如何从元组列表中获取元素列表？

Question

How to get an element inside of a list of tuples? For example

list = [(Tris, 23, 1.40), (Aif, 20 , 1.70)] 

I want all the ages

[23,20]

I know that I can use

age = (_,x,_) = x 

But it only works with tuples and not a tuple inside a list.

Answer 1

You will need to perform a map ping where for each item in the list you map it to its age.

You can do this with the map :: (a -> b) -> [a] -> [b] function , so this will look like:

allAges = map …

where I leave filling in … as an exercise.

Answer 2

you can solve it recursively like:

allages [] = []
allages ((_,x,_):xs) = x : allages xs

Answer 3

I know that I can use

age (_,x,_) = x

But it only works with tuples and not a tuple inside a list.

Why are you saying this? Surely if you know that

age   (_,x,_)  =  x

then you also know that

ages [(_,x,_)] = [x]

and

ages [(_,x,_), (_,y,_)]          = [x,y]
ages [(_,x,_), (_,y,_), (_,z,_)] = [x,y,z]
....

etc. And since we've already established that

ages [         (_,y,_), (_,z,_)] = [  y,z]
ages [                  (_,z,_)] = [    z]

so that surely

ages [                         ] = [     ]

we can just write down these equations,

ages [                       ] = [                               ]
ages ([(_,x,_)] ++ moreTuples) = ages [(_,x,_)] ++ ages moreTuples

and since we know that ages [(_,x,_)] = [x] , it's

ages [                       ] = [                               ]
ages ([(_,x,_)] ++ moreTuples) = [        x   ] ++ ages moreTuples

so now all that's left to make it valid Haskell syntax is to rewrite the first equation's argument using the : operator, along the lines of

      [   a   ] ++ moreTuples ==
          a     :  moreTuples

and you're done.

So in the end you did know how to do this.