如何从Linux的数据文件中删除第一列(实际上是行名)? [英] How to delete the first column ( which is in fact row names) from a data file in linux?

问题描述

我的数据文件有成千上万的列和行.我要删除实际上是行计数器的第一列.我在Linux中使用了以下命令:

I have data file with many thousands columns and rows. I want to delete the first column which is in fact the row counter. I used this command in linux:

cut -d " " -f 2- input.txt > output.txt

，但我的输出没有任何变化.有人知道为什么它不起作用，我该怎么办?

but nothing changed in my output. Does anybody knows why it does not work and what should I do?

这是我的输入文件的样子:

This is what my input file looks like:

col1 col2 col3 col4 ...
     1 0 0 0 1
     2 0 1 0 1
     3 0 1 0 0
     4 0 0 0 0 
     5 0 1 1 1 
     6 1 1 1 0
     7 1 0 0 0 
     8 0 0 0 0
     9 1 0 0 0
     10 1 1 1 1
     11 0 0 0 1
    .
    .
    .

我希望我的输出看起来像这样:

I want my output look like this:

col1 col2 col3 col4 ...
0 0 0 1
0 1 0 1
0 1 0 0
0 0 0 0 
0 1 1 1 
1 1 1 0
1 0 0 0 
0 0 0 0
1 0 0 0
1 1 1 1
0 0 0 1
.
.
.

我还尝试了sed命令:

 sed '1d' input.file > output.file

但是它将删除第一行而不是第一列.

But it deletes the first row not the first column.

有人可以引导我吗?

推荐答案

@Karafka我有CSV文件，因此我添加了，"分隔符(您可以用自己的分隔符替换

@Karafka I had CSV files so I added the "," separator (you can replace with yours

cut -d"," -f2- input.csv  > output.csv

然后，我使用循环遍历目录内的所有文件

Then, I used a loop to go over all files inside the directory

# files are in the directory tmp/
for f in tmp/*
do
    name=`basename $f`
    echo "processing file : $name"
    #kepp all column excep the first one of each csv file 

    cut -d"," -f2- $f > new/$name
    #files using the same names are stored in directory new/  
done

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