When I am running the following statement:
当我运行以下语句时:
@filtered = map {s/ //g} @outdata;
it is returning an empty list instead of the filtered list that I expected. What I am trying to do is remove every occurrence of from an array of string (which is an XML file).
它返回一个空列表而不是我期望的过滤列表。我想要做的是删除每一次出现的 来自一个字符串数组(这是一个XML文件)。
Obviously, I am not understanding something. Can anyone tell me the correct way to do this might be, and why this isn't working for me as is?
显然,我不理解某事。任何人都可以告诉我这样做的正确方法,以及为什么这对我不起作用?
7 个解决方案
#1
10
Try this:
@filtered = map {s/ //g; $_} @outdata;
The problem is the s operator in perl modifies $_ but actually returns the number of changes it made. So, the extra $_ at the end causes perl to return the modified string for each element of @outdata.
问题是perl中的s运算符修改了$ _但实际上返回了它所做的更改次数。因此,最后的额外$ _会导致perl返回@outdata的每个元素的修改后的字符串。
#2
15
Note that map is going to modify your source array as well. So you could either do:
请注意,map也将修改源数组。所以你可以这样做:
map {s/ //g} @outdata;
and skip the @filtered variable altogether, or if you need to retain the originals,
并完全跳过@filtered变量,或者如果你需要保留原件,
@filtered = @outdata;
map {s/ //g} @filtered;
Although, in that case, it might be more readable to use foreach:
虽然,在这种情况下,使用foreach可能更具可读性:
s/ //g foreach @filtered;
#3
9
Greg's answer has the problem that it will modify the original array as the $_ are passed aliased. You need:
Greg的答案有一个问题,它会修改原始数组,因为$ _被传递别名。你需要:
@filtered = map { (my $new = $_) =~ s/ //g; $new} @outdata;
#4
6
To follow up on Tithonium's point, this will also do the trick:
为了跟进Tithonium的观点,这也将起到作用:
@filtered = map {local $_=$_; s/ //g; $_} @outdata;
The "local" ensures you're working on a copy, not the original.
“本地”确保您正在处理副本,而不是原件。
#5
5
In perl 5.14 you could use the /r regex modifier to make non-destructive substitution.
在perl 5.14中,您可以使用/ r regex修饰符进行非破坏性替换。
@filtered = map {s/ //gr} @outdata;
#6
4
use Algorithm::Loops "Filter";
@filtered = Filter { s/ //g } @outdata;
#7
3
As a counterpoint to Greg's answer, you could misuse grep:
作为Greg答案的对照,你可以滥用grep:
@filtered = grep {s/ //g; 1} @outdata;
Don't do this.
不要这样做。
#1
10
Try this:
@filtered = map {s/ //g; $_} @outdata;
The problem is the s operator in perl modifies $_ but actually returns the number of changes it made. So, the extra $_ at the end causes perl to return the modified string for each element of @outdata.
问题是perl中的s运算符修改了$ _但实际上返回了它所做的更改次数。因此,最后的额外$ _会导致perl返回@outdata的每个元素的修改后的字符串。
#2
15
Note that map is going to modify your source array as well. So you could either do:
请注意,map也将修改源数组。所以你可以这样做:
map {s/ //g} @outdata;
and skip the @filtered variable altogether, or if you need to retain the originals,
并完全跳过@filtered变量,或者如果你需要保留原件,
@filtered = @outdata;
map {s/ //g} @filtered;
Although, in that case, it might be more readable to use foreach:
虽然,在这种情况下,使用foreach可能更具可读性:
s/ //g foreach @filtered;
#3
9
Greg's answer has the problem that it will modify the original array as the $_ are passed aliased. You need:
Greg的答案有一个问题,它会修改原始数组,因为$ _被传递别名。你需要:
@filtered = map { (my $new = $_) =~ s/ //g; $new} @outdata;
#4
6
To follow up on Tithonium's point, this will also do the trick:
为了跟进Tithonium的观点,这也将起到作用:
@filtered = map {local $_=$_; s/ //g; $_} @outdata;
The "local" ensures you're working on a copy, not the original.
“本地”确保您正在处理副本,而不是原件。
#5
5
In perl 5.14 you could use the /r regex modifier to make non-destructive substitution.
在perl 5.14中,您可以使用/ r regex修饰符进行非破坏性替换。
@filtered = map {s/ //gr} @outdata;
#6
4
use Algorithm::Loops "Filter";
@filtered = Filter { s/ //g } @outdata;
#7
3
As a counterpoint to Greg's answer, you could misuse grep:
作为Greg答案的对照,你可以滥用grep:
@filtered = grep {s/ //g; 1} @outdata;
Don't do this.
不要这样做。