Question

我正在寻找这个问题的解决方案：我有一个文件（制表符分隔），就像我在下面的blockquote中显示的那样。如你看到的有与第一部分匹配的线条（粗体字段）。

chr4 164440449 165354407 G1 P8002-51-75
   chr1 220871675 220962596 G2 P2368-132-84
   chr1 220871675 220962596 G2 P2369-152-116
   chr1 220871675 220962596 G2 P2371-180-82
   chr1 220871675 220962596 G2 P2372-223-129
   chr1 220871675 220962596 G2 P2373-153-96
   chr1 220871675 220962596 G2 P2370-104-78
   chr5 126198405 126416440 G3 P9333-135-146
   chr5 126198405 126416440 G3 P9334-151-116

使用AWK或PERL如何设法获得以下输出，保留以制表符分隔的格式???一般的概念是尝试根据它的第一部分统一行，并追加最后一个字段

chr4 164440449 165354407 G1 P8002-51-75
   chr1 220871675 220962596 G2 P2368-132-84 P2369-152-116 P2371-180-82 P2372-223-129 P2373-153-96 P2370-104-78
   chr5 126198405 126416440 G3 P9333-135-146 P9334-151-116

一般的概念是尝试根据它的第一部分统一行，并附加最后一个字段

Answer 1

while (<DATA>) {
    ($x, $y) = /^(.*)\s([-\w]+)$/;
    push @{$hash{$x}}, $y;
}
while (($k, $v) = each %hash) {
    print $k, join("\t", @{$v}), "\n";
}
__DATA__
chr4 164440449 165354407 G1 P8002-51-75
chr1 220871675 220962596 G2 P2368-132-84
chr1 220871675 220962596 G2 P2369-152-116
chr1 220871675 220962596 G2 P2371-180-82
chr1 220871675 220962596 G2 P2372-223-129
chr1 220871675 220962596 G2 P2373-153-96
chr1 220871675 220962596 G2 P2370-104-78
chr5 126198405 126416440 G3 P9333-135-146
chr5 126198405 126416440 G3 P9334-151-116

Answer 2

使用perl的一种方式：

perl -ane '
    ## Save all fields but the last one as the key to compare between rows.
    $key = join qq|\t|, @F[ 0 .. $#F - 1 ];

    ## In first line or when current key is equal to previous key, save last
    ## field in an array and stop processing current row.
    if ( $. == 1 || $key eq $pkey ) {
        $pkey = $key;
        push @value, $F[ $#F ];
        next unless eof;
    }

    ## At this point, keys between rows are different, so print previous
    ## key with its values and begin to save the new one.
    printf qq|%s\n|, join qq|\t|, $pkey, @value;
    @value = ();
    push @value, $F[ $#F ];

    ## Exception: Last line with a new key, print it.
    if ( eof && $pkey ne $key ) {
    printf qq|%s\n|, join qq|\t|, $key, @value;
    }

    ## Save previous key.
    $pkey = $key;

' infile

假设infile包含您问题的数据，输出将为：

chr4    164440449       165354407       G1      P8002-51-75
chr1    220871675       220962596       G2      P2368-132-84    P2369-152-116   P2371-180-82    P2372-223-129   P2373-153-96    P2370-104-78
chr5    126198405       126416440       G3      P9333-135-146   P9334-151-116

结合线条元素

2 个答案: