How to remove items from a list not by their position (lreplace) but by the actual content. My solution would be:
proc K { x y } { set x }
proc lremove { listvar string } {
upvar $listvar in
foreach item [K $in [set in [list]]] {
if {[string equal $item $string]} { continue }
lappend in $item
}
}which gives us
% set a [list a b c a b c a b c] a b c a b c a b c % lremove a b % set a a c a c a c
Is there any reason not to do it that way?
Gotisch
RS proposes this built-in way, waving many flags :)
% set a [lsearch -all -inline -not -exact $a b] a c a c a c
Todd A. Jacobs suggests this slightly more flexible version:
# lremove ?-all? list pattern
#
# Removes matching elements from a list, and returns a new list.
proc lremove {args} {
if {[llength $args] < 2} {
puts stderr {Wrong # args: should be "lremove ?-all? list pattern"}
}
set list [lindex $args end-1]
set elements [lindex $args end]
if [string match -all [lindex $args 0]] {
foreach element $elements {
set list [lsearch -all -inline -not -exact $list $element]
}
} else {
# Using lreplace to truncate the list saves having to calculate
# ranges or offsets from the indexed element. The trimming is
# necessary in cases where the first or last element is the
# indexed element.
foreach element $elements {
set idx [lsearch $list $element]
set list [string trim \
"[lreplace $list $idx end] [lreplace $list 0 $idx]"]
}
}
return $list
}
# Test cases.
set foo {1 2 3 4 5 6 7 8 9 10}
puts "Foo: [list $foo]"
puts "Foo: replace one: [lremove $foo $argv]"
puts "Foo: replace all: [lremove -all $foo $argv]"
puts {}
set bar {1 2 3 4 1 6 7 1 9 10}
puts "Bar: [list $bar]"
puts "Bar: replace one: [lremove $bar $argv]"
puts "Bar: replace all: [lremove -all $bar $argv]"dzach ...but then, when I do :
set bar {1 2 3 {} 4 1 6 7 1 9 10 {}}
lremove -all $bar {}I still get:
1 2 3 {} 4 1 6 7 1 9 10 {}See also list