What is str
and __str__
?
The function str
is to return a printable form of the object only! From the docs
str(object)
does not always attempt to return a string that is acceptable toeval()
; its goal is to return a printable string.
The __str__
function in a class is called whenever you call the str
on an object. Again from the documentation
object.__str__(self)
Called by the
str()
built-in function and by the
What is list
function?
The function list
is to create a list from an iterable! Again from the docs
Return a list
whose items are the same and in the same order as
iterable‘s items
Thus, str(list)
gives you a printable form and list(str(list))
will iterate over the string. That is list(str(list))
will give you a list of the individual characters of the printable form of the argument passed.
A small walk-through between the nested calls,
Given list, l = ['a','b']
When you call str(l)
, it returns a printable form of the list l
, that is "['a','b']"
.
Now you can see clearly that "['a','b']"
is a string and is indeed an iterable. Now when you call list
on this i.e. list("['a','b']")
you get a weird list like ['[', "'", 'a', "'", ',', "'", 'b', "'", ']']
. Why does this happen? This happens because the string iterates over its characters, you can test this by using a dummy string,
>>> 'dummy'
'dummy'
>>> list('dummy')
['d', 'u', 'm', 'm', 'y']
Thus when you call the list
on a string you get a list of character. Note that again here, when you call str
on list('dummy')
, you will not get back your original string 'dummy'
, so again you will have to use join
! Thus recalling the same function will NOT get you back your original object!
So, Calling str
over a list calls the builtin __str__
method of the list?
The answer is NO!
What happens internally when you call str
on a list?
Whenever you call str
on an list object, the steps followed are
- Call the
repr
of each of the list element. - Add a fancy
[
at the front and another]
at the end of the list. - Join all of them with a comma.
As you can see from the source code of the list object in cpython on github. Going through the source code of cpython in hg.python, which is more clear, you can see the following three comments.
/* Do repr() on each element. Note that this may mutate the list, so must refetch the list size on each iteration. */ line (382) /* Add "[]" decorations to the first and last items. */ line (398) /* Paste them all together with ", " between. */ line (418)
These correspond to the points I mentioned above.
Now what is repr
?
repr
prints the string representation of all the objects. Again from the documentation
Return a string containing a printable representation of an object.
and also note this sentence!
For many types, this function makes an attempt to return a string
that would yield an object with the same value when passed to eval()
,
otherwise the representation is a string enclosed in angle brackets
that contains the name of the type of the object together with
additional information often including the name and address of the
object.
Does list(str(list))
turn the str(list)
back to the original list?
No
Internally, str(list)
actually creates the repr
representation of the list object. So to get back the list after calling str
on the list, you actually need to do eval
on it and not a list
call.
Workarounds
1. Using literal_eval
But we all know that eval
is evil, so what is the workaround? The work-around would be to use ast.literal_eval
.
ast.literal_eval
is safe unlike the eval
function. The docs itself mention that it is safe --
Safely evaluate an expression node or a string containing a Python literal or container display
2. Using string functions and builtins
Another workaround can be done using str.split
>>> x = ['abc', 'def', 'ghi']
>>> a = str(x)
>>> a[2:-2].split("', '")
['abc', 'def', 'ghi']
This is just a simple way to do that for a list of strings. For a list of integers you will need map
.
>>> x = [1,2,3]
>>> a =str(x)
>>> list(map(int,a[1:-1].split(', '))) # No need for list call in Py2
[1, 2, 3]
Thus unlike literal_eval
these are simple hacks given that you know the elements of the list. If they are heterogeneous in nature like [1, "a", True]
then you will have to loop through the split list and discover the element type and then convert it and append the converted element to a final list.