Question

我是linux和命令行的新手。我试图找到一个命令，这将允许我用除了第一个之外的所有字段的分号替换white space（在.csv文本文件中）。请看下面的例子，感谢任何帮助，我花了很长时间寻找解决方案。如果你确实有答案，请解释一下这个命令，这样我就可以尝试学习如何以及为什么。非常感谢。

输入文字示例：

0 k__Bacteria p__Firmicutes c__Bacilli 1 k__Bacteria p__Firmicutes c__Clostridia 2 k__Bacteria p__Bacteroidetes c__Bacteroidia 3 k__Bacteria p__Bacteroidetes c__Bacteroidia

我需要的是：

0 k__Bacteria;p__Firmicutes;c__Bacilli 1 k__Bacteria;p__Firmicutes;c__Clostridia 2 k__Bacteria;p__Bacteroidetes;c__Bacteroidia 3 k__Bacteria;p__Bacteroidetes;c__Bacteroidia

Answer 1

$ cat file
0   k__Bacteria  p__Firmicutes   c__Bacilli     foo     bar
1   k__Bacteria  p__Firmicutes   c__Clostridia  the   quick     brown
2   k__Bacteria  p__Bacteroidetes    c__Bacteroidia     fox jumped      over
3   k__Bacteria  p__Bacteroidetes    c__Bacteroidia     the lazy dogs back

$ awk -v skip=1 '{match($0,"([^[:space:]]+[[:space:]]+){"skip"}"); head=substr($0,1,RSTART+RLENGTH); tail=substr($0,RSTART+RLENGTH+1); gsub(/[[:space:]]+/,";",tail); print head tail}' file
0   k__Bacteria;p__Firmicutes;c__Bacilli;foo;bar
1   k__Bacteria;p__Firmicutes;c__Clostridia;the;quick;brown
2   k__Bacteria;p__Bacteroidetes;c__Bacteroidia;fox;jumped;over
3   k__Bacteria;p__Bacteroidetes;c__Bacteroidia;the;lazy;dogs;back

$ awk -v skip=2 '{match($0,"([^[:space:]]+[[:space:]]+){"skip"}"); head=substr($0,1,RSTART+RLENGTH); tail=substr($0,RSTART+RLENGTH+1); gsub(/[[:space:]]+/,";",tail); print head tail}' file
0   k__Bacteria  p__Firmicutes;c__Bacilli;foo;bar
1   k__Bacteria  p__Firmicutes;c__Clostridia;the;quick;brown
2   k__Bacteria  p__Bacteroidetes;c__Bacteroidia;fox;jumped;over
3   k__Bacteria  p__Bacteroidetes;c__Bacteroidia;the;lazy;dogs;back

$ awk -v skip=3 '{match($0,"([^[:space:]]+[[:space:]]+){"skip"}"); head=substr($0,1,RSTART+RLENGTH); tail=substr($0,RSTART+RLENGTH+1); gsub(/[[:space:]]+/,";",tail); print head tail}' file
0   k__Bacteria  p__Firmicutes   c__Bacilli;foo;bar
1   k__Bacteria  p__Firmicutes   c__Clostridia;the;quick;brown
2   k__Bacteria  p__Bacteroidetes    c__Bacteroidia;fox;jumped;over
3   k__Bacteria  p__Bacteroidetes    c__Bacteroidia;the;lazy;dogs;back

Answer 2

你可以在python中这样做：

#!/usr/bin/env python
import sys

if __name__ == '__main__':
    for line in sys.stdin:
        cols = line.split()
        print ' '.join([cols[0], ';'.join(cols[1:])])

只需chmod +x script文件并执行./script < input。

请注意，line.split（）将按多个空格分割，'a b\tc'将在['a', 'b', 'c']中生成。

Answer 3

这是解决方案awk。它可能是脏的，有人可以改进它，但它的工作

awk 'OFS=";"{a=$1;$1="";$0=a";"$0}sub(/;;/," ",$0) ' temp.txt

输出

0 k_Bacteria;p_Firmicutes;c_Bacilli
1 k_Bacteria;p_Firmicutes;c_Clostridia
2 k_Bacteria;p_Bacteroidetes;c_Bacteroidia
3 k_Bacteria;p_Bacteroidetes;c_Bacteroidia

cat temp.txt
0 k_Bacteria p_Firmicutes c_Bacilli
1 k_Bacteria p_Firmicutes c_Clostridia
2 k_Bacteria p_Bacteroidetes c_Bacteroidia
3 k_Bacteria p_Bacteroidetes c_Bacteroidia

根据评论

编辑：更新

试试这个awk脚本myawk.sh

 BEGIN { print "Begin Processing "}
   OFS=";"{
       $9=$9"%%"
   b = $0;
   split($0,a,"%%");
   gsub(/;/," ",a[1])
   print a[1]a[2]
   }
  END {print "Process Complete"}

执行awk -f myawk.sh temp.txt，其中$ 9是你要保留空格的变量uptill

Answer 4

awk -v OFS=";" '{$1=$1" "$2;$2="";gsub(/;;/,";",$0);print}' your_file

或者可能是perl：

perl -F -lane 'print join ";",@F' your_file| perl -pe 's/;/ /'

使用awk删除特定的空白区域并用分号替换

4 个答案: