list TCL 从列表中删除一个元素
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TCL remove an element from a list
提问by Narek
How te remove an element from TCL list say:
如何从 TCL 列表中删除元素说：
- which has index = 4
- which has value = "aa"
- 其中索引 = 4
- 其中具有 value = "aa"
I have Googled and have not found any built-in function yet.
我用谷歌搜索过，还没有找到任何内置功能。
回答by drysdam
set mylist {a b c}
puts $mylist
a b c
Remove by index
按索引删除
set mylist [lreplace $mylist 2 2]
puts $mylist
a b
Remove by value
按值删除
set idx [lsearch $mylist "b"]
set mylist [lreplace $mylist $idx $idx]
puts $mylist
a
回答by Donal Fellows
The other way to remove an element is to filter it out. This Tcl 8.5 technique differs from the lsearch&lreplacemethod mentioned elsewhere in that it removes allof a given element from the list.
删除元素的另一种方法是过滤掉它。这种 Tcl 8.5 技术与其他地方提到的lsearch&lreplace方法的不同之处在于它从列表中删除所有给定元素。
set stripped [lsearch -inline -all -not -exact $inputList $elemToRemove]
What it doesn't do is search through nested lists. That's a consequence of Tcl not putting effort into understanding your data structures too deeply. (You can tell it to search by comparing specific elements of the sublists though, via the -indexoption.)
它不做的是搜索嵌套列表。这是 Tcl 没有努力深入理解您的数据结构的结果。（不过，您可以通过-index选项通过比较子列表的特定元素来告诉它进行搜索。）
回答by icanhasserver
Lets say you want to replace element "b":
假设您要替换元素“b”：
% set L {a b c d}
a b c d
You replace the first element 1 and last element 1 by nothing:
您将第一个元素 1 和最后一个元素 1 替换为空：
% lreplace $L 1 1
a c d
回答by Shawn E
regsubmay also be suitable to remove a value from a list.
regsub也可能适合从列表中删除一个值。
set mylist {a b c}
puts $mylist
a b c
regsub b $mylist "" mylist
puts $mylist
a c
llength $mylist
2
回答by Todd Horst
Just wrapped up what others have done
刚刚结束其他人所做的
proc _lremove {listName val {byval false}} {
upvar $listName list
if {$byval} {
set list [lsearch -all -inline -not $list $val]
} else {
set list [lreplace $list $val $val]
}
return $list
}
Then call with
然后用
Inline edit, list lappend
set output [list 1 2 3 20]
_lremove output 0
echo $output
>> 2 3 20
Set output like lreplace/lsearch
set output [list 1 2 3 20]
echo [_lremove output 0]
>> 2 3 20
Remove by value
set output [list 1 2 3 20]
echo [_lremove output 3 true]
>> 1 2 20
Remove by value with wildcar
set output [list 1 2 3 20]
echo [_lremove output "2*" true]
>> 1 3
回答by Antonio Mes
You can also try like this :
你也可以这样尝试：
set i 0
set myl [list a b c d e f]
foreach el $myl {
if {$el in {a b e f}} {
set myl [lreplace $myl $i $i]
} else {
incr i
}
}
set myl
