## Friday, March 24, 2017

### Analysing that Python code snippet

By Vasudev Ram

Some days ago I had written this post:

Analyse this Python code snippet

in which I had shown a snippet of Python code (run in the Python shell), and said:

"Analyse the snippet of Python code below. See what you make of it. I will discuss it in my next post."

I am a few days late in discussing it; sorry about that.

Here is the analysis:

First, here's the the snippet again, for reference:
```>>> a = 1
>>> lis = [a, 2 ]
>>> lis
[1, 2]
>>> lis = [a, 2 ,
... "abc", False ]
>>>
>>> lis
[1, 2, 'abc', False]
>>> a
1
>>> b = 3
>>> lis
[1, 2, 'abc', False]
>>> a = b
>>> a
3
>>> lis
[1, 2, 'abc', False]
>>> lis = [a, 2 ]
>>> lis
[3, 2]
>>>
```

The potential for confusion (at least, as I said, for newbie Pythonistas) lies in these apparent points:

The variable a is set to 1.
Then it is put into the list lis, along with the constant 2.
Then lis is changed to be [a, 2, "abc", False].
One might now think that the variable a is stored in the list lis.
The next line prints its value, which shows it is 1.
All fine so far.
Then b is set to 3.
Then a is set to b, i.e. to the value of b.
So now a is 3.
But when we print lis again, it still shows 1 for the first item, not 3, as some might expect (since a is now set to 3).
Only when we run the next line:
lis = [a, 2]
and then print lis again, do we see that the first item in lis is now 3.

This has to do with the concept of naming and binding in Python.

When a Python statement like:
```a = 1
```
is run, naming and binding happens. The name on the left is first created, and then bound to the (value of the) object on the right of the equals sign (the assignment operator). The value can be any expression, which, when evaluated, results in a value (a Python object [1]) of some kind. In this case it is the int object with value 1.

[1] Almost everything in Python is an object, like almost everything in Unix is a file. [Conditions apply :)]

When that name, a, is used in an expression, Python looks up the value of the object that the name is bound to, and uses that value in the expression, in place of the name.

So when the name a was used inside any of the lists that were bound to the name lis, it was actually the value bound to the name a that was used instead. So, the first time it was 1, so the first item of the list became 1, and stayed as 1 until another binding of some other (list) object to the name lis was done.

But by this time, the name a had been rebound to another object, the int 3, the same one that name b had been earlier bound to just before. So the next time that the name lis was bound to a list, that list now included the value of the current object that name a was now bound to, which was 3.

This is the reason why the code snippet works as it does.

On a related note (also about Python language features, syntax and semantics), I was playing around with the pprint module (Python's pretty-printer) and the Python is operator, and came up with this other snippet:

>>> import pprint
>>> lis = []
>>> for i in range(10):
... lis.append(lis)
...
>>> print lis
[[...], [...], [...], [...], [...], [...], [...], [...], [...], [...]]

>>> pprint.pprint(lis)
[<recursion on list with id=32809968>,
<recursion on list with id=32809968>,
<recursion on list with id=32809968>,
<recursion on list with id=32809968>,
<recursion on list with id=32809968>,
<recursion on list with id=32809968>,
<recursion on list with id=32809968>,
<recursion on list with id=32809968>,
<recursion on list with id=32809968>,
<recursion on list with id=32809968>]

>>> len(lis)
10

>>> lis is lis[0]
True

>>> lis is lis[0] is lis[0][0]
True

>>> lis is lis[0] is lis[0][0] is lis[0][0][0]
True

in which I created a list, appended it to itself, and then used pprint.pprint on it. Also used the Python is operator between the list and its 0th item, recursively, and was interested to see that the is operator can be used in a chain. I need to look that up (pun intended).

Enjoy.

- Vasudev Ram - Online Python training and consulting

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