在Bash Shell脚本中创建.csv
我正在尝试用bashshell脚本生成一个.csv文件。请考虑此代码:在Bash Shell脚本中创建.csv,bash,shell,Bash,Shell,我正在尝试用bashshell脚本生成一个.csv文件。请考虑此代码: #!/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 当然,除了每行以逗号结
#!/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
当然,除了每行以逗号结尾外,这是有效的。以下是示例输出:
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,
省略后面的逗号最有效的方法是什么?以下方法足够有效吗
echo -n `pesq +8000 $f1 $f2 | grep Prediction | rev | cut -b1-5 | rev` | sed 's/,$//'
它用于从中删除行$.结尾之前的最后一个
foo=`pesq +8000 $f1 $f2 | grep Prediction | rev | cut -b1-5 | rev`
echo -n ${foo%?}