How to make a curry version of map that always returns list using toolz

If I import toolz using

from toolz.curried import *

then map will automatically becomes curried form, so

map(func,[[1,2],[3,4]])

can be written as

map(func)([[1,2],[3,4]])

but curried map always return an iterable. I way to define an curried lmap which always return list. But simple try

lmap=compose(list,map)

will not work, for example

lmap(len)([[1,2],[3,4]])

will give

---------------------------------------------------------------------------
TypeError Traceback (most recent call last) in () ----> 1 lmap(len)([[1,2],[3,4]])

C:\ProgramData\Anaconda3\lib\site-packages\toolz\functoolz.py in
call(self, *args, **kwargs)
466 ret = self.first(*args, **kwargs)
467 for f in self.funcs:
--> 468 ret = f(ret)
469 return ret
470

TypeError: 'curry' object is not iterable

So how to define a curried lmap?

You're calling it wrong way. map being passed to compose is curried, but not a whole expression. When you call it like

lmap(len)([[1,2],[3,4]])

it passes len to lmap it returns toolz.functoolz.curry equivalent to

map(len)

and then tries to call list on it:

list(map(len))

which obviously cannot work. If it didn't fail complete expression would be equivalent to:

list(map(len))([[1,2],[3,4]])

while the call you're looking for is:

list(map(len), [[1,2],[3,4]])

So in fact currying doesn't make much sense here.

You probably want something around these lines:

def lmap(g): return compose(list, map(g)) 

which would be callable as you want:

>>> lmap(len)([[1,2],[3,4]])
[2, 2]

but to honest the problem looks a bit artificial - the biggest advantage of toolz in uniform laziness. Converting to list throws away most of that.