Question

我有一个数据集，其中119列用空格分隔，喜欢选择列（3,7,11，...，47,51），（52,53,54，...，118,119）并打印他们到另一个文件。我怎么能用awk做到这一点？

Input file 

c1 c2 c3...c119

Output file

c3 c7 c11 ... c51 c52 c53 c54 ... c118 c119

感谢您的帮助

Answer 1

awk '{for(i=3;i<=51;i+=4) printf "%s ",$i ;for(i=52;i<=119;i++) {printf "%s ",$i} ;print ""}' file

Answer 2

{a=""; 
 for (i=0 ;i<=12; i++) {a = a  $(3+4 * i) " "};
 for (i=52 ;i<=119; i++) {a = a  $(i) " "};
 print a}

HTH！

Answer 3

这是你要找的吗？

awk '{s=""; for (i = 3; i <= 51; i+=4) {s = s $i " "}; for (i = 52; i <= 119; i++) {s = s $i " "}; print s}' inputfile

Answer 4

awk 'for(i=3;i<=119;i+a) 
     {
     if (i<52)
     {
     printf "%s ",$i};
     a=4;
     }
     else
     {
     printf "%s ",$i};
     a=1;
     }
     }' file

如何打印每第4列到第n列，从第（n + 1）列打印到最后使用awk？

4 个答案: