Why Isn't Assigning to an Empty List (E.G. [] = "") an Error

Why isn't assigning to an empty list (e.g. [] = ) an error?

You are not comparing for equality. You are assigning.

Python allows you to assign to multiple targets:

foo, bar = 1, 2

assigns the two values to foo and bar, respectively. All you need is a sequence or iterable on the right-hand side, and a list or tuple of names on the left.

When you do:

[] = ""

you assigned an empty sequence (empty strings are sequences still) to an empty list of names.

It is essentially the same thing as doing:

[foo, bar, baz] = "abc"

where you end up with foo = "a", bar = "b" and baz = "c", but with fewer characters.

You cannot, however, assign to a string, so "" on the left-hand side of an assignment never works and is always a syntax error.

See the Assignment statements documentation:

An assignment statement evaluates the expression list (remember that this can be a single expression or a comma-separated list, the latter yielding a tuple) and assigns the single resulting object to each of the target lists, from left to right.

and

Assignment of an object to a target list, optionally enclosed in parentheses or square brackets, is recursively defined as follows.

Emphasis mine.

That Python doesn't throw a syntax error for the empty list is actually a bit of a bug! The officially documented grammar doesn't allow for an empty target list, and for the empty () you do get an error. See bug 23275; it is considered a harmless bug:

The starting point is recognizing that this has been around for very long time and is harmless.

Also see Why is it valid to assign to an empty list but not to an empty tuple?

Why is it valid to assign to an empty list but not to an empty tuple?

The comment by @user2357112 that this seems to be coincidence appears to be correct. The relevant part of the Python source code is in Python/ast.c:

switch (e->kind) {
    # several cases snipped
    case List_kind:
        e->v.List.ctx = ctx;
        s = e->v.List.elts;
        break;
    case Tuple_kind:
        if (asdl_seq_LEN(e->v.Tuple.elts))  {
            e->v.Tuple.ctx = ctx;
            s = e->v.Tuple.elts;
        }
        else {
            expr_name = "()";
        }
        break;
    # several more cases snipped
}
/* Check for error string set by switch */
if (expr_name) {
    char buf[300];
    PyOS_snprintf(buf, sizeof(buf),
                  "can't %s %s",
                  ctx == Store ? "assign to" : "delete",
                  expr_name);
    return ast_error(c, n, buf);
}

tuples have an explicit check that the length is not zero and raise an error when it is. lists do not have any such check, so there's no exception raised.

I don't see any particular reason for allowing assignment to an empty list when it is an error to assign to an empty tuple, but perhaps there's some special case that I'm not considering. I'd suggest that this is probably a (trivial) bug and that the behaviors should be the same for both types.

Create an empty list with certain size in Python

You cannot assign to a list like xs[i] = value, unless the list already is initialized with at least i+1 elements. Instead, use xs.append(value) to add elements to the end of the list. (Though you could use the assignment notation if you were using a dictionary instead of a list.)

Creating an empty list:

>>> xs = [None] * 10
>>> xs
[None, None, None, None, None, None, None, None, None, None]

Assigning a value to an existing element of the above list:

>>> xs[1] = 5
>>> xs
[None, 5, None, None, None, None, None, None, None, None]

Keep in mind that something like xs[15] = 5 would still fail, as our list has only 10 elements.

range(x) creates a list from [0, 1, 2, ... x-1]

# 2.X only. Use list(range(10)) in 3.X.
>>> xs = range(10)
>>> xs
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Using a function to create a list:

>>> def display():
...     xs = []
...     for i in range(9): # This is just to tell you how to create a list.
...         xs.append(i)
...     return xs
... 
>>> print display()
[0, 1, 2, 3, 4, 5, 6, 7, 8]

List comprehension (Using the squares because for range you don't need to do all this, you can just return range(0,9) ):

>>> def display():
...     return [x**2 for x in range(9)]
... 
>>> print display()
[0, 1, 4, 9, 16, 25, 36, 49, 64]

Why does example = list(...) result in TypeError: 'list' object is not callable?

Seems like you've shadowed the builtin name list, which points at a class, by the same name pointing at an instance of it. Here is an example:

>>> example = list('easyhoss')  # here `list` refers to the builtin class
>>> list = list('abc')  # we create a variable `list` referencing an instance of `list`
>>> example = list('easyhoss')  # here `list` refers to the instance
Traceback (most recent call last):
  File "<string>", line 1, in <module>
TypeError: 'list' object is not callable

I believe this is fairly obvious. Python stores object names (functions and classes are objects, too) in namespaces (which are implemented as dictionaries), hence you can rewrite pretty much any name in any scope. It won't show up as an error of some sort. As you might know, Python emphasizes that "special cases aren't special enough to break the rules". And there are two major rules behind the problem you've faced:

Namespaces. Python supports nested namespaces. Theoretically you can endlessly nest them. As I've already mentioned, they are basically dictionaries of names and references to corresponding objects. Any module you create gets its own "global" namespace, though in fact it's just a local namespace with respect to that particular module.
Scoping. When you reference a name, the Python runtime looks it up in the local namespace (with respect to the reference) and, if such name does not exist, it repeats the attempt in a higher-level namespace. This process continues until there are no higher namespaces left. In that case you get a NameError. Builtin functions and classes reside in a special high-order namespace __builtins__. If you declare a variable named list in your module's global namespace, the interpreter will never search for that name in a higher-level namespace (that is __builtins__). Similarly, suppose you create a variable var inside a function in your module, and another variable var in the module. Then, if you reference var inside the function, you will never get the global var, because there is a var in the local namespace - the interpreter has no need to search it elsewhere.

