Why does creating a list with the same generator only work the first time?

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4

In Python 3.6.8, when I materialize the list, it is shown and erases. This is a bug or so it is?

>>> vet_neg
['EH01', 'EH02', 'EH03']
Categories (3, object): ['EH01', 'EH02', 'EH03']
>>> cenarios
[0, 1]
>>> vet_neg_cenarios = itertools.product(vet_neg, cenarios, cenarios)
>>> vet_neg_cenarios
<itertools.product object at 0x7f6fa16a65e8>
>>> list(vet_neg_cenarios)
[('EH01', 0, 0), ('EH01', 0, 1), ('EH01', 1, 0), ('EH01', 1, 1), ('EH02', 0, 0), ('EH02', 0, 1), ('EH02', 1, 0), ('EH02', 1, 1), ('EH03', 0, 0), ('EH03', 0, 1), ('EH03', 1, 0), ('EH03', 1, 1)]
>>> list(vet_neg_cenarios)
[]
>>>
python python-3.x list

2 answers

5


The way you did it, that’s the way it is.

Internally, the function itertools.product work with generators and, by definition, a generator can only be consumed once.

When you do list(vet_neg_cenarios) the first time, the generator will be consumed completely so that all generated values are stored in your list. After exhausted - consumed completely - the generator will no longer have values to generate and therefore, list(vet_neg_cenarios) will be an empty list.

If you really need to keep in memory, assign the result of the first list(vet_neg_cenarios) to an object.

  • No need to keep. I had put a print before a loop for future debugging and it came up.

1

What you have is a Generator and the value in question erases from memory after you use it. For example, the elements of the variable:

pares=(k for k in range(10) if k%2==0)

can be accessed by the method pares.__next__(), but after calling 5 times it stops working. See:

In [2]: pares.__next__()
Out[2]: 0

In [3]: pares.__next__()
Out[3]: 2

In [4]: pares.__next__()
Out[4]: 4

In [5]: pares.__next__()
Out[5]: 6

In [6]: pares.__next__()
Out[6]: 8

In [7]: pares.__next__()
---------------------------------------------------------------------------
StopIteration                             Traceback (most recent call last)
<ipython-input-7-fa4a719088da> in <module>
----> 1 pares.__next__()

StopIteration:

Note that in your case, if you run print(vet_neg_cenarios.__next__()) the program will print the first element of the list. When you called the method list in Enerator, you consumed all the elements at once.

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