How to find the longst element list of lists?
I think, there should be an easier way then this:
s1 = ["q", "e", "d"]
s2 = ["a", "b"]
s3 = ["a", "b", "c", "d"]
if len(s1) >= len(s2) and len(s1) >= len(s3):
sx1=s1 ## s1 ist längster
if len(s2) >= len(s3):
sx2=s2
sx3=s3
else:
sx2=s3
sx3=s2
if len(s2) >= len(s3) and len(s2) >= len(s1):
sx1=s2 ## s2 ist längster
if len(s3) >= len(s1):
sx2=s3
sx3=s1
else:
sx2=s1
sx3=s3
if len(s3) >= len(s1) and len(s3) >= len(s2):
sx1=s3 ## s3 ist längster
if len(s1) >= len(s2):
sx2=s1
sx3=s2
else:
sx2=s2
sx3=s1
After, the list ist sorted:
sx1 = ["a", "b", "c", "d"]
sx2 = ["q", "e", "d"]
sx3 = ["a", "b"] 16 2604
Michael M. kirjoitti:
How to find the longst element list of lists?
I think, there should be an easier way then this:
s1 = ["q", "e", "d"]
s2 = ["a", "b"]
s3 = ["a", "b", "c", "d"]
<snip>
After, the list ist sorted:
sx1 = ["a", "b", "c", "d"]
sx2 = ["q", "e", "d"]
sx3 = ["a", "b"]
s1 = ["q", "e", "d"]
s2 = ["a", "b"]
s3 = ["a", "b", "c", "d"]
ss = ((len(s1), s1), (len(s2), s2), (len(s3), s3))
sx = [y for (x, y) in sorted(ss)[::-1]]
print sx
sx1, sx2, sx3 = sx
print sx1, sx2, sx3
Cheers,
Jussi
Michael M. a écrit :
How to find the longst element list of lists?
For what definition of "find" ? You want the lenght of the longest
sublist, it's index, or a reference to it ?
I think, there should be an easier way then this:
s1 = ["q", "e", "d"]
s2 = ["a", "b"]
s3 = ["a", "b", "c", "d"]
Err... this makes three distinct lists, not a list of lists.
if len(s1) >= len(s2) and len(s1) >= len(s3):
sx1=s1 ## s1 ist längster
if len(s2) >= len(s3):
sx2=s2
sx3=s3
else:
sx2=s3
sx3=s2
(snip repeated code)
Looks like it would be time to learn how to factor out repetitions...
After, the list ist sorted:
sx1 = ["a", "b", "c", "d"]
sx2 = ["q", "e", "d"]
sx3 = ["a", "b"]
This is still not a list of lists. Now for the answer, sorted() is your
friend:
print sorted([s1, s2, s3], key=list.__len_ _, reverse=True)
=[['a', 'b', 'c', 'd'], ['q', 'e', 'd'], ['a', 'b']]
# Or if you really want sx1, sx2 and sx3:
sx1, sx2, sx3 = sorted([s1, s2, s3], key=list.__len_ _, reverse=True)
Is that easier enough ?-)
On 1/7/07, Michael M. <mi*****@mustun .chwrote:
How to find the longst element list of lists?
s1 = ["q", "e", "d"]
s2 = ["a", "b"]
s3 = ["a", "b", "c", "d"]
s = [s1, s2, s3]
s.sort(key=len, reverse=True)
print s[0] is s3
print s[1] is s1
print s[2] is s2
sx1, sx2, sx3 = s
print 'sx1:', sx1
print 'sx2:', sx2
print 'sx3:', sx3
--
Felipe.
Michael M. schrieb:
How to find the longst element list of lists?
I think, there should be an easier way then this:
s1 = ["q", "e", "d"]
s2 = ["a", "b"]
s3 = ["a", "b", "c", "d"]
if len(s1) >= len(s2) and len(s1) >= len(s3):
sx1=s1 ## s1 ist längster
if len(s2) >= len(s3):
sx2=s2
sx3=s3
else:
sx2=s3
sx3=s2
if len(s2) >= len(s3) and len(s2) >= len(s1):
sx1=s2 ## s2 ist längster
if len(s3) >= len(s1):
sx2=s3
sx3=s1
else:
sx2=s1
sx3=s3
if len(s3) >= len(s1) and len(s3) >= len(s2):
sx1=s3 ## s3 ist längster
if len(s1) >= len(s2):
sx2=s1
sx3=s2
else:
sx2=s2
sx3=s1
After, the list ist sorted:
sx1 = ["a", "b", "c", "d"]
sx2 = ["q", "e", "d"]
sx3 = ["a", "b"]
I don't really get that. You have three lists, you want to sort them
after their length. You should put them into one list.
I think you should rather implement this as:
>>list = [a1, s2, s3] list.sort(lam bda x,y: cmp(len(y), len(x))) list
[['a', 'b', 'c', 'd'], ['q', 'e', 'd'], ['a', 'b']]
Thomas
>
Err... this makes three distinct lists, not a list of lists.
Sure. Logically spoken. Not in Python code. Or a number of lists.
Sure not [[ bla... ] [bla.]] etc.
Michael M. schrieb:
>Err... this makes three distinct lists, not a list of lists.
Sure. Logically spoken. Not in Python code. Or a number of lists.
Sure not [[ bla... ] [bla.]] etc.
???????
Thomas
Bruno Desthuilliers wrote:
>
Err... this makes three distinct lists, not a list of lists.
Sure. Logically spoken. Not in Python code. Or a number of lists.
Sure not [[ bla... ] [bla.]] etc.
On Sun, 07 Jan 2007 22:23:22 +0100,
"Michael M." <mi*****@mustun .chwrote:
How to find the longst element list of lists?
I think, there should be an easier way then this:
s1 = ["q", "e", "d"]
s2 = ["a", "b"]
s3 = ["a", "b", "c", "d"]
[ snip ]
One more thing to think about : if your list of lists grows (i.e., if
you end up with thousands of lists instead of just three), then sorting
may not be the way to go. Assuming that list_of_lists is your list of
lists, then something like this:
longest_list, longest_length = list_of_lists[ 0 ], len( longest_list )
for a_list in list_of_lists[ 1 : ]:
a_length = len( a_list )
if a_length longest_length:
longest_list, longest_length = a_list, a_length
will run faster than sorting the list just to pick off one element (O(n)
vs. O(n log n) for all of you Big-Oh notation fans out there; you know
who you are!).
Regards,
Dan
--
Dan Sommers
<http://www.tombstoneze ro.net/dan/>
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