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Advanced Nested List Comprehension Syntax

Asked 2010-09-22T06:07:19.837
48

I was playing around with list comprehensions to get a better understanding of them and I ran into some unexpected output that I am not able to explain. I haven't found this question asked before, but if it /is/ a repeat question, I apologize.

I was essentially trying to write a generator which generated generators. A simple generator that uses list comprehension would look like this:

(x for x in range(10) if x%2==0) # generates all even integers in range(10)

What I was trying to do was write a generator that generated two generators - the first of which generated the even numbers in range(10) and the second of which generated the odd numbers in range(10). For this, I did:

>>> (x for x in range(10) if x%2==i for i in range(2))
<generator object <genexpr> at 0x7f6b90948f00>

>>> for i in g.next(): print i
... 
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 1, in <genexpr>
UnboundLocalError: local variable 'i' referenced before assignment
>>> g.next()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration
>>> g = (x for x in range(10) if x%2==i for i in range(2))
>>> g
<generator object <genexpr> at 0x7f6b90969730>
>>> g.next()
Traceback (most recent call last):
      File "<stdin>", line 1, in <module>
      File "<stdin>", line 1, in <genexpr>
    UnboundLocalError: local variable 'i' referenced before assignment

I don't understand why 'i' is being referenced before assignment

I thought it might have had something to do with i in range(2), so I did:

>>> g = (x for x in range(10) if x%2==i for i in [0.1])
>>> g
<generator object <genexpr> at 0x7f6b90948f00>
>>> g.next()
Traceback (most recent call la
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1 Answer

11

Lie Ryan's for-loop equivalent leads me to the following, which does seem to work just fine:

[x for i in range(2) for x in range(10) if i == x%2]

outputs

[0, 2, 4, 6, 8, 1, 3, 5, 7, 9]
answered 2013-02-04T22:00:41.897

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