[Tutor] how to understand unhashable type: 'list'
list1
[['61', '34', '61', '34'], ['61', '35', '61', '70', '61'], ['61',
'70', '61', '34'], ['34', '58', '34', '58']]
weight={}
weight{list1[0]}=1
SyntaxError: invalid syntax
weight[list1[0]]=1
Traceback (most recent call last):
File pyshell#292, line 1, in module
weight[list1[0]]=1
TypeError: unhashable type: 'list'
I wonder how to count the occurence of the list of lists.
Thanks, ^_^
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Re: [Tutor] how to understand unhashable type: 'list'
On 2011/11/17 11:59 AM, lina wrote:
list1
[['61', '34', '61', '34'], ['61', '35', '61', '70', '61'], ['61',
'70', '61', '34'], ['34', '58', '34', '58']]
weight={}
weight{list1[0]}=1
SyntaxError: invalid syntax
weight[list1[0]]=1
Traceback (most recent call last):
File pyshell#292, line 1, inmodule
weight[list1[0]]=1
TypeError: unhashable type: 'list'
I wonder how to count the occurence of the list of lists.
Thanks, ^_^
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sum(1 if type(elem) == list else 0 for elem in list1) not work for you
if all you want to do is count how many lists you have in your main list ?
--
Christian Witts
Python Developer
//
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Re: [Tutor] how to understand unhashable type: 'list'
On Thu, Nov 17, 2011 at 6:09 PM, Christian Witts cwi...@compuscan.co.za wrote:
On 2011/11/17 11:59 AM, lina wrote:
list1
[['61', '34', '61', '34'], ['61', '35', '61', '70', '61'], ['61',
'70', '61', '34'], ['34', '58', '34', '58']]
weight={}
weight{list1[0]}=1
SyntaxError: invalid syntax
weight[list1[0]]=1
Traceback (most recent call last):
File pyshell#292, line 1, in module
weight[list1[0]]=1
TypeError: unhashable type: 'list'
I wonder how to count the occurence of the list of lists.
Thanks, ^_^
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sum(1 if type(elem) == list else 0 for elem in list1) not work for you if
all you want to do is count how many lists you have in your main list ?
not count how many sublists in the main list.
wanna count the occurence of the sublists,
I mean, something like
dictionary[list1[1]]=occurence
Traceback (most recent call last):
File pyshell#298, line 1, in module
dictionary[list1[1]]=1
TypeError: unhashable type: 'list'
--
dictionary[list1[1]]=1
Traceback (most recent call last):
File pyshell#298, line 1, in module
dictionary[list1[1]]=1
TypeError: unhashable type: 'list'
Christian Witts
Python Developer
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Re: [Tutor] how to understand unhashable type: 'list'
On 2011/11/17 12:26 PM, lina wrote:
On Thu, Nov 17, 2011 at 6:09 PM, Christian Wittscwi...@compuscan.co.za wrote:
On 2011/11/17 11:59 AM, lina wrote:
list1
[['61', '34', '61', '34'], ['61', '35', '61', '70', '61'], ['61',
'70', '61', '34'], ['34', '58', '34', '58']]
weight={}
weight{list1[0]}=1
SyntaxError: invalid syntax
weight[list1[0]]=1
Traceback (most recent call last):
File pyshell#292, line 1, inmodule
weight[list1[0]]=1
TypeError: unhashable type: 'list'
I wonder how to count the occurence of the list of lists.
Thanks, ^_^
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sum(1 if type(elem) == list else 0 for elem in list1) not work for you if
all you want to do is count how many lists you have in your main list ?
not count how many sublists in the main list.
