AWK 用前一行的列值替换空列

Question

我有一系列具有不同列宽的字段分隔文件(符号 | ):

文件 1:

2p              | 2P*    |  1/2 |    2 |                 10.19880615024                       |
2s              | 2S     |  1/2 |    2 |                 10.19881052514816                    |
2               |        |      |    8 |                 10.1988358                           |
2p              | 2P*    |  3/2 |    4 |                 10.19885151459                       |
3p              | 2P*    |  1/2 |    2 |                 12.0874936591                        |
3s              | 2S     |  1/2 |    2 |                 12.0874949611                        |
3               |        |      |   18 |                 12.0875052                           |
3d              | 2D     |  3/2 |    4 |                 12.0875070783                        |

文件 2:

2s2.2p3            | 4S*    |  3/2 |    4 |             0.0000000               |
2s2.2p3            | 2D*    |  5/2 |    6 |             2.3835298               |
2s2.2p3            |        |  3/2 |    4 |             2.3846100               |
2s2.2p3            | 2P*    |  1/2 |    2 |             3.5755704               |
2s2.2p3            |        |  3/2 |    4 |             3.5756182               |
2s2.2p2.(3P).3s    | 4P     |  1/2 |    2 |            10.3259087               |
2s2.2p2.(3P).3s    |        |  3/2 |    4 |            10.3300944               |
2s2.2p2.(3P).3s    |        |  5/2 |    6 |            10.3358956               |
2s2.2p2.(3P).3s    | 2P     |  1/2 |    2 |            10.6796703               |
2s2.2p2.(3P).3s    |        |  3/2 |    4 |            10.6899808               |

我想用上面最后一个非 NULL 列中的相应值填充第二列中的空白:文件 2 示例:

2s2.2p3            | 4S*    |  3/2 |    4 |             0.0000000               |
2s2.2p3            | 2D*    |  5/2 |    6 |             2.3835298               |
2s2.2p3            | 2D*    |  3/2 |    4 |             2.3846100               |
2s2.2p3            | 2P*    |  1/2 |    2 |             3.5755704               |
2s2.2p3            | 2P*    |  3/2 |    4 |             3.5756182               |
2s2.2p2.(3P).3s    | 4P     |  1/2 |    2 |            10.3259087               |
2s2.2p2.(3P).3s    | 4P     |  3/2 |    4 |            10.3300944               |
2s2.2p2.(3P).3s    | 4P     |  5/2 |    6 |            10.3358956               |
2s2.2p2.(3P).3s    | 2P     |  1/2 |    2 |            10.6796703               |
2s2.2p2.(3P).3s    | 2P     |  3/2 |    4 |            10.6899808               |

这个问题与here中的问题类似我可以在第一列中通过以下命令使用 awk 执行此操作:

awk -F"|"'/^/{$0=(x)substr($0,length(x)+1)}{x=$1}1' "$file"

但是我的知识不足以使其在第二列(或与此相关的任何任意列)中工作。

有人可以帮忙吗？提前致谢!

Answer 1

以您的 file2 为例，您可以使用此 awk 命令:

awk -v c=2 'BEGIN {FS=OFS="|"} $c ~ /^[ \t]*$/ {$c = p} {p = $c} 1' file

2s2.2p3            | 4S*    |  3/2 |    4 |             0.0000000               |
2s2.2p3            | 2D*    |  5/2 |    6 |             2.3835298               |
2s2.2p3            | 2D*    |  3/2 |    4 |             2.3846100               |
2s2.2p3            | 2P*    |  1/2 |    2 |             3.5755704               |
2s2.2p3            | 2P*    |  3/2 |    4 |             3.5756182               |
2s2.2p2.(3P).3s    | 4P     |  1/2 |    2 |            10.3259087               |
2s2.2p2.(3P).3s    | 4P     |  3/2 |    4 |            10.3300944               |
2s2.2p2.(3P).3s    | 4P     |  5/2 |    6 |            10.3358956               |
2s2.2p2.(3P).3s    | 2P     |  1/2 |    2 |            10.6796703               |
2s2.2p2.(3P).3s    | 2P     |  3/2 |    4 |            10.6899808               |

要填充第 3 列，只需传递 c=3 代替 c=2。