连续的行，无需更改顺序

Question

猫文件1

487

cat file2

4.999286 12.669064 0.0000005.999343 12.753258 0.0000006.999401 12.654514 0.0000007.999458 12.774485 0.0000008.999515 12.662147 0.0000009.999572 12.700071 0.00000010.999629 12.717721 0.000000

我希望以相同的顺序和另外两个连续的行来grep file1。我试过了：

grep -A 2 -Ff file1 file2

我想要这样的输出：

4.999286 12.669064 0.0000005.999343 12.753258 0.0000006.999401 12.654514 0.0000008.999515 12.662147 0.0000009.999572 12.700071 0.00000010.999629 12.717721 0.0000007.999458 12.774485 0.0000008.999515 12.662147 0.0000009.999572 12.700071 0.000000

Answer 1

grep快速进行少量文件扫描

$ while IFS= read -r n; do grep -A2 "^$n." file2; done < file1

4.999286 12.669064 0.000000
5.999343 12.753258 0.000000
6.999401 12.654514 0.000000
8.999515 12.662147 0.000000
9.999572 12.700071 0.000000
10.999629 12.717721 0.000000
7.999458 12.774485 0.000000
8.999515 12.662147 0.000000
9.999572 12.700071 0.000000

Answer 2

[请您尝试以下操作，这将以与$1在Input_file1中出现的相同顺序给出结果。

awk '
BEGIN{
  s1="\""
}
FNR==NR{
  a[$0]
  next
}
(int($1) in a){
  system("grep -A2 " s1 $0 s1 OFS FILENAME)
}
'  Input_file1  Input_file2

输出将如下。

4.999286 12.669064 0.000000
5.999343 12.753258 0.000000
6.999401 12.654514 0.000000
7.999458 12.774485 0.000000
8.999515 12.662147 0.000000
9.999572 12.700071 0.000000
8.999515 12.662147 0.000000
9.999572 12.700071 0.000000
10.999629 12.717721 0.000000