☰

替换方括号之间的数据并忽略其他发生的情况

tag-icon • tag-icon text-processing
替换方括号之间的数据并忽略其他发生的情况

我有一个包含以下数据的 csv 文件

输入

"01","5","Male","[""No.**"",""**kg""]","","",""  
"02","6","FeMale","[""No.""]","","",""

我想替换方括号之间的“，”字符串，但不替换方括号之外的数据。

输出：

"01","5","Male","[""No.**,**kg""]","","",""  
"02","6","FeMale","[""No.""]","","",""

我尝试了下面的命令，但它确实替换了方括号之外的数据，但不适用于所有行。

sed '/\[/,/\]/s/"",""/,/'

答案1

sed '/\[/,/\]/s/"",""/,/'

这将查找带有 a 的行[，然后查找带有 a 的行]，并在这些行之间替换每行中遇到的第一个"",""。它并没有真正关注它们[..]在一行中的位置。

作为零阶尝试，如下所示：

$ cat x
"01","5","Male","[""No.**"",""**kg""]","","","" 
$ sed 's/\(\[.*\)"",""\(.*\]\)/\1,\2/g' x
"01","5","Male","[""No.**,**kg""]","","",""

该模式匹配 a [、任何内容"",""、任何内容和 a ]，同时捕获除 the 之外的所有内容，"","" 以便它可以将各个部分重新组合在一起。

这会破坏诸如[..],"","",[..]（括号在"",""看到 a 之前关闭，模式搜索后面的]）和[.."","".."",""..]（括号内有多个"",""序列，仅删除一个）之类的内容。

对于 Perl 来说更通用一些，尽管这是一个可怕的替换中的替换技巧。您可能应该使用适当的解析器来代替：

$ cat y
no removal here: [...],"","",[...]
double removal here: [  "",""  "",""  ]
[""remove"",""here""],""not"",""here"",[""also"",""here""]

$ perl -pe 'sub x {$a = shift; $a =~ s/"",""/,/g; return $a;}
     s/(\[.*?\])/ x($1)  /eg  ' y
no removal here: [...],"","",[...]
double removal here: [  ,  ,  ]
[""remove,here""],""not"",""here"",[""also,here""]

（.*?是非贪婪匹配，它会尽快停止，即]在本例中是第一次停止。）

答案2

sed '/"Male/s/"",""/,/1' filename

输出：

"01","5","Male","[""No.**,**kg""]","","",""
"02","6","FeMale","[""No.""]","","",""

相关内容