使用 bash-script 修改 CSV 中的值
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modify value's in CSV with bash-script
提问by Mirco Schmidt
I've the following CSV file's:
我有以下 CSV 文件：
2012-07-12 15:30:09; 353.2
2012-07-12 15:45:08; 347.4
2012-07-12 16:00:08; 197.6
2012-07-12 16:15:08; 308.2
2012-07-12 16:30:09; 352.6
What I want to do is modify the value in the 2nd column...
我想要做的是修改第二列中的值...
What I already can do is extract the value and modify it this way:
我已经可以做的是提取值并以这种方式修改它：
#!/bin/bash
cut -d ";" -f2 > .tmp.csv
for num in $(cat .tmp.csv)
do
(echo "scale=2;$num/5" | bc -l >> .tmp2.csv)
done
rm .tmp.csv
rm .tmp2.csv
But I need to have column1 in that file too...
但我也需要在那个文件中有 column1 ......
I hope one of you can give me a hint, I'm just stuck!
我希望你们中的一个人能给我一个提示，我只是被卡住了！
采纳答案by Debaditya
From your code, this is what I understood
从你的代码，这就是我的理解
Input
输入
2012-07-12 15:30:09; 353.2
2012-07-12 15:45:08; 347.4
2012-07-12 16:00:08; 197.6
2012-07-12 16:15:08; 308.2
2012-07-12 16:30:09; 352.6
Awk code
awk 代码
awk -F ";" '{print $1 ";" $2/5}' input
awk -F ";" '{打印 $1";" $2/5}' 输入
Output
输出
2012-07-12 15:30:09;70.64
2012-07-12 15:45:08;69.48
2012-07-12 16:00:08;39.52
2012-07-12 16:15:08;61.64
2012-07-12 16:30:09;70.52
回答by Steve
One way, using awk:
一种方法，使用awk：
awk '{ $NF = $NF/5 }1' file.txt
Results:
结果：
2012-07-12 15:30:09; 70.64
2012-07-12 15:45:08; 69.48
2012-07-12 16:00:08; 39.52
2012-07-12 16:15:08; 61.64
2012-07-12 16:30:09; 70.52
HTH
HTH
回答by chepner
Here's an almost-pure bashsolution, without temp files:
这是一个几乎纯粹的bash解决方案，没有临时文件：
#!/bin/bash
while IFS=$';' read col1 col2; do
echo "$col1; $(echo "scale=2;$col2/5" | bc -l)"
done
回答by Birei
Try with awk:
尝试awk：
awk '
BEGIN {
## Split fields with ";".
FS = OFS = "; "
}
{
= sprintf( "%.2f", /5 )
print 2012-07-12 15:30:09; 70.64
2012-07-12 15:45:08; 69.48
2012-07-12 16:00:08; 39.52
2012-07-12 16:15:08; 61.64
2012-07-12 16:30:09; 70.52
}
' infile
Output:
输出：
##代码##