Please check my understanding...
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 ** -- http://mail.python.org/mailman/listinfo/python-list
Re: Please check my understanding...
On Jul 1, 12:35 pm, Tobiah [EMAIL PROTECTED] 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. 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Please check my understanding...
On 1 juil, 21:35, Tobiah [EMAIL PROTECTED] wrote: 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 -- http://mail.python.org/mailman/listinfo/python-list
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.TypeError' Traceback (most recent call last) /home/bj/ipython console in module() type 'exceptions.TypeError': 'int' object is not iterable See, both ways need something iterable. Ciao, Marc 'BlackJack' Rintsch -- http://mail.python.org/mailman/listinfo/python-list