wanna count the occurence of the sublists,
I mean, something like
dictionary[list1[1]]=occurence
Traceback (most recent call last):
File pyshell#298, line 1, inmodule
dictionary[list1[1]]=1
TypeError: unhashable type: 'list'
dictionary[list1[1]]=1
Traceback (most recent call last):
File pyshell#298, line 1, inmodule
dictionary[list1[1]]=1
TypeError: unhashable type: 'list'
For that you'll need to convert your list to a hash-able type if you
want to use dictionaries for your store, easiest would probably to use
the string representation for it so you could do
list1 = [['61', '34', '61', '34'], ['61', '35', '61', '70', '61'],
['61', '70', '61', '34'], ['34', '58', '34', '58']]
weight = {}
for elem in list1:
... if elem.__repr__() in weight:
... weight[elem.__repr__()] += 1
... else:
... weight[elem.__repr__()] = 1
...
weight
{['34', '58', '34', '58']: 1, ['61', '70', '61', '34']: 1, ['61',
'34', '61
', '34']: 1, ['61', '35', '61', '70', '61']: 1}
or
from collections import defaultdict
list1 = [['61', '34', '61', '34'], ['61', '35', '61', '70', '61'],
['61', '7
0', '61', '34'], ['34', '58', '34', '58']]
weight = defaultdict(int)
for elem in list1:
... weight[elem.__repr__()] += 1
...
weight
defaultdict(type 'int', {['34', '58', '34', '58']: 1, ['61', '70',
'61', '34']: 1, ['61', '34', '61', '34']: 1, ['61', '35', '61',
'70', '61']: 1})
Hope that helps.
--
Christian Witts
Python Developer
//
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Re: [Tutor] how to understand unhashable type: 'list'
lina wrote:
list1
[['61', '34', '61', '34'], ['61', '35', '61', '70', '61'], ['61',
'70', '61', '34'], ['34', '58', '34', '58']]
You have a list of lists.
weight={}
weight{list1[0]}=1
SyntaxError: invalid syntax
In Python, {} are used for dicts, and in Python 3, sets. They aren't
used for anything else. obj{...} is illegal syntax.
weight[list1[0]]=1
Traceback (most recent call last):
File pyshell#292, line 1, in module
weight[list1[0]]=1
TypeError: unhashable type: 'list'
You are trying to store a list as a key inside a dict. This cannot be
done because lists (like all mutable types) can't be hashed.
If your list of lists is small, say, less than a few thousand items, it
would be fast enough to do this:
counts = []
seen = []
for sublist in list1:
if sublist in seen:
# No need to do it again.
continue
seen.append(sublist)
counts.append( (sublist, list1.count(sublist)) )
If I run this code on your list1, I get these counts:
[(['61', '34', '61', '34'], 1), (['61', '35', '61', '70', '61'], 1),
(['61', '70', '61', '34'], 1), (['34', '58', '34', '58'], 1)]
That is, each list is unique.
However, if you have many thousands of items, the above may be too slow.
It is slow because of all the linear searches: sublist in seen
searches seen for the sublist, one item at a time, and
list1.count(sublist) searches list1 for sublist, also one item at a
time. If there are many items, this will be slow.
To speed it up, you can do this:
(1) Keep the seen list sorted, and use binary search instead of linear
search. See the bisect module for help.
(2) Sort list1, and write your own smarter count() function that doesn't
have to travel along the entire list.
All that involves writing a lot of code though. Probably much faster
would be to make a temporary copy of list1, with the inner lists
converted to tuples:
list2 = [tuple(l) for l in list1]
counts = {}
for t in list2:
counts[t] = 1 + counts.get(t, 0)
for t, n in counts.items():
print list(t), n
--
Steven
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Re: [Tutor] how to understand unhashable type: 'list'
snip
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sum(1 if type(elem) == list else 0 for elem in list1) not work for you if
all you want to do is count how many lists you have in your main list ?
not count how many sublists in the main list.
wanna count the occurence of the sublists,
I mean, something like
dictionary[list1[1]]=occurence
Traceback (most recent call last):
File pyshell#298, line 1, in module
dictionary[list1[1]]=1
TypeError: unhashable type: 'list'
dictionary[list1[1]]=1
Traceback (most recent call last):
File pyshell#298, line 1, in module
dictionary[list1[1]]=1
TypeError: unhashable type: 'list'
For that you'll need to convert your list to a hash-able type if you want to
use dictionaries for your store, easiest would probably to use the string
representation for it so you could do
list1 = [['61', '34', '61', '34'], ['61', '35', '61', '70', '61'],
['61', '70', '61', '34'], ['34', '58', '34', '58']]
weight = {}
for elem in list1:
... if elem.__repr__() in weight:
This is cool.
