Question

我有很多* .csv文件。我要删除特定行之后的内容。我将删除20031231之后的所有行如何使用Shell脚本的某些行解决此问题？

Test,20031231,000107,0.74843,0.74813
Test,20031231,000107,0.74838,0.74808
Test,20031231,000108,0.74841,0.74815
Test,20031231,000108,0.74835,0.74809
Test,20031231,000110,0.74842,0.74818

Test,20040101,000100,0.73342,0.744318

Answer 1

快速又脏，但没有其他有关约束的信息

sed '1,/20031231/p;d' YourFile

Answer 2

如果要使用Shell脚本，最好使用awk。这将达到目的：

awk 'BEGIN {FS=","} {if ($2 == "20031231") print $0}' input.csv > output.csv

此代码将仅将具有20031231的行写入不同的文件。

Answer 3

忽略空行和不匹配的数据

awk文件：

$ cat awk.awk 
{
    if($2<="20031231" && $0!=""){
        print $0
    }else{
        next
    }
}

执行：

$ awk -F',' -f awk.awk input 
Test,20031231,000107,0.74843,0.74813
Test,20031231,000107,0.74838,0.74808
Test,20031231,000108,0.74841,0.74815
Test,20031231,000108,0.74835,0.74809
Test,20031231,000110,0.74842,0.74818

一个班轮：

$ awk -F',' '{if($2<="20031231" && $0!=""){print $0}else{next}}' input
Test,20031231,000107,0.74843,0.74813
Test,20031231,000107,0.74838,0.74808
Test,20031231,000108,0.74841,0.74815
Test,20031231,000108,0.74835,0.74809
Test,20031231,000110,0.74842,0.74818

Answer 4

与米勒（http://johnkerl.org/miller/doc/）

mlr --nidx --fs "," filter '$2>20031231' input

给你

Test,20040101,000100,0.73342,0.744318

Answer 5

使用awk，请尝试：

awk -F, '$2<=20031231' input.csv

在特定日期之后删除所有行

5 个答案: