I would switch to doing this in two passes:
First get the length of the longest sublist:
Maximum(List(l, x -> Length(x));
Then get the lists of that length:
Filtered(l, x -> Length(x)=4);
If you aren't familiar with these notations (they are worth learning), you
could write out some loops instead:
maxlen := 0;
for i in [1..Length(l)] do
maxlen := Maximum(maxlen, Length(l[i]));
od;
k := [];
for i in [1..Length(l)] do
if Length(l[i]) = maxlen then
Add(k, l[i]);
fi;
od;
Or, we could loop directly over the members of l:
maxlen := 0;
for i in l do
maxlen := Maximum(maxlen, Length(i));
od;
k := [];
for i in l do
if Length(i) = maxlen then
Add(k, i);
fi;
od;
On 10/10/2014 13:59, "Siddiqua Mazhar (PGR)" <s.maz...@newcastle.ac.uk>
wrote:
>Dear Sir/Madam,
> Here I made changes in my previous email, sorry for inconvinience
>
>If I have a list let say
>a:=[[1,2,3],[4,5],[6,7,8,9],[10,11,12,],[13,14],[15,16,17,18]];
>
>I want to find the list of maximum length in it, or collection of list of
>max length from above
>
>For instance, result:=[[6,7,8,9],[15,16,17,18]]; how can i find this
>result?
>
>here is one algorithm that I tried
>
>>Comparison:=function(a)
>>local i,k;
>>k:=[];
>> for i in [1..(Length(a)-1)] do
>> if Length(a[i])<Length(a[i+1]) then
>> Add(k,a[i+1]);
>> elif not a[i] in k then
>> Add(k,a[i]);
>> fi;
>> od;
>>return k;
>>end;
>
>This program works, however , what if two list are of same length next to
>each other?
>for instance a:=[[1,2,3],[4,5],[6,7],[8,9,10]]?
>
>want this result:=[1,2,3],[8,9,10]
>
>or if a:=[[1,2],[3,4],[5,6]] in this case the result wouldl be the same
>as a because every list are of same length.
>
>Many thanks.
>
>Kind regards
>Siddiqua
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