List dilema

Replies to This Discussion

Permalink Reply by Florian Frank on February 13, 2013 at 10:29am

http://www.peterbe.com/plog/uniqifiers-benchmark

Permalink Reply by Miguel Rus on February 13, 2013 at 10:59am

Hi Florian,

That simplifies the list. What i want to do is if a number gets repeated remove both numbers from the list. eg. [1,2,3,3,3] >> [1,2]

 

Permalink Reply by Miguel Rus on February 13, 2013 at 11:52am

it is a bit of a workaround but I managed to do it.

Any other ideas are welcome.

this is what i did:

trial = [1,2,3,4,5,6,7,3,4,5,6,8,9,10]
newTrial = sorted(trial)
print newTrial
list = []
listLength = len(newTrial)

for i in range(0,25):
for j in range(0,listLength):
if i == newTrial[j]:
if j == 0:
if newTrial[j] != newTrial[j+1]:
list.append(newTrial[j])
elif j == listLength-1:
if newTrial[j] != newTrial[j-1]:
list.append(newTrial[j])

else:
if newTrial[j] != newTrial[j+1] and newTrial[j] != newTrial[j-1]:
print newTrial[j]
list.append(newTrial[j])

print newTrial[j]
list.append(newTrial[j])

thanks,

M

Permalink Reply by Anders Holden Deleuran on February 13, 2013 at 2:53pm

You could use Python sets for this, like so:

myList = [1,2,3,4,4,5,5,5,6,7]

myListCullDup = list(set(myList))

Permalink Reply by Anders Holden Deleuran on February 13, 2013 at 3:03pm

Just understood what you're trying to do. How about something like this (untested):

inputList= [1,2,3,4,4,5,5,5,6,7]
outputList = []
for item in inputList:
    if inputList.count(item) == 1:
        outputList.append(item)

Permalink Reply by Anders Holden Deleuran on February 14, 2013 at 5:28am

Just tested the code above and it works as expected. It could probably be optimized if you're running it on very large lists with lots of duplicate values. Also, here's an even simpler version using list comprehension:

inputList = [1,2,3,4,4,5,5,5,6,7]
outputList = [i for i in inputList if inputList.count(i) == 1]

Can't really beat one line of code as far as simplicity goes :)

Permalink Reply by Miguel Rus on February 14, 2013 at 11:22am

Hi Andres,

That is exactly what I needed.

Thanks,

M

Permalink Reply by Anders Holden Deleuran on February 15, 2013 at 2:36am

Cool, glad it helped..

Permalink Reply by David Rutten on February 14, 2013 at 5:13am

Three approaches that spring to mind:

1. Iterate over all items in the collection (except the last one). Then iterate over all subsequent items. If two of these items match then you have a duplicate value, cache this value. Iterate over all items in the collection and remove them if they equal the cache.
2. Iterate over all items in the collection. Iterate over all other items in the collection. If there is no equality, copy the item into a new list.
3. Iterate over all items in the collection (except the last one). Then iterate over all subsequent items. If the subsequent item matches the current item, remove the subsequent item and set a deletion boolean. When you're done iterating over all subsequent items, check the deletion bool and remove the current item if it's true.

(1) in VB:

For i As Int32 = 0 To list.Count - 2

  If (i >= list.Count-1) Then Exit For

  Dim cache As Int32 = list(i)

  Dim delete As Boolean = False

  For j As Int32 = i+1 To list.Count - 1

    If (list(i) = list(j)) Then

      delete = True

      Exit For

    End If

  Next

  If (delete) Then

    For j As Int32 = list.Count - 1 To 0 Step -1

      If (list(j) = cache) Then

        list.RemoveAt(j)

    Next

  End If

Next

(3) in C#:

for (int i = 0; i < list.Count; i++)

{

  bool deleteItem = false;

  for (int j = list.Count-1; j > i; j--)

  {

    if (list[i] = list[j])

    {

      list.RemoveAt(j);

      deleteItem = true;

    }

  }

  if (deleteItem)

    list.RemoveAt(i);

}

There are all kinds of clever tricks you can use with Linq and I'm sure that python has some clever list handling as well. But from a pure no-tricks and no-optimizations point of view the above would be my first approach.

--

David Rutten

david@mcneel.com

Poprad, Slovakia

Permalink Reply by Anton Bakker on December 17, 2013 at 10:20am

Miguel,

I'm looking for the same, do you have a GH module you could share?

Anton