NoneType object not iterable

[tom]

Hi!
My code is

> db = {}
>
def display():
keyList = db.keys()
sortedList = keyList.sort()
for name in sortedList:
line = 'Name: %s, Number: %s' % (name, db[name])
print line.replace('\r', '')

And it gives following error:

for name in sortedList:
TypeError: 'NoneType' object is not iterable

How can sortedList variable turn into NoneType? I just don't get it...

 

[Stargaming]

Hi!
My code is

db = {}

def display():
keyList = db.keys()
sortedList = keyList.sort()
for name in sortedList:
line = 'Name: %s, Number: %s' % (name, db[name])
print line.replace('\r', '')

And it gives following error:

for name in sortedList:
TypeError: 'NoneType' object is not iterable

How can sortedList variable turn into NoneType? I just don't get it...

It does not turn into something. The `sort()` method just works "in
place", i. e. it will mutate the list it has been called with. It
returns None (because there is no other sensible return value).

For you, that means: You don't have to distinguish between keyList and
sortedList. Just call ``.sort()`` on keyList and it will just work.

HTH,
Stargaming

P.S. same information is in the docs:
http://docs.python.org/lib/typesseq-mutable.html

 

[Daniel]

Hi!
My code is

db = {}

def display():
keyList = db.keys()
sortedList = keyList.sort()
for name in sortedList:
line = 'Name: %s, Number: %s' % (name, db[name])
print line.replace('\r', '')

And it gives following error:

for name in sortedList:
TypeError: 'NoneType' object is not iterable

How can sortedList variable turn into NoneType? I just don't get it...

db is out of scope, you have to pass it to the function:

def display(db):
keyList = db.keys()
sortedList = keyList.sort()
for name in sortedList:
line = 'Name: %s, Number: %s' % (name, db[name])
print line.replace('\r', '')

 

[Bjoern Schliessmann]

db is out of scope, you have to pass it to the function:

What's wrong about module attributes?

Regards,


Björn

 

[Daniel]

What's wrong about module attributes?

I made a mistake

 

[Paul Rubin]

It does not turn into something. The `sort()` method just works "in
place", i. e. it will mutate the list it has been called with. It
returns None (because there is no other sensible return value).

For you, that means: You don't have to distinguish between keyList and
sortedList. Just call ``.sort()`` on keyList and it will just work.

Alternatively, use
sortedList = sorted(keyList)