how to remove multiple occurrences of a string within a list?

[bahoo]

Hi,

I have a list like ['0024', 'haha', '0024']
and as output I want ['haha']

If I
myList.remove(' 0024')

then only the first instance of '0024' is removed.

It seems like regular expressions is the rescue, but I couldn't find
the right tool.

Thanks!
bahoo

[mik3l3374]

7stud wrote:

On Apr 3, 3:53 pm, "bahoo" <b83503...@yaho o.comwrote:

target = "0024"
l = ["0024", "haha", "0024"]

for index, val in enumerate(l):
if val==target:
del l[index]

print l

This latter suggestion (with the for loop) seems to be buggy: if there
are multiple items in the list "l" equal to "target", then only the
first one will be removed!

Thanks anyways.

Prove it.

here's something:

>>l = ["0024", "haha", "0024", "0024", "sfs"]
target = "0024"
for index, val in enumerate(l):

.... print "index: " , index , "value: " , val
.... del l[index]
.... print "after del: " , l
....
index: 0 value: 0024
after del: ['haha', '0024', '0024', 'sfs']
index: 1 value: 0024
after del: ['haha', '0024', 'sfs']
index: 2 value: sfs
after del: ['haha', '0024']

>>l

['haha', '0024']

[mik3l3374]

On Apr 4, 2:20 am, "bahoo" <b83503...@yaho o.comwrote:

Hi,

I have a list like ['0024', 'haha', '0024']
and as output I want ['haha']

If I
myList.remove(' 0024')

then only the first instance of '0024' is removed.

It seems like regular expressions is the rescue, but I couldn't find
the right tool.

Thanks!
bahoo

how about this:

>>target = "0024"
l = ["0024", "haha", "0024", "0024", "sfs"]
result = [ item for item in l if item != target]
result

['haha', 'sfs']

[=?ISO-8859-1?Q?Ma=EBl_Benjamin_Mettler?=]

How about:

list(frozenset(['0024', 'haha', '0024']))

mi*******@gmail .com schrieb:

On Apr 4, 2:20 am, "bahoo" <b83503...@yaho o.comwrote:

>Hi,

I have a list like ['0024', 'haha', '0024']
and as output I want ['haha']

If I
myList.remove( '0024')

then only the first instance of '0024' is removed.

It seems like regular expressions is the rescue, but I couldn't find
the right tool.

Thanks!
bahoo

how about this:

>>>target = "0024"
l = ["0024", "haha", "0024", "0024", "sfs"]
result = [ item for item in l if item != target]
result

['haha', 'sfs']

[Bruno Desthuilliers]

bahoo a écrit :

Hi,

I have a list like ['0024', 'haha', '0024']
and as output I want ['haha']

If I
myList.remove(' 0024')

then only the first instance of '0024' is removed.

It seems like regular expressions is the rescue,

Nope. re wouldn't help here - it works on strings, not on lists (you
need to understand that in Python, strings are *not* 'lists of chars').

Use list comps:
mylist = [item for item in mylist if item != '0024']

or filter() :
mylist = filter(lambda item: item != '0024', mylist)

but I couldn't find
the right tool.

May I second Grant Edward's suggestion in previous thread ? You'd really
do yourself a favor by following a couple Python tutorials. I'd say the
one in the official doc, and Dive into Python.

[Goldfish]

I don't think I would use sets at all. They change the semantic
meaning of the original list. What if you have duplicate entries that
you want to keep? Converting to a set and back will strip out the
duplicates of that.

[7stud]

On Apr 4, 3:08 am, Steven D'Aprano <s...@REMOVEME. cybersource.com .au>
wrote:

On Wed, 04 Apr 2007 00:59:23 -0700, 7stud wrote:

On Apr 3, 3:53 pm, "bahoo" <b83503...@yaho o.comwrote:

target = "0024"
l = ["0024", "haha", "0024"]

for index, val in enumerate(l):
if val==target:
del l[index]

print l

This latter suggestion (with the for loop) seems to be buggy: if there
are multiple items in the list "l" equal to "target", then only the
first one will be removed!

Thanks anyways.

Prove it.

Try replacing l = ["0024", "haha", "0024"]
with

l = ["0024", "0024", "haha"]

and re-running the code.

Actually, the description of the bug isn't quite right. The behaviour of
the for loop isn't defined -- sometimes it will work, sometimes it won't,
depending on how many items there are, and which of them are equal to the
target.

Thank you for the explanation.

[Ayaz Ahmed Khan]

"kyosohma" typed:

If you want to get really fancy, you could do a list comprehension
too:

your_list = ["0024","haha"," 0024"]
new_list = [i for i in your_list if i != '0024']

Or, just:

In [1]: l = ["0024","haha"," 0024"]
In [2]: filter(lambda x: x != "0024", l)
Out[2]: ['haha']

--
Ayaz Ahmed Khan

Do what comes naturally now. Seethe and fume and throw a tantrum.

[Steven Bethard]

Ayaz Ahmed Khan wrote:

"kyosohma" typed:

>If you want to get really fancy, you could do a list comprehension
too:

your_list = ["0024","haha"," 0024"]
new_list = [i for i in your_list if i != '0024']

Or, just:

In [1]: l = ["0024","haha"," 0024"]
In [2]: filter(lambda x: x != "0024", l)
Out[2]: ['haha']

Only if you want to make your code harder to read and slower::

$ python -m timeit -s "L = ['0024', 'haha', '0024']"
"[i for i in L if i != '0024']"
1000000 loops, best of 3: 0.679 usec per loop

$ python -m timeit -s "L = ['0024', 'haha', '0024']"
"filter(lam bda i: i != '1024', L)"
1000000 loops, best of 3: 1.38 usec per loop

There really isn't much use for filter() anymore. Even in the one place
I would have expected it to be faster, it's slower::

$ python -m timeit -s "L = ['', 'a', '', 'b']" "filter(Non e, L)"
1000000 loops, best of 3: 0.789 usec per loop

$ python -m timeit -s "L = ['', 'a', '', 'b']" "[i for i in L if i]"
1000000 loops, best of 3: 0.739 usec per loop

STeVe

[Michael J. Fromberger]

In article <11************ **********@n59g 2000hsh.googleg roups.com>,
"bahoo" <b8*******@yaho o.comwrote:

Hi,

I have a list like ['0024', 'haha', '0024']
and as output I want ['haha']

If I
myList.remove(' 0024')

then only the first instance of '0024' is removed.

It seems like regular expressions is the rescue, but I couldn't find
the right tool.

If you know in advance which items are duplicated, then there have been
several simple solutions already proposed. Here's another way to tackle
the problem of removing ANY duplicated item from the list (i.e., any
string that appears 1 time).

def killdups(lst):
"""Filter duplicated elements from the input list, and return the
remaining (unique) items in their original order.
"""
count = {}
for elt in lst:
count[elt] = count.get(elt, 0) + 1
return [elt for elt in lst if count[elt] == 1]

This solution is not particularly tricky, but it has the nice properties
that:

1. It works on lists of any hashable type, not just strings,
2. It preserves the order of the unfiltered items,
3. It makes only two passes over the input list.

Cheers,
-M

--
Michael J. Fromberger | Lecturer, Dept. of Computer Science
http://www.dartmouth.edu/~sting/ | Dartmouth College, Hanover, NH, USA

[Alex Martelli]

Amit Khemka <kh********@gma il.comwrote:

On 3 Apr 2007 11:20:33 -0700, bahoo <b8*******@yaho o.comwrote:

Hi,

I have a list like ['0024', 'haha', '0024']
and as output I want ['haha']

If I
myList.remove(' 0024')

then only the first instance of '0024' is removed.

To remove all items with multiple occurances in myList:

list(set(myList ) - set([x for x in myList if myList.count(x) >1]))

This approach is unfortunately quite inefficient, because each call to
myList.count is O(N), and there are O(N) such calls, so the whole thing
is O(N squared). [also, the inner square brackets are not needed, and
cause more memory to be consumed than omitting them would].
A "we-don't-need-no-stinkin'-one-liners" more relaxed approach:

import collections
d = collections.def aultdict(int)
for x in myList: d[x] += 1
list(x for x in myList if d[x]==1)

yields O(N) performance (give that dict-indexing is about O(1)...).
Collapsing this kind of approach back into a "one-liner" while keeping
the O(N) performance is not easy -- whether this is a fortunate or
unfortunate ocurrence is of course debatable. If we had a "turn
sequence into bag" function somewhere (and it might be worth having it
for other reasons):

def bagit(seq):
import collections
d = collections.def aultdict(int)
for x in seq: d[x] += 1
return d

then one-linerness might be achieved in the "gimme nonduplicated items"
task via a dirty, rotten trick...:

list(x for bag in [bagit(myList)] for x in myList if bag[x] == 1)

....which I would of course never mention in polite company...
Alex