May I ask which role the __repr__ plays here?
... weight[elem.__repr__()] += 1
... else:
... weight[elem.__repr__()] = 1
...
weight
{['34', '58', '34', '58']: 1, ['61', '70', '61', '34']: 1, ['61', '34',
'61
', '34']: 1, ['61', '35', '61', '70', '61']: 1}
or
from collections import defaultdict
list1 = [['61', '34', '61', '34'], ['61', '35', '61', '70', '61'],
['61', '7
0', '61', '34'], ['34', '58', '34', '58']]
weight = defaultdict(int)
for elem in list1:
... weight[elem.__repr__()] += 1
...
weight
defaultdict(type 'int', {['34', '58', '34', '58']: 1, ['61', '70',
'61', '34']: 1, ['61', '34', '61', '34']: 1, ['61', '35', '61', '70',
'61']: 1})
Hope that helps.
--
Christian Witts
Python Developer
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Re: [Tutor] how to understand unhashable type: 'list'
On Thu, Nov 17, 2011 at 8:17 PM, Steven D'Aprano st...@pearwood.info wrote:
lina wrote:
list1
[['61', '34', '61', '34'], ['61', '35', '61', '70', '61'], ['61',
'70', '61', '34'], ['34', '58', '34', '58']]
You have a list of lists.
weight={}
weight{list1[0]}=1
SyntaxError: invalid syntax
In Python, {} are used for dicts, and in Python 3, sets. They aren't used
for anything else. obj{...} is illegal syntax.
weight[list1[0]]=1
Traceback (most recent call last):
File pyshell#292, line 1, in module
weight[list1[0]]=1
TypeError: unhashable type: 'list'
You are trying to store a list as a key inside a dict. This cannot be done
because lists (like all mutable types) can't be hashed.
I checked online dictionary, still confused about hashed. is it equal
to mix together or mess together?
If your list of lists is small, say, less than a few thousand items, it
would be fast enough to do this:
counts = []
seen = []
for sublist in list1:
if sublist in seen:
# No need to do it again.
continue
seen.append(sublist)
counts.append( (sublist, list1.count(sublist)) )
Thanks, can I print the counts based on its value?
If I run this code on your list1, I get these counts:
[(['61', '34', '61', '34'], 1), (['61', '35', '61', '70', '61'], 1), (['61',
'70', '61', '34'], 1), (['34', '58', '34', '58'], 1)]
That is, each list is unique.
However, if you have many thousands of items, the above may be too slow. It
is slow because of all the linear searches: sublist in seen searches seen
for the sublist, one item at a time, and list1.count(sublist) searches
list1 for sublist, also one item at a time. If there are many items, this
will be slow.
To speed it up, you can do this:
(1) Keep the seen list sorted, and use binary search instead of linear
search. See the bisect module for help.
(2) Sort list1, and write your own smarter count() function that doesn't
have to travel along the entire list.
All that involves writing a lot of code though. Probably much faster would
be to make a temporary copy of list1, with the inner lists converted to
tuples:
list2 = [tuple(l) for l in list1]
counts = {}
for t in list2:
counts[t] = 1 + counts.get(t, 0)
for t, n in counts.items():
print list(t), n
--
Steven
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Re: [Tutor] how to understand unhashable type: 'list'
lina wrote: May I ask which role the __repr__ plays here? ... weight[elem.__repr__()] += 1 ... else: ... weight[elem.__repr__()] = 1 ... You should never call elem.__repr__(), any more than you would call elem.__len__() or elem.__str__(). (Well, technically there *are* rare uses for calling such double underscore special methods directly, but this is not one of them!) Instead, you should call repr(elem), len(elem), str(elem). The purpose here is to convert the sublist, elem, which is unhashable, into something which is hashable. Strings are hashable, and so if you use repr() to convert it into a string, you can use it in a dictionary. I don't see any advantage to using repr() instead of str(). Given the data you are working with, there is no difference: py str([['1', '2'], ['33', '44']]) [['1', '2'], ['33', '44']] py repr([['1', '2'], ['33', '44']]) [['1', '2'], ['33', '44']] In my opinion, turning objects into strings unnecessarily is not good practice. Better to use a tuple than a string. -- Steven ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] how to understand unhashable type: 'list'
On 2011/11/17 03:56 PM, lina wrote:
snip
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sum(1 if type(elem) == list else 0 for elem in list1) not work for you if
all you want to do is count how many lists you have in your main list ?
