lreplace replaces or deletes elements in a list, and can also prepend elements to a list.
lreplace returns a new list formed by replacing one or more elements of list with the element arguments. first gives the zero-based index in list of the first element to be replaced. If first is less than zero then it is interpreted as zero. first must correspond to the index of an existing element in list. Last gives the index in list of the last element to be replaced. If last is less than first then no elements are deleted; the new elements are simply inserted before first. first or last may be end (or any abbreviation of it), which denotes the last element of the list.
Each element is added to the list in place of those that are deleted. If element arguments are not specified, then the elements between first and last are simply deleted.
% lreplace {a b c} 0 0 @
@ b c
% lreplace {a b c} 1 1
a c
% lreplace {a b c} 1 1 hello there
a hello there cIf lreplace is handed an unshared list, it will grab the opportunity to modify the list in place, avoiding the processing overhead of making the copy, and the additional memory usage of the copy. The problem with the following example is that $l is shared:
set l [lreplace $l $j $j $newvalue]
The value of $l is bound to $l in the caller of lreplace, making it shared, so lreplace must copy the value before it can modify it. To unshare the value, set $l to the empty string after obtaining its value, and before lreplace is actually invoked, i.e., sometime later on the command line:
set l [lreplace $l[set l {}] $j $j $newvalue]Tcl substitutes the value $l, and then appends the empty string to that value, except that it doesn't, because appending the empty string is a no-op. In the meantime, thanks to set, $l is no longer bound to that value, so that value is unshared. When lreplace is finally called, it the value is no longer shared. unset l would also work, but would be slower because of the additional overhead of unsetting a variable.
In older versions of Tcl, before appending the empty string was a no-op, K was used:
set l [lreplace [K $l [set l {}]] $j $j $newvalue]Set also Shuffle a list and lset
The following procedure accomplishes the same thing:
proc lipreplace {name args} {
upvar 1 $name list
set list [lreplace $list[set list {}] {*}$args]
}% set A [list 1 2 3] 1 2 3 % lipreplace A 1 end c d 1 c d
It's more expensive to replace items toward the front of the list. If lreplace causes the size of the list to change, all remaining items in the list must be shifted up or down to accomodate the new size. Therefore, when iterating through a list of lists and deleting items, designing the process so that items toward the end of the list are replaced first can have a big speed benefit. For example, the list could be iterated from back to front:
for {set i [expr {[llength $list]-1}]} {$i >= 0} {incr i -1} {
set list [lreplace $list[set list {}] $i $i]
}
AMG: While lreplace can be used to prepend elements to the beginning of a list, it cannot be used to append elements to the end of a list.
lreplace {0 1 2} 0 -1 x y z
→ x y z 0 1 2
lreplace {0 1 2} end -1 x y z
→ 0 1 x y z 2
lreplace {0 1 2} end+1 -1 x y z
→ error: list doesn't contain element end+1The documentation says "if last is less than first, then any specified elements will be inserted into the list at the point specified by first with no elements being deleted." Like lindex, it refers to string index for an explanation of the index notation and interpretation, so end is the last element, which is 2 in this case.
Clearly, the insertion takes place just before the first element, even when first is end-relative. This is inconsistent with linsert, which (despite the documentation) inserts just after the indexth element when index is end'-relative.
It is also inconsistent that lreplace accepts indexes before the beginning of the list but rejects indexes after the end of the list. linsert accepts out-of-bounds indexes in either direction.
By the way, to prepend and append individual elements or entire lists to a list, several combinations of concat, list, and {*} can be used. Or if in-place modification is desired, use lappend with or without {*}, or use lset one element at a time. For example, all of the following combinations return x y 0 1:
set elem1 0; set elem2 1
set list1 {x y}; set list2 {0 1}| Combine lists and elements using concat and list | concat $list1 [list $elem1] [list $elem2] |
| Combine lists using concat | concat $list1 $list2 |
| Combine lists and elements using list and {*} | list {*}$list1 $elem1 $elem2 |
| Combine lists using list and {*} | list {*}$list1 {*}$list2 |
| Append elements to a list using lappend | lappend list1 $elem1 $elem2 |
| Append a list to a list using lappend and {*} | lappend list1 {*}$list2 |
| Append elements to a list using lset | lset list1 end+1 $elem1; lset list1 end+1 $elem2 |
PYK 2016-04-14: (Note: The documentation has been changed to say "... before the point specified by first with no elements being deleted."). Another way to describe the behaviour of linsert is to say that when it's given an end-relative index, it approaches the list from the end, and from that perspective, inserting a value just before the index'th element of list means inserting it prior to the first item it encounters, which is last item in the list. If anything, linsert is the "inconsistent" one among the list commands, but it's a useful inconsistency.
lreplace is more oriented towards ensuring that no elements beyond the last index it's given are replaced. That "beyond" idea makes switching orientation like linsert does less appealing. In the case of a range like end -1, appending to the list wouldn't be very consistent with its behaviour for more common ranges that specify a non-negative width, replacing the last element wouldn't be very consistent either, since the range has no width, and replacing the entire list is unlikely to be what was desired, so it does what seems most sensible.
escargo 2005-09-15: Perhaps there is a better place for the following discussion, but this is where I saw a specific feature mentioned that I want to raise an issue about.
List commands bounds checking and associated behavior
The description above states (in part): "If first is less than zero then it refers to the first element of list."
I see two problems with this:
Personally, I see allowing negative indexes (first or last) to be errors, and ones that are silent failures at that. (This is unlike Icon where at least some negative indexes have semantically valid meaning.) Perhaps list operations that do indexing should have a -strict option so that if indexes are outside the proper ranges, errors would be thrown.
On the second issue, lindex allows negative indexes (and indexes greater than the length of the list), but returns no values. The negative index is not coerced to zero. Thus the same index value (a negative one) refers to no element with lindex but the zeroth element with lreplace. Perhaps an item for the Tcl 9.0 WishList would be consistent behavior for list and string indexing semantics.
If the last argument of lreplace was made optional, one could essentially just alias ldelete to lreplace CmCc
AMG: I assume your suggestion is that omitting the last argument (plus all the element arguments to follow, of course) be taken to mean that the element at index first be removed from the output list. This seems reasonable to me.
My kneejerk complaint is that the name lreplace doesn't seem suitable for this usage, but this complaint already applies to the current case of omitting all element arguments, so no new naming problem is introduced.