Here is a simple illustration.

>>> example = list("abc")  # Works fine
>>> 
>>> # Creating name "list" in the global namespace of the module
>>> list = list("abc")
>>> 
>>> example = list("abc")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'list' object is not callable
>>> # Python looks for "list" and finds it in the global namespace,
>>> # but it's not the proper "list".
>>> 
>>> # Let's remove "list" from the global namespace
>>> del list
>>> # Since there is no "list" in the global namespace of the module,
>>> # Python goes to a higher-level namespace to find the name. 
>>> example = list("abc")  # It works.

So, as you see there is nothing special about Python builtins. And your case is a mere example of universal rules. You'd better use an IDE (e.g. a free version of PyCharm, or Atom with Python plugins) that highlights name shadowing to avoid such errors.

You might as well be wondering what is a "callable", in which case you can read this post. list, being a class, is callable. Calling a class triggers instance construction and initialisation. An instance might as well be callable, but list instances are not. If you are even more puzzled by the distinction between classes and instances, then you might want to read the documentation (quite conveniently, the same page covers namespaces and scoping).

If you want to know more about builtins, please read the answer by Christian Dean.

P.S. When you start an interactive Python session, you create a temporary module.

SyntaxError: cannot assign to operator

Python is upset because you are attempting to assign a value to something that can't be assigned a value.

((t[1])/length) * t[1] += string

When you use an assignment operator, you assign the value of what is on the right to the variable or element on the left. In your case, there is no variable or element on the left, but instead an interpreted value: you are trying to assign a value to something that isn't a "container".

Based on what you've written, you're just misunderstanding how this operator works. Just switch your operands, like so.

string += str(((t[1])/length) * t[1])

Note that I've wrapped the assigned value in str in order to convert it into a str so that it is compatible with the string variable it is being assigned to. (Numbers and strings can't be added together.)

Why does Python allow unpacking an empty iterable?

Based on the relevant Python issue discussion, assigning to an empty list [] was actually possible for a long time, but not documented. This was considered "a fairly harmless quirk" until it was documented in 3.5. Then assigning to an empty tuple () was added for consistency.

In the same thread, Martin Panter provides a possible use case:

[...] I have used this syntax on purpose as a concise way to ensure that a generator is exhaused with no more yields:
>>> def gen():
...     yield "partial computation"
...     print("computation allowed to complete")
... 
>>> g = gen()
>>> next(g)
'partial computation'
>>> [] = g
computation allowed to complete

Why does append() always return None in Python?

append is a mutating (destructive) operation (it modifies the list in place instead of of returning a new list). The idiomatic way to do the non-destructive equivalent of append would be

>>> l = [1,2,3]
>>> l + [4]
[1,2,3,4]
>>> l
[1,2,3]

to answer your question, my guess is that if append returned the newly modified list, users might think that it was non-destructive, ie they might write code like

m = l.append("a")
n = l.append("b")

and expect n to be [1,2,3,"b"]

The default 'List' constructor isn't available when null safety is enabled. Try using a list literal, 'List.filled' or 'List.generate'

Short answer:

Instead of the pre-null-safety operations

var foo = List<int>();  // Now error
var bar = List<int>(n); // Now error
var baz = List<int>(0); // Now error

use the following:

var foo = <int>[];           // Always the recommended way.
var bar = List.filled(1, 0); // Not filled with `null`s.
var baz = List<int>.empty();

Long answer:

The List constructor had two uses:

new List() to create an empty growable list, equivalent to [].
new List(n) to create a fixed-length list of length n filled with null values

With null safety, the second use was unsound most of the time, and there was no good way to fix it. It's possible to force a type argument to be non-nullable, but List<T>(4) only works when T is nullable. There is no way to enforce that.

So, the List(n) mode needed to go (replaced by List.filled(n, value) which forces you to provide a fill-value).
That left List(), which doesn't really carry its own weight. You can just use [] instead (and you should!), so it was decided to remove the constructor entirely - all uses of it was either unsafe or useless.
(Also, it was a weird constructor already, because if we wanted to properly make it null safe, it would have an optional parameter with a non-nullable type and no default value.)

By removing it completely, it makes it possible to, potentially, introduce a new List constructor in the future, perhaps as a shorter alias for List.filled. One can hope.

Why Isn't Assigning to an Empty List (E.G. [] = "") an Error