Question

我有一个大约有500行和480K列的文件，我需要在末尾移动2,3和4列。我的文件是逗号分隔文件，有没有更快的方法来安排使用awk或sed？

Answer 1

您可以尝试以下解决方案 -

perl -F"," -lane 'print "@F[0]"," ","@F[4..$#F]"," ","@F[1..3]"' input.file

Answer 2

您可以轻松复制列，480K列移动时间过长。

$ awk 'BEGIN{FS=OFS=","} {print $0,$2,$3,$4}' input.file > output.file

这是什么样的数据格式？

Answer 3

使用5个字段进行测试：

$ cat foo
1,2,3,4,5
a,b,c,d,e
$ cat program.awk
{
    $6=$2 OFS $3 OFS $4 OFS $1  # copy fields to the end and $1 too
    sub(/^([^,],){4}/,"")       # remove 4 first columns
    $1=$5 OFS $1                # catenate current $5 (was $1) to $1 
    NF=4                        # reduce NF
} 1                             # print

运行它：

$ awk -f program.awk FS=, OFS=, foo
1,5,2,3,4
a,e,b,c,d

理论上说这应该有用：

{
    $480001=$2 OFS $3 OFS $4 OFS $1
    sub(/^([^,],){4}/,"")
    $1=$480000 OFS $1
    NF=479999 
} 1

编辑：它确实有效。

Answer 4

也许perl：

perl -F, -lane 'print join(",", @F[0,4..$#F,1,2,3])' file

或

perl -F, -lane '@x = splice @F, 1, 3; print join(",", @F, @x)' file

另一种方法：正则表达式

perl -lpe 's/^([^,]+)(,[^,]+,[^,]+,[^,]+)(.*)/$1$3$2/' file

使用500行文件对其进行定时，每行包含480,000个字段

$ time perl -F, -lane 'print join(",", @F[0,4..$#F,1,2,3])' file.csv > file2.csv
40.13user 1.11system 0:43.92elapsed 93%CPU (0avgtext+0avgdata 67960maxresident)k
0inputs+3172752outputs (0major+16088minor)pagefaults 0swaps

$ time perl -F, -lane '@x = splice @F, 1, 3; print join(",", @F, @x)' file.csv > file2.csv
34.82user 1.18system 0:38.47elapsed 93%CPU (0avgtext+0avgdata 52900maxresident)k
0inputs+3172752outputs (0major+12301minor)pagefaults 0swaps

纯文本操作是赢家

$ time perl -lpe 's/^([^,]+)(,[^,]+,[^,]+,[^,]+)(.*)/$1$3$2/' file.csv > file2.csv
4.63user 1.36system 0:20.81elapsed 28%CPU (0avgtext+0avgdata 20612maxresident)k
0inputs+3172752outputs (0major+149866minor)pagefaults 0swaps

Answer 5

另一种技术，只是bash：

while IFS=, read -r a b c d e; do
    echo "$a,$e,$b,$c,$d"
done < file

使用awk或sed排列列？

5 个答案: