我有一个空格分隔的文件,看起来像:
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)。有没有人知道这个sed或awk命令?
随着削减:
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
)将删除重复的字符。所以如果你想要替换所有重复的空格,你可以写:
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
此AWK用单个空格替换所有出现的连续空格:
$ 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空格:
$ 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
使用GNU sed
,在第一列之后用单个空格替换多个空格字符
sed -E 's/^(\S+)\s+/\1 /' ip.txt
对于其他版本,请使用
[[:space:]]
为\s
[^[:space:]]
为\S
或:blank:
(空格和制表符)而不是:space:
(空白字符)