4

我的输入（例如，来自ifconfig run0 scanOpenBSD）有一些由空格分隔的字段，但一些字段本身包含空格（幸运的是，这些包含空格的字段总是用引号引起来）。

我需要区分引号内的空格和分隔符空格。这个想法是用下划线替换引号内的空格。

样本数据：

%cat /tmp/ifconfig_scan | fgrep nwid | cut -f3
nwid Websense chan 6 bssid 00:22:7f:xx:xx:xx 59dB 54M short_preamble,short_slottime
nwid ZyXEL chan 8 bssid cc:5d:4e:xx:xx:xx 5dB 54M privacy,short_slottime
nwid "myTouch 4G Hotspot" chan 11 bssid d8:b3:77:xx:xx:xx 49dB 54M privacy,short_slottime

最终没有按照我想要的方式处理，因为我还没有用下划线替换引号内的空格：

%cat /tmp/ifconfig_scan | fgrep nwid | cut -f3 |\
    cut -s -d ' ' -f 2,4,6,7,8 | sort -n -k4
"myTouch Hotspot" 11 bssid d8:b3:77:xx:xx:xx
ZyXEL 8 cc:5d:4e:xx:xx:xx 5dB 54M
Websense 6 00:22:7f:xx:xx:xx 59dB 54M
4

5 回答 5

4

试试这个：

awk -F'"' '{for(i=2;i<=NF;i++)if(i%2==0)gsub(" ","_",$i);}1' OFS="\"" file

它适用于一行中的多引号部分：

echo '"first part" foo "2nd part" bar "the 3rd part comes" baz'| awk -F'"' '{for(i=2;i<=NF;i++)if(i%2==0)gsub(" ","_",$i);}1' OFS="\"" 
"first_part" foo "2nd_part" bar "the_3rd_part_comes" baz

编辑替代形式：

awk 'BEGIN{FS=OFS="\""} {for(i=2;i<NF;i+=2)gsub(" ","_",$i)} 1' file
于 2013-02-16T23:35:57.787 回答
4

对于sed-only 解决方案（我不一定提倡），请尝试：

echo 'a b "c d e" f g "h i"' |\
sed ':a;s/^\(\([^"]*"[^"]*"[^"]*\)*[^"]*"[^"]*\) /\1_/;ta'
a b "c_d_e" f g "h_i"

翻译：

  • 从行首开始。
  • 寻找模式junk"junk"，重复零次或多次，其中junk没有引号，然后是junk"junk space.
  • 用 替换最后的空格_。
  • 如果成功，则跳回到开头。
于 2013-02-17T01:57:36.380 回答
4

另一个 awk 尝试：

awk '!(NR%2){gsub(FS,"_")}1' RS=\" ORS=\"

删除引号：

awk '!(NR%2){gsub(FS,"_")}1' RS=\" ORS=

在@steve 完成的早期测试的基础上，使用三倍大小的测试文件进行了一些额外的测试。我不得不sed稍微改变一下语句，以便非 GNUsed也可以处理它。我包括awk( bwk)gawk3和gawk4:mawk

$ for i in {1..1500000}; do echo 'a b "c d e" f g "h i" j k l "m n o "p q r" s t" u v "w x" y z' ; done > test
$ time perl -pe 's:"[^"]*":($x=$&)=~s/ /_/g;$x:ge' test >/dev/null

real    0m27.802s
user    0m27.588s
sys 0m0.177s
$ time awk 'BEGIN{FS=OFS="\""} {for(i=2;i<NF;i+=2)gsub(" ","_",$i)} 1' test >/dev/null

real    0m6.565s
user    0m6.500s
sys 0m0.059s
$ time gawk3 'BEGIN{FS=OFS="\""} {for(i=2;i<NF;i+=2)gsub(" ","_",$i)} 1' test >/dev/null

real    0m21.486s
user    0m18.326s
sys 0m2.658s
$ time gawk4 'BEGIN{FS=OFS="\""} {for(i=2;i<NF;i+=2)gsub(" ","_",$i)} 1' test >/dev/null

real    0m14.270s
user    0m14.173s
sys 0m0.083s
$ time mawk 'BEGIN{FS=OFS="\""} {for(i=2;i<NF;i+=2)gsub(" ","_",$i)} 1' test >/dev/null

real    0m4.251s
user    0m4.193s
sys 0m0.053s
$ time awk '!(NR%2){gsub(FS,"_")}1' RS=\" ORS=\" test >/dev/null

real    0m13.229s
user    0m13.141s
sys 0m0.075s
$ time gawk3 '!(NR%2){gsub(FS,"_")}1' RS=\" ORS=\" test >/dev/null

real    0m33.965s
user    0m26.822s
sys 0m7.108s
$ time gawk4 '!(NR%2){gsub(FS,"_")}1' RS=\" ORS=\" test >/dev/null

real    0m15.437s
user    0m15.328s
sys 0m0.087s
$ time mawk '!(NR%2){gsub(FS,"_")}1' RS=\" ORS=\" test >/dev/null

real    0m4.002s
user    0m3.948s
sys 0m0.051s
$ time sed -e :a -e 's/^\(\([^"]*"[^"]*"[^"]*\)*[^"]*"[^"]*\) /\1_/;ta' test > /dev/null

real    5m14.008s
user    5m13.082s
sys 0m0.580s
$ time gsed -e :a -e 's/^\(\([^"]*"[^"]*"[^"]*\)*[^"]*"[^"]*\) /\1_/;ta' test > /dev/null

real    4m11.026s
user    4m10.318s
sys 0m0.463s

mawk呈现最快的结果...

