AWK或sed的在文件中改变列值(awk or sed to change column value

我有一个数据csv文件如下

16:47:07,3,r-4-VM,230000000.,0.466028518635,131072,0,0,0,60,0
16:47:11,3,r-4-VM,250000000.,0.50822578824,131072,0,0,0,0,0
16:47:14,3,r-4-VM,240000000.,0.488406067907,131072,0,0,32768,0,0
16:47:17,3,r-4-VM,230000000.,0.467893525702,131072,0,0,0,0,0

我想缩短在第5列的值。

所需的输出

16:47:07,3,r-4-VM,230000000.,0.46,131072,0,0,0,60,0
16:47:11,3,r-4-VM,250000000.,0.50,131072,0,0,0,0,0
16:47:14,3,r-4-VM,240000000.,0.48,131072,0,0,32768,0,0
16:47:17,3,r-4-VM,230000000.,0.46,131072,0,0,0,0,0

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Answer 1:

awk '{$5=sprintf( "%.2g", $5)} 1' OFS=, FS=, input

这将圆化打印.47 ，而不是.46的第一线，但也许这是可取的。

Answer 2:

试试这个：

cat filename | sed 's/\(^.*\)\(0\.[0-9][0-9]\)[0-9]*\(,.*\)/\1\2\3/g'

到目前为止，输出为GNU / Linux的标准输出，所以

cat filename | sed 's/\(^.*\)\(0\.[0-9][0-9]\)[0-9]*\(,.*\)/\1\2\3/g' > out_filename

将发送所需的结果out_filename

Answer 3:

如果舍入是不期望的，即， 0.466028518635需要被打印为0.46 ，使用：

cat <input> | awk -F, '{$5=sprintf( "%.4s", $5)} 1' OFS=,

（这可以的另一示例无用使用猫 ）

Answer 4:

你想在Perl中，这是它：

perl -F, -lane '$F[4]=~s/^(\d+\...).*/$1/g;print join ",",@F' your_file

下面的测试：

> cat temp
16:47:07,3,r-4-VM,230000000.,0.466028518635,131072,0,0,0,60,0
16:47:11,3,r-4-VM,250000000.,10.50822578824,131072,0,0,0,0,0
16:47:14,3,r-4-VM,240000000.,0.488406067907,131072,0,0,32768,0,0
16:47:17,3,r-4-VM,230000000.,0.467893525702,131072,0,0,0,0,0
> perl -F, -lane '$F[4]=~s/^(\d+\...).*/$1/g;print join ",",@F' temp
16:47:07,3,r-4-VM,230000000.,0.46,131072,0,0,0,60,0
16:47:11,3,r-4-VM,250000000.,10.50,131072,0,0,0,0,0
16:47:14,3,r-4-VM,240000000.,0.48,131072,0,0,32768,0,0
16:47:17,3,r-4-VM,230000000.,0.46,131072,0,0,0,0,0

Answer 5:

sed -r 's/^(([^,]+,){4}[^,]{4})[^,]*/\1/' file.csv

Answer 6:

这可能会为你工作（GNU SED）：

sed -r 's/([^,]{,4})[^,]*/\1/5' file

这取代非逗号的不超过4个字符长度的第五occurence。

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