def qs3(ls):
'''
And we can also take away the call to p.
'''
pivot = middle(ls)
(ls, iff(len, (qs3, {_ < pivot}) + [pivot] + (qs3, {_ > pivot})))
def qs2(ls):
'''
Now, we take away the return statement for space.
'''
pivot = middle(ls)
p(ls, iff(len, (qs2, {_ < pivot}) + [pivot] + (qs2, {_ > pivot})))
def qs4(ls):
'''
And we can even take away the accum. Notice, however, the final comma, making
the statement a tuple.
'''
pivot = middle(ls)
iff(len, (qs4, {_ < pivot}) + [pivot] + (qs4, {_ > pivot})),
def qs1(ls):
'''
Now, we are going to apply the iff macro. So for a conditional fArg cond, and a
statement, iff(cond,statement) evaluates as:
{cond:statement,
'else':_}
.
So, if len evaluates as True (nonzero), then we return the statement, otherwise
we return the empty list.
'''
pivot = middle(ls)
return p(ls, iff(len, (qs1, {_ < pivot}) + [pivot] + (qs1, {_ > pivot})))
def qs0(ls):
'''
middle takes the middle element of the list. Let's explore the expression:
{len:(qs0,{_ < pivot}) + [pivot] + (qs0,{_ > pivot}),
'else':_}
First, this is a switch dict, meaning that if one of the keys evaluates as true,
then the corresponding expression also evaluates as true. So if ls is not
empty, then len evaluates as true, and we apply the function recursively. Otherwise
we return the emply ls.
(qs0,{_ < pivot})
This applies a filter to ls, taking anything below the pivot. Then we recursively
apply the function to this list.
[pivot]
This constructs a singleton list with pivot.
(qs0,{_ > pivot})
This applies a filter to ls, taking anything above the pivot. Then we recursively
apply the function to this list.
(qs0,{_ < pivot}) + [pivot] + (qs0,{_ > pivot})
We concatenate these three lists, and return.
'''
pivot = middle(ls)
return p(ls, {
len: (qs0, {_ < pivot}) + [pivot] + (qs0, {_ > pivot}),
'else': _
})