从第二列的开头删除空格

remove whitespace from beginning of second column

我有一个以空格分隔的文件,如下所示:

12  12.57428314.57490104 ENSG00000065361 rs2271194 rs61939899
2  2.198148577.198835577 ENSG00000065413 rs4524134 rs2697288 rs6738721
6  6.84279922.84407274 ENSG00000065609 rs2016358 rs35791305
10  10.104585135.104956335 ENSG00000065613 rs72811696

我想从第二列中删除前导空格(有两个空格分隔第 1 列和第 2 列,而不是一个空格)。有谁知道这方面的 sed 或 awk 命令吗?

有删减:

cut -d " " -f 1,3- file

输出:

12 12.57428314.57490104 ENSG00000065361 rs2271194 rs61939899
2 2.198148577.198835577 ENSG00000065413 rs4524134 rs2697288 rs6738721
6 6.84279922.84407274 ENSG00000065609 rs2016358 rs35791305
10 10.104585135.104956335 ENSG00000065613 rs72811696

tr -s(或tr --squeeze-repeats)将删除重复字符。所以你如果要替换所有重复的空格,你可以这样写:

tr -s ' '   < input-file   > output-file

输入:

12  12.57428314.57490104 ENSG00000065361 rs2271194 rs61939899
2  2.198148577.198835577 ENSG00000065413 rs4524134 rs2697288 rs6738721
6  6.84279922.84407274 ENSG00000065609 rs2016358 rs35791305
10  10.104585135.104956335 ENSG00000065613 rs72811696

输出:

12.57428314.57490104 ENSG00000065361 rs2271194 rs61939899
2 2.198148577.198835577 ENSG00000065413 rs4524134 rs2697288 rs6738721
6 6.84279922.84407274 ENSG00000065609 rs2016358 rs35791305
10 10.104585135.104956335 ENSG00000065613 rs72811696

使用 GNU sed,在第一列

后用单个 space 替换多个白色-space 字符
sed -E 's/^(\S+)\s+/ /' ip.txt

对于其他版本,使用

  • [[:space:]] 对于 \s
  • [^[:space:]] 对于 \S

:blank:(space和制表符)而不是:space:(白色-space字符)

此 AWK 将所有出现的连续 space 替换为单个 space:

$ awk 'gsub(/ +/," ")' file 
12 12.57428314.57490104 ENSG00000065361 rs2271194 rs61939899
2 2.198148577.198835577 ENSG00000065413 rs4524134 rs2697288 rs6738721
6 6.84279922.84407274 ENSG00000065609 rs2016358 rs35791305
10 10.104585135.104956335 ENSG00000065613 rs72811696

只需删除每行的第一个 space:

$ sed 's/ //' file
12 12.57428314.57490104 ENSG00000065361 rs2271194 rs61939899
2 2.198148577.198835577 ENSG00000065413 rs4524134 rs2697288 rs6738721
6 6.84279922.84407274 ENSG00000065609 rs2016358 rs35791305
10 10.104585135.104956335 ENSG00000065613 rs72811696