is preserved and it always takes the first one that appeared in the input list.
154
If you can just ignore the middle entry, then:
val l = List(('a,'e,1), ('b,'f,2), ('a,'g,3), ('b,'h,4))
l.groupBy(_._1).mapValues(_.map(_._3).sum)
// Map('b -> 6, 'a -> 4)
If you have to keep the middle entry around:
l.groupBy(_._1).map {
case (_, values) =>
val (a,b,_) = values.head
(a, b, values.map(_._3).sum)
}
// List(('b,'f,6), ('a,'e,4))
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