[Python] Can you yield from a lambda function?

I have a generator function in a class:

self.s = [['a',1],['b',2],['c',3]

def generator(self):
for r in self.s:
yield r


In another function I initalize it as a variable:

var = self.generator()


And it's yielded as necessary:

>>> next(var) # returns ['a',1]
>>> next(var) # returns ['b',2]
>>> next(var) # returns ['c',3]


Can defining the generator be done in one line, however? I've considered the below:

var = lambda: [(yield r) for r in self.s]
>>> next(var) # SyntaxError: 'yield' inside list comprehension


Here's a minimal code I'm working with:

class Foo():
def __init__(self):
self.s = {}

def generator(self):
for r in self.s:
yield r

def fetch(self):
if not self.s:
self.fetch_gen = self.generator()
self.s['a'] = 1
self.s['b'] = 2
self.s['c'] = 3
try:
var = self.s.get(next(self.fetch_gen))
except StopIteration:
return None
return var

BAR = Foo()
while True:
OUT = BAR.fetch()
if OUT is None:
break
print(OUT)


Output is below:

1
2
3


I just wanted to see if I could get rid of Foo.generator and instead declare the generator in one line.

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