[英]How to replace the nth space before a string on each line in a file using sed
I am trying to replace the space before the surname on each line of a file with a comma using sed. 我正在尝试使用sed用逗号替换文件每一行上的姓氏前的空格。
Example Source: 来源示例:
George W Heong§New York§USA
Elizabeth Black§Sheffield, Yorkshire§England
Lucy Jones§Cardiff§Wales
James G K Shackleton§Dallas, Texas§USA
Carl Seddon§Canberra,Australia
Example Ouput: 示例输出:
George W,Heong§New York§USA
Elizabeth,Black§Sheffield, Yorkshire§England
Lucy,Jones§Cardiff§Wales
James G K,Shackleton§Dallas, Texas§USA
Carl,Seddon§Canberra,Australia
I think I've worked out a method to obtain the index of the relevant space as follows: 我想我已经设计出一种方法来获取相关空间的索引,如下所示:
int idx$ = str.indexOf("§");
int nthSpace = str.lastIndexOf(" ", idx$);
but I haven't been able to work out how to replace the nth instance with the variable nthSpace. 但是我还无法弄清楚如何用变量nthSpace替换第n个实例。 This is what have got so far:
到目前为止,这是:
sed "s/$nthSpace" "/,/" datain.txt > dataout.txt
Any asistance would be appreciated. 任何协助将不胜感激。
With gensub
, available in GNU awk
, you can do this: 使用
gensub
(可在GNU awk
,您可以执行以下操作:
awk 'BEGIN{FS=OFS="§"} {$1=gensub(/[[:blank:]]([^[:blank:]]+)$/, ",\\1", 1, $1)} 1' file
Output: 输出:
George W,Heong§New York§USA
Elizabeth,Black§Sheffield, Yorkshire§England
Lucy,Jones§Cardiff§Wales
James G K,Shackleton§Dallas, Texas§USA
Carl,Seddon§Canberra,Australia
With sed : 与sed:
sed 's/ \([^ ]*§\)/,\1/' sourcefile
The pattern looks for the first occurence of : 该模式查找以下项的首次出现:
The name is captured in a group that is used in the substitution to be prefixed with a ,
该名称是在替换中使用的组中捕获的,前缀为
,
UPDATE : 更新:
To prevent strings as name §
to be matched, you can preprocess the first substitution with s/ +§/§/
. 为了防止匹配
name §
字符串,可以使用s/ +§/§/
预处理第一次替换。 The final command will be : 最终命令将是:
sed 's/ +§/§/;s/ \([^ ]*§\)/,\1/' sourcefile
As noticed in question comments, multipart surnames (separated with spaces) will be split if not rewritten manually. 如问题注释中所注意到的,如果不手动重写,多部分姓(以空格分隔)将被拆分。
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