parsing list in python

Question

I have list in python which has following entries

name-1

name-2

name-3

name-4

name-1

name-2

name-3

name-4

name-1

name-2

name-3

name-4

I would like remove name-1 from list except its first appearance -- resultant list should look like

name-1

name-2

name-3

name-4

name-2

name-3

name-4

name-2

name-3

name-4

How to achieve this ?

Just name-1, or all duplicates? Is preserving order important? I'm guessing it is, but the problem is much easier if it isn't. — Marcelo Cantos
– Marcelo Cantos, Commented Apr 14, 2010 at 13:00
the first item should remain as "name-1" - other than that order of remaining items is not important. — webminal.org
– webminal.org, Commented Apr 14, 2010 at 13:16

Jochen Ritzel · Accepted Answer · 2010-04-14 13:06:35Z

3

def remove_but_first( lst, it):
    first = lst.index( it )
    # everything up to the first occurance of it, then the rest of the list without all it
    return lst[:first+1] + [ x for x in lst[first:] if x != it ]

s = [1,2,3,4,1,5,6]
print remove_but_first( s, 1)

answered Apr 14, 2010 at 13:06

Jochen Ritzel

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2 Comments

kame Over a year ago

lst[:first+1] What does this mean?

fortran Over a year ago

a slice from the beginning to the element first+1

Marcelo Cantos · Accepted Answer · 2010-04-14 21:38:10Z

2

Assuming name-1 denotes "the first element":

[names[0]] + [n for n in names[1:] if n != names[0]]

EDIT: If the overall goal is to de-duplicate the entire list, just use set(names).

edited Apr 14, 2010 at 21:38

answered Apr 14, 2010 at 13:03

Marcelo Cantos

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Comments

systempuntoout · Accepted Answer · 2010-04-14 13:43:19Z

2

Based on Marcelo's solution:

[name for cnt,name in enumerate(names) if (name != names[0] or cnt > 0)]

answered Apr 14, 2010 at 13:43

systempuntoout

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Comments

A. Levy · Accepted Answer · 2010-04-14 13:10:18Z

1

Find the index of the first element you wish to remove, then filter the rest of the list. The following works in Python 2.5:

def removeAllButFirst(elem, myList):
    idx = myList.index(elem)
    return myList[0:idx+1] + filter(lambda x: x != elem, myList[idx+1:])

edited Apr 14, 2010 at 13:10

answered Apr 14, 2010 at 13:02

A. Levy

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3 Comments

ChristopheD Over a year ago

First three lines could be idx = myList.index(elem)

ChristopheD Over a year ago

Also myList = myList[0:idx] + ... should probably be myList = myList[0:idx+1] + ... (we need to include the first occurence of 'name1').

A. Levy Over a year ago

Yeah, I just typed in the first thing that came to my head. I couldn't remember the method to find something in a list. I've tested the revised function above, and it does indeed work.

Dave · Accepted Answer · 2010-04-14 13:20:47Z

1

mylist = ['name-1', 'name-2', 'name-3', 'name-4', 'name-1', 'name-2', 'name-3', 'name-4', 'name-1', 'name-2', 'name-3', 'name-4']
newlist = filter(lambda x: x != 'name-1', mylist)
newlist.insert(mylist.index('name-1'), 'name-1')
print newlist
['name-1', 'name-2', 'name-3', 'name-4', 'name-2', 'name-3', 'name-4', 'name-2', 'name-3', 'name-4']

answered Apr 14, 2010 at 13:20

Dave

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