not count how many sublists in the main list.
wanna count the occurence of the sublists,
I mean, something like
dictionary[list1[1]]=occurence
Traceback (most recent call last):
File pyshell#298, line 1, inmodule
dictionary[list1[1]]=1
TypeError: unhashable type: 'list'
dictionary[list1[1]]=1
Traceback (most recent call last):
File pyshell#298, line 1, inmodule
dictionary[list1[1]]=1
TypeError: unhashable type: 'list'
For that you'll need to convert your list to a hash-able type if you want to
use dictionaries for your store, easiest would probably to use the string
representation for it so you could do
list1 = [['61', '34', '61', '34'], ['61', '35', '61', '70', '61'],
['61', '70', '61', '34'], ['34', '58', '34', '58']]
weight = {}
for elem in list1:
... if elem.__repr__() in weight:
This is cool.
May I ask which role the __repr__ plays here?
__repr__ is the string representation of a Python object [1] so I'm
using it to create something that is hash-able to be used as the
dictionary key value.
[1] http://docs.python.org/library/functions.html#repr
--
Christian Witts
Python Developer
//
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Re: [Tutor] how to understand unhashable type: 'list'
Hi Lina On 17 November 2011 14:04, lina lina.lastn...@gmail.com wrote: Traceback (most recent call last): File pyshell#292, line 1, in module weight[list1[0]]=1 TypeError: unhashable type: 'list' You are trying to store a list as a key inside a dict. This cannot be done because lists (like all mutable types) can't be hashed. I checked online dictionary, still confused about hashed. is it equal to mix together or mess together? No you need to think programming/computer science where hash/hashing has a different meaning. See wikipedia: http://en.wikipedia.org/wiki/Hash_function http://en.wikipedia.org/wiki/Hash_table Also see the description for hashable in the Python glossary: http://docs.python.org/glossary.html#hashable Basically hashing is a way to convert something (often a string) into an identifying number (not neccesarily but usually unique) in order to use this number as a token for the original thing. Aside, you can infer that that because the dict code complains about hashing implies that dicts themselves are essentially implemented as hash tables... ;) HTH, Walter ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] how to understand unhashable type: 'list'
Thanks for all. Everything new to me is so amazing. Well, still a remaining question, how to sort based on the value of the key. Actually I googled hours ago, wanna see more ways of doing it. Best regards, On 17 November 2011 14:04, lina lina.lastn...@gmail.com wrote: Traceback (most recent call last): File pyshell#292, line 1, in module weight[list1[0]]=1 TypeError: unhashable type: 'list' You are trying to store a list as a key inside a dict. This cannot be done because lists (like all mutable types) can't be hashed. I checked online dictionary, still confused about hashed. is it equal to mix together or mess together? No you need to think programming/computer science where hash/hashing has a different meaning. See wikipedia: http://en.wikipedia.org/wiki/Hash_function http://en.wikipedia.org/wiki/Hash_table Also see the description for hashable in the Python glossary: http://docs.python.org/glossary.html#hashable Basically hashing is a way to convert something (often a string) into an identifying number (not neccesarily but usually unique) in order to use this number as a token for the original thing. Aside, you can infer that that because the dict code complains about hashing implies that dicts themselves are essentially implemented as hash tables... ;) HTH, Walter ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] how to understand unhashable type: 'list'
