Any difference between dir() and locals() in Python?

Question

According to Python documentation, both dir() (without args) and locals() evaluates to the list of variables in something called local scope. First one returns list of names, second returns a dictionary of name-value pairs. Is it the only difference? Is this always valid?

assert dir() == sorted( locals().keys() )

Answer 1

The output of dir() when called without arguments is almost same as locals(), but dir() returns a list of strings and locals() returns a dictionary and you can update that dictionary to add new variables.

dir(...)
    dir([object]) -> list of strings

    If called without an argument, return the names in the current scope.


locals(...)
    locals() -> dictionary

    Update and return a dictionary containing the current scope's local variables.

---

Type:

>>> type(locals())
<type 'dict'>
>>> type(dir())
<type 'list'>

---

Update or add new variables using locals():

In [2]: locals()['a']=2

In [3]: a
Out[3]: 2

using dir(), however, this doesn't work:

In [7]: dir()[-2]
Out[7]: 'a'

In [8]: dir()[-2]=10

In [9]: dir()[-2]
Out[9]: 'a'

In [10]: a
Out[10]: 2