求教：重复字符的删除

jason680

回复 10# yinyuemi

some bug

# echo 'AQBATBABCCDD' |perl -nle 'print join "\t",map{s/(\w).*\1/\1/g;$_}split;'
ABCD
   

yizhengming

1. echo 'ABABCCDD' |perl -nle 's/(\w)(?=.*\1)//g; print'
   
2. ABCD
   
3. echo 'CCDDENG' |perl -nle 's/(\w)(?=.*\1)//g; print'
   
4. CDENG
   
5. echo 'AQBATBABCCDD' |perl -nle 's/(\w)(?=.*\1)//g; print'
   
6. QTABCD

复制代码

yinyuemi

回复 11# jason680


    thx，
echo 'AQBATBABCCDD' |perl -nle 'print join "\t",map{while(s/(\w)(.*)\1/\1\2/g){};$_}split;'
AQBTCD

yizhengming

原来while(s///g){} 和while(m//g){} 是有区别的  一个是每次循环都从字符串首开始全局替换，整个while结束条件是下一次循环开始没有可替换的文本。
另一个是每次循环从上一次替换末位置开始，直到整个字符串末结束。

inchonline

回复 1# frankhyk

echo "AABBCCDD          EEFFGG
> HHIIJJKKLLMM    NNOOPPQQRRSS
> TTUU                  VVWWXXYYZZaabbcc
>
> "|perl -lpe 'tr/a-zA-Z//s;'
ABCD          EFG
HIJKLM    NOPQRS
TU                  VWXYZabc


   

inchonline

回复 2# jason680
这里的\1 与$1有什么区别？

   

jason680

回复 16# inchonline

it hard to say what are different between $1 and \1

simple rule for normal using.
use \1 on the left side  of s///.   ex: s/(..)\1/.../
use $1 on the right side of s///. ex: s/....../$1/

please refer the "perlre" for detail information or below:

$ perldoc perlre
NAME
       perlre - Perl regular expressions
DESCRIPTION
       This page describes the syntax of regular expressions in Perl.
    ...

       Capture buffers

       The bracketing construct "( ... )" creates capture buffers. To refer to the
       current contents of a buffer later on, within the same pattern, use \1 for the
       first, \2 for the second, and so on.  Outside the match use "$" instead of "\".
       (The \<digit> notation works in certain circumstances outside the match.  See
       "Warning on \1 Instead of $1" below for details.)  Referring back to another part
       of the match is called a backreference.

   ...
   Warning on \1 Instead of $1
       Some people get too used to writing things like:

           $pattern =~ s/(\W)/\\\1/g;

       This is grandfathered (for \1 to \9) for the RHS of a substitute to avoid
       shocking the sed addicts, but it's a dirty habit to get into.  That's because in
       PerlThink, the righthand side of an "s///" is a double-quoted string.  "\1" in
       the usual double-quoted string means a control-A.  The customary Unix meaning of
       "\1" is kludged in for "s///".  However, if you get into the habit of doing that,
       you get yourself into trouble if you then add an "/e" modifier.

           s/(\d+)/ \1 + 1 /eg;        # causes warning under -w

       Or if you try to do

           s/(\d+)/\1000/;

       You can't disambiguate that by saying "\{1}000", whereas you can fix it with
       "${1}000".  The operation of interpolation should not be confused with the
       operation of matching a backreference.  Certainly they mean two different things
       on the left side of the "s///".

gaoyang636

There is a special form of this construct, called \K , which causes the regex engine to "keep" everything it had matched prior to the \K and not include it in $& .  For various reasons \K may be significantly more efficient than the equivalent (?<=...) construct, and it is especially useful in situations where you want to efficiently remove something following something else in a string. For instance:
  s/(foo)bar/$1/g;
can be rewritten as the much more efficient:
  s/foo\Kbar//g;

sjdy521

回复 17# jason680


    \1是posix所支持的正则形式，$1是perl的正则捕获变量，就这么简单，虽然perl出于好心，在你想用$1的时候错误写成了\1，它也不会抱怨，但我们不应该把简单的问题复杂化

jason680

回复 19# sjdy521

Would you please tell us what's "Certainly they mean two different things on the left side of the "s///"" on "Warning on \1 Instead of $1" section in perlre



Warning on \1 Instead of $1
       ...
       You can't disambiguate that by saying "\{1}000", whereas you can fix it with
       "${1}000".  The operation of interpolation should not be confused with the
       operation of matching a backreference.  Certainly they mean two different things
       on the left side of the "s///".