Question

我正在尝试使用paste命令循环遍历文件对，将它们粘贴在一起并将每个文件作为唯一文件输出。我尝试了很多东西，这里有几个：

 for i in *_temp4.csv; do paste *_temp4.csv *_temp44.csv > ${i}_out.csv; done
 #Each output contains each input file (rather than pairs). Obviously this is because of the * wildcard

 for i in *_temp2.csv_temp4.csv; do paste $_temp2_temp4.csv $_temp3_temp44.csv > ${i}_out.csv; done

没有错误，空输出文件

 for i in *_temp2.csv_temp4.csv; do paste ${_temp2_temp4.csv} ${_temp3_temp44.csv} > ${i}_out.csv; done

输出：

 combo15.awk: line 12: ${_temp2_temp4.csv}: bad substitution

我想我必须遗漏一些关于如何使用$的基本信息，但我一直在谷歌上搜索无效。

我的整个代码，对于上下文，虽然我不明白为什么前面的行会影响到这一点。

 for i in *.dat; do awk 'NR > 23 { print }' ${i} > ${i}_temp1.csv; done

 for i in *_temp1.csv; do awk 'BEGIN{OFS=FS=","}$2==0{$2="between"}BEGIN{OFS=FS=","}$2==1{$2="lego"}BEGIN{OFS=FS=","}$2==2{$2="pin"}BEGIN{OFS=FS=","}$2==3{$2="dice"}BEGIN{OFS=FS=","}$2==4{$2="jack"}BEGIN{OFS=FS=","}$2==8{$2="escape"}{print}'  ${i} > ${i}_temp2.csv; done

 for i in *_temp2.csv; do awk -v OFS="," '{$4 = $1 - prev1; prev1 = $1; print;}' ${i} > ${i}_temp3.csv; done  

 for i in *_temp2.csv; do awk -F "," 'BEGIN{print "new line"}{print $2}' ${i} > ${i}_temp4.csv; done

 for i in *_temp3.csv; do awk -F "," '{print $5}' ${i} > ${i}_temp44.csv; done

 for i in *_temp2.csv_temp4.csv; do paste $_temp2_temp4.csv    $_temp3_temp44.csv > ${i}_out.csv; done

Answer 1

您的问题是，您的文件名称无法控制地增长。这种变化应该可以解决这个问题：

for i in *.dat; do awk 'NR > 23 { print }' ${i} > ${i}_temp1.csv; done

for i in *.dat; do awk 'BEGIN{OFS=FS=","}$2==0{$2="between"}BEGIN{OFS=FS=","}$2==1{$2="lego"}BEGIN{OFS=FS=","}$2==2{$2="pin"}BEGIN{OFS=FS=","}$2==3{$2="dice"}BEGIN{OFS=FS=","}$2==4{$2="jack"}BEGIN{OFS=FS=","}$2==8{$2="escape"}{print}'  ${i}_temp1.csv > ${i}_temp2.csv; done

for i in *.dat; do awk -v OFS="," '{$4 = $1 - prev1; prev1 = $1; print;}' ${i}_temp2.csv > ${i}_temp3.csv; done

for i in *.dat; do awk -F "," 'BEGIN{print "new line"}{print $2}' ${i}_temp2.csv > ${i}_temp4.csv; done

for i in *.dat; do awk -F "," '{print $5}' ${i}_temp3.csv > ${i}_temp44.csv; done

for i in *.dat; do paste ${i}_temp4.csv    ${i}_temp44.csv > ${i}_out.csv; done

bash paste：循环通过基于通配符的文件对，生成单独的输出文件

1 个答案: