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Please check my understanding...

P: n/a
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 **
Jul 1 '08 #1
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3 Replies

P: n/a
On Jul 1, 12:35*pm, Tobiah <t...@tobiah.orgwrote:
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.com**
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
Jul 1 '08 #2

P: n/a
On 1 juil, 21:35, Tobiah <t...@tobiah.orgwrote:
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(iterable) is that it is equivalent
to:

for item in iterable:
list.append(item)

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
Jul 1 '08 #3

P: n/a
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.TypeError' Traceback (most recent call last)

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

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

See, both ways need something iterable.

Ciao,
Marc 'BlackJack' Rintsch
Jul 1 '08 #4

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