[Tutor] Looking for duplicates within a list [SOLVED]

[Ken G.]

Jose Amoreira wrote:
> On Friday, June 11, 2010 02:57:34 pm Ken G. wrote:
>   
>> I have been working on this problem for several days and I am not making
>> any progress.  I have a group of 18 number, in ascending order, within a
>> list.  They ranged from 1 to 39.  Some numbers are duplicated as much as
>> three times or as few as none.
>>
>> I started with one list containing the numbers.  For example, they are
>> listed as like below:
>>
>> a = [1, 2, 3, 3, 4]
>>
>> I started off with using a loop:
>>
>>     for j in range (0, 5):
>>     x = a[0] # for example, 1
>>
>> How would I compare '1' with 2, 3, 3, 4?
>>
>> Do I need another duplicated list such as b = a and compare a[0] with
>> either b[0], b[1], b[2], b[3], b[4]?
>>
>> Or do I compare a[0] with a[1], a[2], a[3], a[4]?
>>
>> In any event, if a number is listed more than once, I would like to know
>> how many times, such as 2 or 3 times.  For example, '3' is listed twice
>> within a list.
>>
>> TIA,
>>
>>     
>
> I would do it with a dictionary:
> def reps(lst):
> 	dict = {}
> 	for item in lst:
> 		if item in dict:
> 			dict[item] += 1
> 		else:
> 			dict[item] = 1
> 	return dict
>
> This function returns a dictionary with of the number of times each value in 
> the list is repeated. Even shorter using dict.setdefault:
>
> def reps(lst):
> 	dict={}
> 	for item in lst:
> 		dict[item] = dict.setdefault(item,0) + 1
> 	return dict
>
> For instance, if lst=[1,2,2,2,4,4,5], then reps(lst) returns
> {1: 1, 2: 3, 4: 2, 5: 1}
>
> Using the fact that the list is ordered, one can design a more efficient 
> solution (go through the list; if this item is equal to the previous, then it 
> is repeated, else, it is a new value). But you list is short enough for this 
> direct approach to work.
> Hope this helps. Cheers,
> Jose
>
>   
Thanks.  I will be studying your approach.  Thanks all.

Ken
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