如何多次复制列?
e.g。
输入
1 4771131 4772199 ENSMUSG00000103922 0 + 0.670011
1 4773206 4785739 ENSMUSG00000033845 0 - 95.0352
1 4778063 4779212 ENSMUSG00000102275 0 - 0.1806
1 4807788 4848410 ENSMUSG00000025903 0 + 110.078
输出
1 4771131 4772199 ENSMUSG00000103922 0 + 0.670011 0.670011 x 998 times
1 4773206 4785739 ENSMUSG00000033845 0 - 95.0352 95.0352 x 998 times
1 4778063 4779212 ENSMUSG00000102275 0 - 0.1806 0.1806 x 998 times
1 4807788 4848410 ENSMUSG00000025903 0 + 110.078 110.078 x 998 times
谢谢!
答案 0 :(得分:1)
使用简单的for循环并打印线条您想要的次数:
awk '{printf $0;for(i=1;i<=998;i++){printf("%s%s",$NF,i==998?"":" ")};print ""}' Input_file
答案 1 :(得分:1)
使用awk
,根据需要更改变量n=<your_interest>
值。
<强>一衬垫:强>
awk -v col=1 -v n=2 'function repeat(v, n,i){for(i=1; i<=n; i++)printf("%s%s",(i==1?"":OFS),v)}{for(i=1; i<=NF; i++)printf("%s%s",(i==col?repeat($i,n):$i),i==NF?RS:OFS)}' infile
<强>输入:强>
$ cat infile
1 4771131 4772199 ENSMUSG00000103922 0 + 0.670011
1 4773206 4785739 ENSMUSG00000033845 0 - 95.0352
1 4778063 4779212 ENSMUSG00000102275 0 - 0.1806
1 4807788 4848410 ENSMUSG00000025903 0 + 110.078
时
col=7 and v=5
$ awk -v col=7 -v n=5 'function repeat(v, n,i){for(i=1; i<=n; i++)printf("%s%s",(i==1?"":OFS),v)}{for(i=1; i<=NF; i++)printf("%s%s",(i==col?repeat($i,n):$i),i==NF?RS:OFS)}' infile
1 4771131 4772199 ENSMUSG00000103922 0 + 0.670011 0.670011 0.670011 0.670011 0.670011
1 4773206 4785739 ENSMUSG00000033845 0 - 95.0352 95.0352 95.0352 95.0352 95.0352
1 4778063 4779212 ENSMUSG00000102275 0 - 0.1806 0.1806 0.1806 0.1806 0.1806
1 4807788 4848410 ENSMUSG00000025903 0 + 110.078 110.078 110.078 110.078 110.078
假设您设置的第一列是
col=1
,那么
$ awk -v col=1 -v n=5 'function repeat(v, n,i){for(i=1; i<=n; i++)printf("%s%s",(i==1?"":OFS),v)}{for(i=1; i<=NF; i++)printf("%s%s",(i==col?repeat($i,n):$i),i==NF?RS:OFS)}' infile
1 1 1 1 1 4771131 4772199 ENSMUSG00000103922 0 + 0.670011
1 1 1 1 1 4773206 4785739 ENSMUSG00000033845 0 - 95.0352
1 1 1 1 1 4778063 4779212 ENSMUSG00000102275 0 - 0.1806
1 1 1 1 1 4807788 4848410 ENSMUSG00000025903 0 + 110.078
更好的可读性:
awk -v col=7 -v n=5 '
function repeat(v, n,i)
{
for(i=1; i<=n; i++)
printf("%s%s",(i==1?"":OFS),v)
}
{
for(i=1; i<=NF; i++)
printf("%s%s",(i==col?repeat($i,n):$i),i==NF?RS:OFS)
}
' infile
答案 2 :(得分:1)
Awk 解决方案(单print
次操作):
awk '{ n=998;r=$NF; while(--n) r=r FS $NF; print $0,r}' OFS='\t' file
答案 3 :(得分:1)
凭借格式化字符串的强大功能, awk 非常容易。
对于前。
$ awk -v count=3 '{s=sprintf("%0*s",count,""); gsub(/ /," "$NF,s); printf $0 s "\n"}' file
1 4771131 4772199 ENSMUSG00000103922 0 + 0.670011 0.670011 0.670011 0.670011
1 4773206 4785739 ENSMUSG00000033845 0 - 95.0352 95.0352 95.0352 95.0352
1 4778063 4779212 ENSMUSG00000102275 0 - 0.1806 0.1806 0.1806 0.1806
1 4807788 4848410 ENSMUSG00000025903 0 + 110.078 110.078 110.078 110.078
您可以将其修改为count=999
以获得所需的输出。