Building unique comma-delimited list?

[Roy Smith]

I've got a silly little problem that I'm solving in C++, but I got to
thinking about how much easier it would be in Python. Here's the
problem:

You've got a list of words (actually, they're found by searching a
data structure on the fly, but for now let's assume you've got them as
a list). You need to create a comma-delimited list of these words.
There might be duplicates in the original list, which you want to
eliminate in the final list. You don't care what order they're in,
except that there is a distinguised word which must come first if it
appears at all.

Some examples ("foo" is the distinguised word):

["foo"] => "foo"
["foo", "bar"] => "foo, bar"
["bar", "foo"] => "foo, bar"
["bar", "foo", "foo", "baz", "bar"] => "foo, bar, baz" or "foo, baz, bar"

The best I've come up with is the following. Can anybody think of a
simplier way?

--------------------
words = ["foo", "bar", "baz", "foo", "bar", "foo", "baz"]

# Eliminate the duplicates; probably use set() in Python 2.4
d = dict()
for w in words:
d[w] = w

if d.has_key ("foo"):
newWords = ["foo"]
del (d["foo"])
else:
newWords = []

for w in d.keys():
newWords.append (w)

s = ', '.join (newWords)
print s
--------------------

[Berthold Höllmann]

ro*@panix.com (Roy Smith) writes:

I've got a silly little problem that I'm solving in C++, but I got to
thinking about how much easier it would be in Python. Here's the
problem:

You've got a list of words (actually, they're found by searching a
data structure on the fly, but for now let's assume you've got them as
a list). You need to create a comma-delimited list of these words.
There might be duplicates in the original list, which you want to
eliminate in the final list. You don't care what order they're in,
except that there is a distinguised word which must come first if it
appears at all.

Some examples ("foo" is the distinguised word):

["foo"] => "foo"
["foo", "bar"] => "foo, bar"
["bar", "foo"] => "foo, bar"
["bar", "foo", "foo", "baz", "bar"] => "foo, bar, baz" or "foo, baz, bar"

The best I've come up with is the following. Can anybody think of a
simplier way?

....

How about:

..>>> words = ["foo", "bar", "baz", "foo", "bar", "foo", "baz"]
..>>> ', '.join(dict( ( (w,w) for w in words ) ).keys())
'baz, foo, bar'
..>>> words = ["foo",]
..>>> ', '.join(dict( ( (w,w) for w in words ) ).keys())
'foo'

or with Python 2.3 or higher:

..>>> import sets
..>>> words = ["foo", "bar", "baz", "foo", "bar", "foo", "baz"]
..>>> ', '.join(sets.Set (words))
'baz, foo, bar'
..>>> words = ["foo",]
..>>> ', '.join(sets.Set (words))
'foo'
Kind regards
Berthold
--
be******@xn--hllmanns-n4a.de / <http://höllmanns.de/>
bh***@web.de / <http://starship.python .net/crew/bhoel/>

[Mark McEahern]

Roy Smith wrote:

You've got a list of words (actually, they're found by searching a
data structure on the fly, but for now let's assume you've got them as
a list). You need to create a comma-delimited list of these words.
There might be duplicates in the original list, which you want to
eliminate in the final list. You don't care what order they're in,
except that there is a distinguised word which must come first if it
appears at all.

Some examples ("foo" is the distinguised word):

["foo"] => "foo"
["foo", "bar"] => "foo, bar"
["bar", "foo"] => "foo, bar"
["bar", "foo", "foo", "baz", "bar"] => "foo, bar, baz" or "foo, baz, bar"

The best I've come up with is the following. Can anybody think of a
simplier way?

Who knows whether this is "simpler", but it does demonstrate that you
can customize the sort of a list:

#!/usr/bin/env python

def makesorter(firs t):
"""Return a sort function that sorts first to the top."""
def sorter(x, y):
if x == first:
return -1
elif y == first:
return 1
else:
return 0
return sorter

words = ["foo", "bar", "baz", "foo", "bar", "foo", "baz"]
first = 'foo'
sorter = makesorter(firs t)
unique = {}
for word in words:
unique[word] = word
keys = unique.keys()
keys.sort(sorte r)
print ', '.join(keys)

[Scott David Daniels]

How about (for 2.4 or 2.3 using "from collections import Set as set":

def combine(source, special='foo'):
parts = set(source)
if special in parts:
return ', '.join([special] + list(parts - set([special])))
return ', '.join(parts)

--Scott David Daniels
Sc***********@A cm.Org

[Duncan Booth]

Roy Smith wrote:

The best I've come up with is the following. Can anybody think of a
simplier way?

--------------------
words = ["foo", "bar", "baz", "foo", "bar", "foo", "baz"]

# Eliminate the duplicates; probably use set() in Python 2.4
d = dict()
for w in words:
d[w] = w

if d.has_key ("foo"):
newWords = ["foo"]
del (d["foo"])
else:
newWords = []

for w in d.keys():
newWords.append (w)

s = ', '.join (newWords)
print s
--------------------

You need to make the dictionary and list types work harder for you. They
have a variety of methods you might find useful.

words = ["foo", "bar", "baz", "foo", "bar", "foo", "baz"]
distinguished = ["foo"]
d = dict.fromkeys(w ords, True)
newwords = [ w for w in distinguished if d.pop(w, False) ]
newwords.extend (d.keys())
newwords ['foo', 'baz', 'bar']

[Bengt Richter]

On 5 Jan 2005 17:05:40 -0500, ro*@panix.com (Roy Smith) wrote:

I've got a silly little problem that I'm solving in C++, but I got to
thinking about how much easier it would be in Python. Here's the
problem:

You've got a list of words (actually, they're found by searching a
data structure on the fly, but for now let's assume you've got them as
a list). You need to create a comma-delimited list of these words.
There might be duplicates in the original list, which you want to
eliminate in the final list. You don't care what order they're in,
except that there is a distinguised word which must come first if it
appears at all.

Some examples ("foo" is the distinguised word):

["foo"] => "foo"
["foo", "bar"] => "foo, bar"
["bar", "foo"] => "foo, bar"
["bar", "foo", "foo", "baz", "bar"] => "foo, bar, baz" or "foo, baz, bar"

The best I've come up with is the following. Can anybody think of a
simplier way?
(Not tested beyond what you see ;-)
python 2.4:

words = ["foo", "bar", "baz", "foo", "bar", "foo", "baz"]
w2 = list(t[1] for t in sorted((w!='foo ', w) for w in set(words)))
w2 ['foo', 'bar', 'baz']

Gets you a sort in the bargain ;-)
--------------------
words = ["foo", "bar", "baz", "foo", "bar", "foo", "baz"]

# Eliminate the duplicates; probably use set() in Python 2.4 Yup, but 2.3 can be a one-liner too:
words = ["foo", "bar", "baz", "foo", "bar", "foo", "baz"]
w2 = ('foo' in words and ['foo'] or []) + [w for w in dict(zip(words, words)) if w!='foo']
w2

['foo', 'baz', 'bar']

Not sorted, but foo is out front.
d = dict()
for w in words:
d[w] = w

if d.has_key ("foo"):
newWords = ["foo"]
del (d["foo"])
else:
newWords = []

for w in d.keys():
newWords.append (w)

s = ', '.join (newWords)
print s
--------------------

Regards,
Bengt Richter