Different results from yield vs return

python, recursion, yield

Solution

The problem is you're iterating over generators instead of lists in the yield version, specifically secondG which is exhausted after one loop. Change the line to this and it works:

firstG, secondG = f(first), list(f(second))

Or, you can change your loop:

for e in ("(%s%s%s)" % (e1, op, e2) for e1 in f(first) for e2 in f(second)):
#                               new generator object every loop  ^^^^^^^^^

The non-yield version works because you return lists, which can be iterated over again, unlike generators. Also note you only iterate over firstG once, so it's not affected.

Remember that this:

r = [v for a in A for b in B]

Is equivalent to:

r = []
for a in A:
  for b in B:
    r.append(v)

Which more clearly shows the repeated loop over B.

Another example:

def y():
  yield 1
  yield 2
  yield 3
def r():
  return [1, 2, 3]

vy = y()
for v in vy:
  print v
for v in vy:
  print v

print "---"

vr = r()
for v in vr:
  print v
for v in vr:
  print v

Problem

I don't really understand how `yield` statement works in this situation. The problem says that given an expression without parentheses, write a function to generate all possible fully parenthesized (FP) expressions. Say, the input is `'1+2+3+4'` which should be generated to 5 FP expressions: - (1+(2+(3+4))) - (1+((2+3)+4)) - ((1+2)+(3+4)) - ((1+(2+3))+4) - (((1+2)+3)+4) My code is as follows. ``` OPS = ('+', '-', '*', '/') def f(expr): """ Generates FP exprs Recursive formula: f(expr1[op]expr2) = (f(expr1) [op] f(expr2)) """ if expr.isdigit(): yield expr # return [expr] # ret = [] first = '' i = 0 while i < len(expr): if expr[i] not in OPS: first += expr[i] i += 1 else: op = expr[i] i += 1 second = expr[i:] firstG, secondG = f(first), f(second) for e in ('(' + e1 + op + e2 + ')' for e1 in firstG for e2 in secondG): yield e # ret.append(e) first += op # return ret ``` If I use `return` statement (the commented out lines), then the code works as expected. However, when I change to `yield` statement as the code shows, I only get the first 4 results. If the number of operands of the input expression is increased, then of course more results will be lost. For example, for the input `'1+2+3+4+5'`, I only get 8 instead of 14. I finally figure out the way to make the code work by commenting out the line `firstG, secondG = f(first), f(second)` and replace the line `for e in ('(' + e1 + op + e2 + ')' for e1 in firstG for e2 in secondG):` by `for e in ('(' + e1 + op + e2 + ')' for e1 in f(first) for e2 in f(second)):` That means some 'information' of the generator is lost because of the line `firstG, secondG = f(first), f(second)` but I can't figure out the real reason. Could you guys give me some ideas?

Original source