添加字符的专栏中，我需要添加角色“在一列在文件中。在bash
London,20170616,16,Fri,0,21.5,11.0,3466,16.120,3.71,0,37,11
London,20170617,17,Sat,0,20.8,10.8,3998,342.55,6.89,0,0,0
London,20170618,18,Sun,0,20.1,7.5,3853,335.29,4.62,0,0,0
London,20170619,19,Mon,0,19.4,4.1,3525,255.07,0.49,0,3,0
London,20170620,20,Tue,0.1,22.4,8.7,3632,327.02,1.91,0,0,0
London,20170621,21,Wed,0.3,21.5,11.3,3594,322.33,4.20,0,3,3.2
在第4，第9和第10列，我要补充”，例如...“星期五”
我该怎么做？
谢谢
您可以使用AWK：
awk 'BEGIN{FS=OFS=","}{print $1,$2,$3,"\""$4"\"",$5,$6,$7,$8,"\""$9"\"","\""$10"\"",$11,$12,$13}' yourfile > outfile
这里awk是：
FS和输出字段分隔符OFS到一个逗号（BEGIN{FS=OFS=","}）print $1,$2,$3,"\""$4"\"",$5,$6,$7,$8,"\""$9"\"","\""$10"\"",$11,$12,$13）awk 'BEGIN{FS=OFS=","}{$4="\""$4"\"";$9="\""$9"\"";$10="\""$10"\"";}1' file
字段分隔符和输出字段分隔围绕双引号是逗号。 在awk，因为1永远是真实的，它执行默认操作{print $0}，因此打印存储在$0
尽管此代码可能会回答问题，但提供有关如何解决问题和/或为何解决问题的其他上下文可以提高答案的长期价值。 – Badacadabra
是的。我编辑过。 – tso
awk -v q=\" 'BEGIN{FS=OFS=","}{$4=q$4q;$9=q$9q;$10=q$10q}1' file
London,20170616,16,"Fri",0,21.5,11.0,3466,"16.120","3.71",0,37,11
London,20170617,17,"Sat",0,20.8,10.8,3998,"342.55","6.89",0,0,0
London,20170618,18,"Sun",0,20.1,7.5,3853,"335.29","4.62",0,0,0
London,20170619,19,"Mon",0,19.4,4.1,3525,"255.07","0.49",0,3,0
London,20170620,20,"Tue",0.1,22.4,8.7,3632,"327.02","1.91",0,0,0
London,20170621,21,"Wed",0.3,21.5,11.3,3594,"322.33","4.20",0,3,3.2
当前行是否要修改同一个文件，或者你想创建一个不同的文件？ –
这是一个需求转储，并显示你绝对没有努力。这不是如何工作。 –
@MadPhysicist我认为这是完全合理的。他们已经获得了样本数据，期望的结果以及对转化的可理解的解释。它可以用简短的答案回答，并且没有模糊的事情发生。我不喜欢的唯一的事情是它显示的研究工作很少，因为这可能在命令行上以10种不同的方式完成。 – JNevill