Question

我有以下文件

cat file.txt

ID      Location
MNS1    NC_000004.12:g.d.a144120555T>C;NC_001423.23:c.a144120513G<C
MNS2    NC_000142.12:g.a144120552C,N>D
MNS3    NC_000142.12:g.a144120559C>N

我想用这种方式替换输入：

ID      Location
MNS1    NC_000004.12:144120555;NC_001423.23:144120513
MNS2    NC_000142.12:144120552
MNS3    NC_000142.12:144120559

我希望删除除:和;之间的数字以外的所有内容

例如，我尝试过：

echo "NC_000004.12:g.d.a144120555T>C;" | sed 's/:[^0-9]*/:/g; s/[^0-9]*;/;/g; s/[^0-9]*$//g'

期望的输出

NC_000004.12:144120555

Answer 1

这可能会成功！

sed -i.bak 's/g\.//g; s/\w>\w//g' filename

对于(NC.*?): concat，关于终端输出的位说明将有所帮助，尽管这可能会起作用：

s/NC[0-9]?:/:/

Answer 2

如果您可以选择Perl：

cat file.txt 
ID      Location
MNS1    NC_000004.12:g.d.a144120555T>C;NC_001423.23:c.a144120513G<C
MNS2    NC_000142.12:g.a144120552C,N>D
MNS3    NC_000142.12:g.a144120559C>N

perl -ape 's/:\D+(\d+).*?(?=;|$)/:$1/g' file.txt 
ID      Location
MNS1    NC_000004.12:144120555;NC_001423.23:144120513
MNS2    NC_000142.12:144120552
MNS3    NC_000142.12:144120559

说明：

s/            # substitute
    :           # colon
    \D+         # 1 or more non digits
    (\d+)       # group 1,, 1 or more digit
    .*?         # 0 or more any character but bewline, not greedy
    (?=;|$)     # positive lookahead, make sure we have semi-colon or end of line
/             # with
    :           # colon
    $1          # content of group 1 (i.e. the digits)
/g            # end, global

正则表达式操作和字符串替换-替代sed

2 个答案: