[Tutor] how to understand unhashable type: 'list'

[lina]

On Thu, Nov 17, 2011 at 8:17 PM, Steven D'Aprano <steve at 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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