Please check my understanding...

[Tobiah]

list.append([1,2]) will add the two element list as the next
element of the list.

list.extend([1,2]) is equivalent to list = list + [1, 2]
and the result is that each element of the added list
becomes it's own new element in the original list.

Is that the only difference?

From the manual:

s.extend(x) | same as s[len(s):len(s)] = x

But: (python 2.5.2)

>>a

[1, 2, 3]

>>a[len(a):len(a)] = 4

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: can only assign an iterable

>>>

Also, what is the difference between list[x:x] and list[x]?

>>a[3:3] = [4]
>>a

[1, 2, 3, 4]
** Posted from http://www.teranews.com **

[Matimus]

Re: Please check my understanding.. .

On Jul 1, 12:35 pm, Tobiah <t...@tobiah.or gwrote:

list.append([1,2]) will add the two element list as the next
element of the list.
>
list.extend([1,2]) is equivalent to list = list + [1, 2]
and the result is that each element of the added list
becomes it's own new element in the original list.
>
Is that the only difference?
>
From the manual:
>
s.extend(x)  |  same as s[len(s):len(s)] = x
>
But: (python 2.5.2)
>

>a

[1, 2, 3]

>a[len(a):len(a)] = 4

>
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: can only assign an iterable
>
>
>
Also, what is the difference between list[x:x] and list[x]?
>

>a[3:3] = [4]
>a

>
[1, 2, 3, 4]
** Posted fromhttp://www.teranews.co m**

In this example:

s.extend(x) | same as s[len(s):len(s)] = x

x _must_ be iterable. As the error states, `4` is not iterable.

the s[start:stop] notation is called slicing:

>>x = range(10)
>>x[0]

0

>>x[1]

1

>>x[0:1]

[0]

>>x[0:2]

[0, 1]

>>x[0:3]

[0, 1, 2]

>>x[1:3]

[1, 2]

>>x[5:-1]

[5, 6, 7, 8]

>>x[5:]

[5, 6, 7, 8, 9]


In general `x[len(x):len(x)] = seq` is a stupid way to extend a list,
just use .extend or +=.

Matt

[bruno.desthuilliers@gmail.com]

Re: Please check my understanding.. .

On 1 juil, 21:35, Tobiah <t...@tobiah.or gwrote:

list.append([1,2]) will add the two element list as the next
element of the list.

list.append(obj ) will add obj as the last element of list, whatever
type(obj) is.

list.extend([1,2]) is equivalent to list = list + [1, 2]

Not quite. The second statement rebinds the name list (a very bad name
BTW but anyway...) to a new list object composed of elements of the
list object previously bound to the name list and the elements of the
anonymous list object [1, 2], while the first expression modifies the
original list object in place. The results will compare equal (same
type, same content), but won't be identical (not the same object).

A better definition for list.extend(ite rable) is that it is equivalent
to:

for item in iterable:
list.append(ite m)

The difference is important if list is bound to other names. A couple
examples:

a = [1, 2, 3}
b = a
# b and a points to the same list object
b is a
=True

a.append(4)
print b
=[1, 2, 3, 4]

b.extend([5, 6])
print a
=[1, 2, 3, 4, 5, 6]

a = a + [7, 8]
print b
=[1, 2, 3, 4, 5, 6]

print a
=[1, 2, 3, 4, 5, 6, 7, 8]

a is b
=False

def func1(lst):
lst.extend([9, 10])
print lst

def func2(lst):
lst = lst + [11, 12]
print lst

func1(a)
=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
print a
=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

func2(a)
=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
print a
=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Is that the only difference?

cf above.

From the manual:
>
s.extend(x) | same as s[len(s):len(s)] = x
>
But: (python 2.5.2)
>

>a

[1, 2, 3]

>a[len(a):len(a)] = 4

>
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: can only assign an iterable

And if you try with extend, you'll also have a TypeError:

a.extend(4)
=Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'int' object is not iterable


list.extend expects an iterable, and so does slice assignment.

You want:

a[len(a):len(a)] = [4]

>
Also, what is the difference between list[x:x] and list[x]?

The first expression refers to the *sublist* starting at x and ending
one element before x. Of course, if x == x, then it refers to an empty
list !-)

>>a[3:3]

[]

>>a[1:3]

[2, 3]

>>a[0:2]

[1, 2]

>>a[0:1]

[1]

>>>

The second expression refers to the *element* at index x.

HTH

[Marc 'BlackJack' Rintsch]

Re: Please check my understanding.. .

On Tue, 01 Jul 2008 12:35:01 -0700, Tobiah wrote:

list.append([1,2]) will add the two element list as the next
element of the list.
>
list.extend([1,2]) is equivalent to list = list + [1, 2]
and the result is that each element of the added list
becomes it's own new element in the original list.

It's not 100% equivalent because `list.extend()` mutates the original list
while ``+`` creates a new list object:

In [8]: a = [1, 2, 3]

In [9]: b = a

In [10]: b.extend([4, 5])

In [11]: b
Out[11]: [1, 2, 3, 4, 5]

In [12]: a
Out[12]: [1, 2, 3, 4, 5]

In [13]: b = b + [6, 7]

In [14]: b
Out[14]: [1, 2, 3, 4, 5, 6, 7]

In [15]: a
Out[15]: [1, 2, 3, 4, 5]

Is that the only difference?
>
From the manual:
>
s.extend(x) | same as s[len(s):len(s)] = x
>
But: (python 2.5.2)
>

>>>a

[1, 2, 3]

>>>a[len(a):len(a)] = 4

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: can only assign an iterable

>>>>

Have you tried `extend()` with the same value?

In [15]: a
Out[15]: [1, 2, 3, 4, 5]

In [16]: a.extend(6)
---------------------------------------------------------------------------
<type 'exceptions.Typ eError' Traceback (most recent call last)

/home/bj/<ipython consolein <module>()

<type 'exceptions.Typ eError'>: 'int' object is not iterable

See, both ways need something iterable.

Ciao,
Marc 'BlackJack' Rintsch