"'int' object is not iterable" iterating over a dict

[mmiikkee13]

>>a_list = range(37)

>>list_as_dict = dict(zip(range(len(a_list)), [str(i) for i in a_list]))
for k, v in list_as_dict:

.... print k, v
....
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'int' object is not iterable

What 'int' object is this referring to? I'm iterating over a dict, not
an int.

[Alex_Gaynor]

On Oct 5, 8:11*pm, mmiikkee13 <[e-mail]:

>a_list = range(37)
list_as_dict = dict(zip(range(len(a_list)), [str(i) for i in a_list]))
for k, v in list_as_dict:

... * * print k, v
...
Traceback (most recent call last):
* File "<stdin>", line 1, in <module>
TypeError: 'int' object is not iterable

What 'int' object is this referring to? I'm iterating over a dict, not
an int.

It thinks you are trying to unpack the first item, you mean to be
doing for k, v in list_as_dict.iteritems()

[Diez B. Roggisch]

mmiikkee13 schrieb:

>>>a_list = range(37)
list_as_dict = dict(zip(range(len(a_list)), [str(i) for i in a_list]))
for k, v in list_as_dict:

... print k, v
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'int' object is not iterable

What 'int' object is this referring to? I'm iterating over a dict, not
an int.

Because

for name in dictionary:
will bind name to the *keys* of dictionary. Thus you end up with an int.

Then the tuple-unpacking occurs:

k, v = <number>

Which is equivalent to

k = <number>[0]
v = <number>[1]

And because <numberisn't iterable, it will fail.

Use

for k, v in dictionary.iteritems():
instead.

Diez