在Bash Shell脚本中创建.csv

Question

我想在Bash shell脚本中生成一个.csv文件。请考虑以下代码：

#!/bin/bash 

for f1 in *_8_kHz.wav 
do 
    for f2 in *_8_kHz.wav 
    do 
     # echo "pesq +8000 $f1 $f2" 
     echo -n `pesq +8000 $f1 $f2 | grep Prediction | rev | cut -b1-5 | rev` 
    done 

    echo 
done 

这个工作除了当然，每行都以逗号结尾。这里是示例输出：

4.500,1.029,1.651,1.475,1.698,1.706, 
1.550,4.500,1.477,1.148,1.788,1.478, 
1.251,0.958,4.500,1.472,2.091,1.800, 
0.961,1.154,1.550,4.500,1.702,1.501, 
1.194,0.974,1.356,1.206,4.500,1.626, 
0.857,0.960,1.091,1.064,2.012,4.500, 

什么是省略这些尾随逗号的最有效方法？

Answer 1

以下是否足够有效？

echo -n `pesq +8000 $f1 $f2 | grep Prediction | rev | cut -b1-5 | rev` | sed 's/,$//' 

它使用sed行$年底前删除最后,。

Answer 2

从 https://stackoverflow.com/a/18262725/2796606

foo=`pesq +8000 $f1 $f2 | grep Prediction | rev | cut -b1-5 | rev` 

echo -n ${foo%?}