AWK：取代他们的平均每n行，每一列 [英] Awk: substituting each n lines with their average, for each column

问题描述

我怎样才能用awk替换文件的每个n线，他们的平均？

这问题答案完全，除了它只能处理一列和 3 不是参数化的事实：
<一href=\"http://stackoverflow.com/questions/8551349/how-to-sum-up-every-10-lines-and-calculate-average-using-awk\">How总结起来每10行，用AWK计算平均？

基本上我想借这样一个文件

  1
2 1
3 1
4 1
5 1
6 1
2.5 2.0
3.5 2.0
4 2.0
 

和获得是这样的：

  2 1
5 1
3.33 2.0
 

解决方案

下面是一个完整的shell脚本：

 的awk -v计数= 3'
    {
        如果（NF＆GT; tot_col）
            tot_col = NF;        CUR = 1;
        而（CUR＆LT; = NF）
        {
            总和[CUR] + = $ CUR;
            CUR ++;
        }        如果（（NR％计）== 0）
        {
            CUR = 1;
            而（CUR＆LT; = tot_col）
            {
                的printf（％0.2F，数额[CUR] /计数）;
                CUR ++;
            }            打印;            删除款项;
            tot_col = 0;
        }
    }'$ @
 

How can I substitute each n lines of a file with their average using awk?

This question answer perfectly, except for the fact that it handles only one column and the 3 is not parametrized: How to sum up every 10 lines and calculate average using AWK?

basically I would like to take a file like this

1     1
2     1
3     1
4     1
5     1
6     1
2.5   2.0
3.5   2.0
4     2.0

and obtain something like this:

2     1   
5     1
3.33  2.0

解决方案

Here's a complete shell script:

awk -v count=3 '
    {
        if ( NF > tot_col )
            tot_col = NF;

        cur = 1;
        while ( cur <= NF )
        {
            sums[cur] += $cur;
            cur++;
        }

        if ( ( NR % count ) == 0 )
        {
            cur = 1;
            while ( cur <= tot_col )
            {
                printf("%0.2f ", sums[cur] / count);
                cur++;
            }

            print "";

            delete sums;
            tot_col = 0;
        }
    }' "$@"

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