lina wrote: You are trying to store a list as a key inside a dict. This cannot be done because lists (like all mutable types) can't be hashed. I checked online dictionary, still confused about hashed. is it equal to mix together or mess together? Almost. In ordinary English, a hash is a mixed up mess. For example, hash browns are fried mashed potato. In computing, a hash got its name by analogy with mixed up. A hash is short for hash table. An ordinary table might look something like this: Position Value 1 Fred 2 Barney 3 George 4 Betty 5 Mary 6 *unused* 7 *unused* 8 *unused* 9 *unused* To find whether Mary is in the table, you have to start at position 1, and inspect each value to see whether it equals Mary: for position 1 to 9: if value at position == Mary: print FOUND If there are many items, this will be slow. But here's a faster way, using a hash table: Position Value 1 *unused* 2 Barney 3 *unused* 4 Betty 5 *unused* 6 *unused* 7 Fred 8 Mary 9 George Take the string Mary, and mix it up into a hash value between 1 and 9. In this case, you will get 8. Look in position 8, and you will find Mary. Take the string Susan and mix it up into a hash value. In this case, you might get 3. Since position 3 is actually unused, you know that Susan is not in the hash table. All the hard work is done in the mixing of the string. This is called a hash function, and Python already has one: py hash(Mary) -1963774307 py hash(Susan) 92926151 If you change just one letter of the string, the hash can change a lot: py hash(abcdef) -33722981 py hash(bbcdef) -508939398 But lists are not hashable, because they are mutable (can be changed): py hash([1, 2, 3]) Traceback (most recent call last): File stdin, line 1, in module TypeError: unhashable type: 'list' Python dictionaries are hash tables. You could have a million items in a dict, and to check whether Mary is one of them with a list would require you to check all one million items. But with a dict, Python hashes the string to get a number, then reduces that number to fit within the range of positions in the dict, then looks immediately in the right spot to find Mary. -- Steven ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] how to understand unhashable type: 'list'
On Thu, Nov 17, 2011 at 10:44 PM, Steven D'Aprano st...@pearwood.info wrote: lina wrote: You are trying to store a list as a key inside a dict. This cannot be done because lists (like all mutable types) can't be hashed. I checked online dictionary, still confused about hashed. is it equal to mix together or mess together? Almost. In ordinary English, a hash is a mixed up mess. For example, hash browns are fried mashed potato. In computing, a hash got its name by analogy with mixed up. A hash is short for hash table. An ordinary table might look something like this: Position Value 1 Fred 2 Barney 3 George 4 Betty 5 Mary 6 *unused* 7 *unused* 8 *unused* 9 *unused* To find whether Mary is in the table, you have to start at position 1, and inspect each value to see whether it equals Mary: for position 1 to 9: if value at position == Mary: print FOUND If there are many items, this will be slow. But here's a faster way, using a hash table: Position Value 1 *unused* 2 Barney 3 *unused* 4 Betty 5 *unused* 6 *unused* 7 Fred 8 Mary 9 George Take the string Mary, and mix it up into a hash value between 1 and 9. In this case, you will get 8. Look in position 8, and you will find Mary. Take the string Susan and mix it up into a hash value. In this case, you might get 3. Since position 3 is actually unused, you know that Susan is not in the hash table. All the hard work is done in the mixing of the string. This is called a hash function, and Python already has one: py hash(Mary) -1963774307 py hash(Susan) 92926151 If you change just one letter of the string, the hash can change a lot: py hash(abcdef) -33722981 py hash(bbcdef) -508939398 But lists are not hashable, because they are mutable (can be changed): py hash([1, 2, 3]) Traceback (most recent call last): File stdin, line 1, in module TypeError: unhashable type: 'list' Python dictionaries are hash tables. You could have a million items in a dict, and to check whether Mary is one of them with a list would require you to check all one million items. But with a dict, Python hashes the string to get a number, then reduces that number to fit within the range of positions in the dict, then looks immediately in the right spot to find Mary. Really appreciated for your detailed explaination. Right now I wanna check which are not hash-able, for strings, set, and tuple (which I don't understand). Seems string is hash-able. Just another quick Q: basket ['apple', 'pineapple', 'apple', 'pear'] hash(basket[1]) -8771120720059735049 hash(basket[2]) 6709291584508918021 print(id(basket)) 29215848 print(id(basket[1])) 27389096 What are those numbers means? Sorry, -- Steven ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] how to understand unhashable type: 'list'
