bash 如何删除vi文件中的列？
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How to delete columns in vi file?
提问by Jarhead
I have a vi file containing hundreds of lines with the following format
我有一个包含数百行的 vi 文件，格式如下
029.inp.log: SCF Done: E(RHF) = -844.790844670 A.U. after 26 cycles
I want to delete all of the columns (separated by spaces/tabs. got messed up when I pasted it here) other than the first (029.inp.log:) and the fifth (-844.790844670). Can anyone help me?
我想删除除第一列 (029.inp.log:) 和第五列 (-844.790844670) 之外的所有列（由空格/制表符分隔。当我在这里粘贴时搞砸了）。谁能帮我？
回答by shafeeq
Press ctrl+vfor block selection. use h, j, k, lkeys to navigate and press key dto delete the selected block. The graphical editor like katealso having capability of the block selection.
按ctrl+v进行块选择。使用h, j, k, l键导航并按 键d删除所选块。图形编辑器kate还具有块选择功能。
回答by gzh
If you are not confined to vi, shell command is a good choice for this task.
如果您不局限于 vi，shell 命令是执行此任务的不错选择。
cat your_filename | cut -f 1,5 > result_filename
cat your_filename | cut -f 1,5 > result_filename
About rectangle selection in vim, you must guarantee that the field in all lines with the same width.
关于vim中的矩形选择，必须保证所有行中的字段宽度相同。
Perhaps vim command
:%!cut -f1,5is OK too, but I can not give it a try now.
也许 vim 命令
:%!cut -f1,5也可以，但我现在不能试一试。
