lists - append - unique and sorted

[rhXX]

hi,

can i append a item to a list using criterias:

- UNIQUE - if there already exist don't append

and/or

- SORTED - INSERT in the correct place using some criteria?

tks in advance

 

[Diez B. Roggisch]

hi,

can i append a item to a list using criterias:

- UNIQUE - if there already exist don't append

- SORTED - INSERT in the correct place using some criteria?

Both can be accomplished using the bisect-module. It will give you the
leftmost/rightmost insertion point for a given item and a list, and then
you have to see if it is contained in between (or something along these
lines, I hope you get the gist of it)

Diez

 

[Neil Cerutti]

hi,

can i append a item to a list using criterias:

- UNIQUE - if there already exist don't append

Consult the Python Docs about sets.

and/or

- SORTED - INSERT in the correct place using some criteria?

Consult the Python Docs about the heapq module.

 

[Josiah Carlson]

Consult the Python Docs about the heapq module.

Heaps (as produced by heapq) are not sorted. This will not produce
correct results unless one then pops everything and de-dupes the output.

As Diez has already said, 'use the bisect module' .

- Josiah

 

[rhXX]

On Jun 6, 6:35 pm, Josiah Carlson <[email protected]>
wrote:
ok, tks to all for ur help and comments!!!

 

[Neil Cerutti]

Heaps (as produced by heapq) are not sorted. This will not
produce correct results unless one then pops everything and
de-dupes the output.

i agree that using bisect and inserting manually clearly meets
the stated requirements, while there isn't enough information to
know if a heapq will meet his requirements.

Thanks for the correction.

 

[Dan Bishop]

hi,

can i append a item to a list using criterias:

- UNIQUE - if there already exist don't append

and/or

- SORTED - INSERT in the correct place using some criteria?

tks in advance

If you don't need the list to be sorted until you're done building it,
you can just use:

lst = sorted(set(lst))

 

[Terry Reedy]

| If you don't need the list to be sorted until you're done building it,
| you can just use:
|
| lst = sorted(set(lst))

?? looks same as
lst.sort()