从多个文件的列中的所有行中删除最后一位数字

Removing the last digit from all rows in a column in several files

在我拥有的所有文件的第 11 列,我想删除最后一位数字 - 0。我不确定如何快速执行此操作?

输入文件示例

chr1    13281   C   G   20  4   16.670  S   28  3   9.680   S   Germline    0.006377004201320391    0.8803370017576622  19  9   2   1
chr1    13302   C   T   12  10  45.450  Y   20  11  35.480  Y   Germline    3.992679813441744E-8    0.8451888548995793  13  7   9   2
chr1    13402   G   C   114 0   00  G   153 2   1.290   S   Somatic 1.0 0.3311046995506439  139 14  1   1
chr1    136048  C   T   19  12  38.710  Y   32  7   17.950  Y   Germline    4.6433264291396187E-7   0.98655920157275    15  17  2   5
chr1    871269  A   C   12  2   14.290  M   26  2   7.140   M   Germline    0.058009817045962773    0.9000000000000099  4   22  0   2
chr1    881627  G   A   0   31  1000    A   0   43  1000    A   Germline    4.280406726905674E-44   1.0 21  22

输出

chr1    13281   C   G   20  4   16.670  S   28  3   9.68    S   Germline    0.006377004201320391    0.8803370017576622  19  9   2   1
chr1    13302   C   T   12  10  45.450  Y   20  11  35.48   Y   Germline    3.992679813441744E-8    0.8451888548995793  13  7   9   2
chr1    13402   G   C   114 0   00  G   153 2   1.29    S   Somatic 1.0 0.3311046995506439  139 14  1   1
chr1    136048  C   T   19  12  38.710  Y   32  7   17.95   Y   Germline    4.6433264291396187E-7   0.98655920157275    15  17  2   5
chr1    871269  A   C   12  2   14.290  M   26  2   7.14    M   Germline    0.058009817045962773    0.9000000000000099  4   22  0   2
chr1    881627  G   A   0   31  1000    A   0   43  100 A   Germline    4.280406726905674E-44   1.0 21  22

你可以使用 awk,

awk '{sub(/[0-9]$/, "", )}1' file
sed 's/\(\([^[:space:]]\{1,\}[[:space:]]\{1,\}\)\{10\}[^[:space:]]*\)0//' YourFile
  • -i 内联编辑选项(在这种情况下 --POSIX 由于 GNU sed)