lina wrote: Right now I wanna check which are not hash-able, for strings, set, and tuple (which I don't understand). Mutable objects are those that can be changed in place: lists, sets, dicts are all mutable, because you can change them: mylist = [1, 2, 3] mylist[1] = 200 print mylist [1, 200, 3 Mutable objects are not hashable. Immutable objects are those that cannot be changed in place: you have to create a new object. Tuples, frozensets, strings, ints, floats are all immutable. They are hashable. Seems string is hash-able. Just another quick Q: basket ['apple', 'pineapple', 'apple', 'pear'] hash(basket[1]) -8771120720059735049 hash(basket[2]) 6709291584508918021 The number doesn't mean anything. It's just a number. It gets used by dictionaries for fast searching, but that's all. This might help you understand. Here is a very simple hash function that takes a string and mixes it up to make a number. It isn't very good, it is a terrible hash function, but it is simple! The Python one is better, but this will give you an idea of how they work: def my_hash(string): num = 102345 # some random number I just made up for c in string: num = (num*17 + ord(c)*9) % 3200 return num Designing a good hash function is very hard. print(id(basket)) 29215848 print(id(basket[1])) 27389096 The id() of an object is just an identity number. Do you have a passport or driver's licence? Or a social security number? That is like your ID. It identifies you. Nobody else can have the same ID at the same time. In CPython, IDs are large numbers, and they are re-usable. If you delete an object, another object can get the same ID later on. In Jython and IronPython, IDs start counting at 1 and increase with every object created. Used IDs are never reused. IDs aren't important. In 15 years of using Python, I don't think I have needed to care about the ID of an object once. -- Steven ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] how to understand unhashable type: 'list'
Steven D'Aprano wrote: Immutable objects are those that cannot be changed in place: you have to create a new object. Tuples, frozensets, strings, ints, floats are all immutable. They are hashable. Tuples are special: there are hashable and unhashable tuples. To be hashable all items in a tuple have to be hashable: a = 1, 2 hash(a) 3713081631934410656 b = 1, [] hash(b) Traceback (most recent call last): File stdin, line 1, in module TypeError: unhashable type: 'list' Instances of custom classes are special, too; by default they are hashable and mutable. That is possible without messing things up too much because instead of the instance's data they use its id() to calculate the hash value (and object equality). This is obvious in older Python versions like Python 2.6.4 (r264:75706, Dec 7 2009, 18:43:55) [GCC 4.4.1] on linux2 Type help, copyright, credits or license for more information. class A: pass ... a = A() id(a) 140312106243352 hash(a) 140312106243352 id(a) == hash(a) True but the principle also applies in 2.7 and 3.1 and above. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] how to understand unhashable type: 'list'
On 17/11/11 15:30, lina wrote: tuple (which I don't understand). You mean you don't understand the term 'tuple'? Basically its just a collection of data. You can get a triple - 3 items, a quadruple - 4 items, quintiple - 5 items etc The generic term for a collection of N items is a tuple. triple, quadruple, etc are all specific types of tuple. It's a math thing... In python a tuple is just lie a list except you can't change it - ie. it is immutable. This makes it perfect for using as a key in a dictionary when you'd like to use a list... As an aside some folks like to distinguish between tuples and lists by suggesting that tuples can hold collections of different types of data: t = (1,'two', 3.0) whereas lists should only hold items of the same type. L = [1,2,3] But that's a semantic nicety, Python doesn't make any such distinction. And for dictionary keys (1,2,3) is the only way to do it... -- Alan G Author of the Learn to Program web site http://www.alan-g.me.uk/ ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
