读取文本文件，更改某些行的列顺序 [英] Reading text file, change order of columns of some lines

问题描述

我的输入文件的格式为:

My input file is of the form:

   0     1     0     0     0     1     1     0     0    0 / 1    0 / 1    0 / 1
   0     1     0 3/4     1     0     0 1/4     0     0    -1 1/2
   0    -1     0 1/4    -1     0     0 3/4     0     0     1 1/2

我想重新排列其中包含分数的行的顺序.目前，我有:

I want to rearrange the order of the lines that have the fraction within them. Currently I have:

#!bin/bash
filename="input.txt"
while ((i++)); read -r line; do
  re='[0-9][/][0-9]';
  if [[ $line =~ $re ]]
    then
      echo $line
  fi
done < "$filename"

将回显第二行和第三行.是否可以使用awk或sed命令来获取这两行内容以将其顺序(保持第一行不变)更改为

which will echo the second and third line. Is there an awk or sed command I could use to get these two lines to change their order (leaving the first as is) to being

$1,$2,$3,$5,$6,$7,$9,$10,$11,$4,$8,$12

这将使我的文件现在看起来像

which would make my file now look like

   0     1     0     0     0     1     1     0     0    0 / 1    0 / 1    0 / 1
   0     1     0     1     0     0     0     0    -1 3/4 1/4 1/2
   0    -1     0    -1     0     0     0     0     1 1/4 3/4 1/2

推荐答案

您可以使用awk轻松做到这一点，尽管如此，我认为定义游戏规则很重要. 在以下假设下:

you can easily do this with awk, nonetheless I think it is important to define the rules of the game a bit. Under the following assumptions :

• 分数是以下任何形式:a/b或a / b或a/ b
• 如果小数出现在第4列或第8列中，请重新排列各列.
• 您要保持格式正确

• A fraction is anything of the form : a/b or a / b or a/ b
• If a fraction appears in column 4 or 8, reshuffle the columns.
• you want to keep the formatting correct

请记住，您可以使用以下awk代码

With this in mind, you can use the following awk code

awk 'BEGIN{format="%4s%6s%6s%6s%6s%6s%6s%6s%6s%7s%7s%7s\n"}
     { gsub(/[[:blank:]]*\/[[:blank:]]*/,"/",$0); $0=$0 }
     ($4 ~ /\//) || ($8 ~ /\//) { 
        $12=$4" "$8" "$12
        $4=""; $8=""
        $0=$0
     }                                           
     { printf format,$1,$2,$3,$4,$5,$6,$7,$8,$9,$10,$11,$12 }
    ' file.txt

这将执行以下操作:

• 用单个/

$0 = $0重新定义字段，即在您的前两行中， 将从18个字段移至12个

$0 = $0 redefines the fields, i.e. in your first two lines, you will move from 18 fields to 12

如果在字段4或8中出现小数(即/)，则重新定义字段12，删除字段4和8，然后再次执行$0=$0.

if a fraction (i.e. a /) appears in field 4 or 8, then redefine field 12, delete field 4 and 8 and do $0=$0 again.

以正确的格式打印.

注意:在上面的示例中，分数具有不同的输出(无空格)

NOTE: in the above example, the fractions have a different output (no spaces)

以上内容将为您提供以下输出:

The above will give you the following output :

   0     1     0     0     0     1     1     0     0    0/1    0/1    0/1
   0     1     0     1     0     0     0     0    -1    3/4    1/4    1/2
   0    -1     0    -1     0     0     0     0     1    1/4    3/4    1/2

如果您不想在第一行中更改分数，那么可以很容易地做到这一点

If you do not want change your fractions in the first line, then you can do it very easy like this

awk 'BEGIN{format="%4s%6s%6s%6s%6s%6s%6s%6s%6s%7s%7s%7s\n"}
     (NF>12) { print; next }
     { 
        $12=$4" "$8" "$12
        $4=""; $8=""
        $0=$0
        printf format,$1,$2,$3,$4,$5,$6,$7,$8,$9,$10,$11,$12
     }
    ' file.txt

您在这里假设

• 如果一行中的字段超过12个，则只需打印

• if a line has more then 12 fields, just print it

否则，请随机排列列

然而，它的鲁棒性较弱，因为所有内容都取决于在第4、8和12列中键入分数的方式. IE.必须 输入空格.输出将如下所示:

This is however less robust as everything depends how the fractions are typed in the 4th,8th and 12th column. I.e. they must be typed without spaces. Output will look like :

   0     1     0     0     0     1     1     0     0    0 / 1    0 / 1    0 / 1
   0     1     0     1     0     0     0     0    -1    3/4    1/4    1/2
   0    -1     0    -1     0     0     0     0     1    1/4    3/4    1/2

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