于 2013-02-18T13:44:56.500 回答
2

你最好用perl. 代码更具可读性和可维护性：

perl -pe 's:"[^"]*":($x=$&)=~s/ /_/g;$x:ge'

根据您的输入，结果是：

a b "c_d_e" f g "h_i"

解释：

-p            # enable printing
-e            # the following expression...

s             # begin a substitution

:             # the first substitution delimiter

"[^"]*"      # match a double quote followed by anything not a double quote any
              # number of times followed by a double quote

:             # the second substitution delimiter

($x=$&)=~s/ /_/g;      # copy the pattern match ($&) into a variable ($x), then 
                       # substitute a space for an underscore globally on $x. The
                       # variable $x is needed because capture groups and
                       # patterns are read only variables.

$x            # return $x as the replacement.

:             # the last delimiter

g             # perform the nested substitution globally
e             # make sure that the replacement is handled as an expression

一些测试：

for i in {1..500000}; do echo 'a b "c d e" f g "h i" j k l "m n o "p q r" s t" u v "w x" y z' >> test; done

time perl -pe 's:"[^"]*":($x=$&)=~s/ /_/g;$x:ge' test >/dev/null

real    0m8.301s
user    0m8.273s
sys     0m0.020s

time awk 'BEGIN{FS=OFS="\""} {for(i=2;i<NF;i+=2)gsub(" ","_",$i)} 1' test >/dev/null

real    0m4.967s
user    0m4.924s
sys     0m0.036s

time awk '!(NR%2){gsub(FS,"_")}1' RS=\" ORS=\" test >/dev/null

real    0m4.336s
user    0m4.244s
sys     0m0.056s

time sed ':a;s/^\(\([^"]*"[^"]*"[^"]*\)*[^"]*"[^"]*\) /\1_/;ta' test >/dev/null

real    2m26.101s
user    2m25.925s
sys     0m0.100s
于 2013-02-17T08:00:41.137 回答
1

不是答案，只是发布 @steve 的 perl 代码的 awk 等效代码以防万一有人感兴趣（并帮助我将来记住这一点）：

@steve 发布：

perl -pe 's:"[^\"]*":($x=$&)=~s/ /_/g;$x:ge'

并且通过阅读@steve 的解释，与该 perl 代码等效的最简短的 awk（不是首选的 awk 解决方案 - 请参阅@Kent 的答案）将是 GNU awk：

gawk '{
   head = ""
   while ( match($0,"\"[^\"]*\"") ) {
      head = head substr($0,1,RSTART-1) gensub(/ /,"_","g",substr($0,RSTART,RLENGTH))
      $0 = substr($0,RSTART+RLENGTH)
   }
   print head $0
}'

我们从具有更多变量的 POSIX awk 解决方案开始：

awk '{
   head = ""
   tail = $0
   while ( match(tail,"\"[^\"]*\"") ) {
      x = substr(tail,RSTART,RLENGTH)
      gsub(/ /,"_",x)
      head = head substr(tail,1,RSTART-1) x
      tail = substr(tail,RSTART+RLENGTH)
   }
   print head tail
}'

并使用 GNU awk 的 gensub() 保存一行：

gawk '{
   head = ""
   tail = $0
   while ( match(tail,"\"[^\"]*\"") ) {
      x = gensub(/ /,"_","g",substr(tail,RSTART,RLENGTH))
      head = head substr(tail,1,RSTART-1) x
      tail = substr(tail,RSTART+RLENGTH)
   }
   print head tail
}'

然后摆脱变量x：

gawk '{
   head = ""
   tail = $0
   while ( match(tail,"\"[^\"]*\"") ) {
      head = head substr(tail,1,RSTART-1) gensub(/ /,"_","g",substr(tail,RSTART,RLENGTH))
      tail = substr(tail,RSTART+RLENGTH)
   }
   print head tail
}'

然后摆脱变量“tail”，如果你不需要 $0，NF 等，在循环后留下来：

gawk '{
   head = ""
   while ( match($0,"\"[^\"]*\"") ) {
      head = head substr($0,1,RSTART-1) gensub(/ /,"_","g",substr($0,RSTART,RLENGTH))
      $0 = substr($0,RSTART+RLENGTH)
   }
   print head $0
}'
于 2013-02-17T12:41:44.150 回答