Ubuntu：如果制表符分隔文件包含特定字符串，我如何仅提取特定列？

Question

我想从 FC305JN_s_1_eland_result.txt 文件中提取带有 chr6.fa 的行。然后，我只想从此子文件中提取第 1、2、7、8 和 9 列。

grep -E "chr6.fa" FC305JN_s_1_eland_result.txt > out.txt
awk -F, '{OFS=",";print , , , , }' out.txt > outfile.txt
My out.txt is exactly the same as outfile.txt.

FC305JN_s_1_eland_result.txt 文件的小样本：

>FC305JN_20080525:1:15:1412:166 GTGAATCCTTATTCCGATATATATNNNN    U0  1   0   0   chrX.fa 45974622    R   ..
>FC305JN_20080525:1:15:944:72   GATGACTTCCTTAATTTTCTTTATNNNN    U0  1   0   0   chr6.fa 7200804 R   ..
>FC305JN_20080525:1:15:1049:473 GAATGGCAACACAAACAGGGCTGANNNN    R2  0   0   4
>FC305JN_20080525:1:15:1196:1959    GGGAGAAGCCTCCCCGCCTCGGCCNNNN    U2  0   0   1   chr17.fa    38386704    F   ..  17A 23T
>FC305JN_20080525:1:15:1034:505 GAAAATGTTTCAAATCAATTTCTANNNN    U0  1   0   0   chr2.fa 183305566   R   ..
>FC305JN_20080525:1:15:983:126  GGATAGAGAGTTTGCACTGAGTTGNNNN    U0  1   0   0   chrX.fa 92367529    F   ..
>FC305JN_20080525:1:15:1799:100 TTCAGCTTATTGATAAAGAAGCACNNNN    U0  1   0   0   chr6.fa 20979453    R   ..
>FC305JN_20080525:1:15:743:1028 GAATGGAATGGAATGGAAAGAAACNNNN    R1  0   33  255
>FC305JN_20080525:1:15:771:1076 GAGTTCACTAAACAAAAGAGTGTCNNNN    U2  0   0   1   chr6.fa 136877852   R   ..  7A  13G

当前输出outfile.txt（示例）：

>FC305JN_20080525:1:15:944:72   GATGACTTCCTTAATTTTCTTTATNNNN    U0  1   0   0   chr6.fa 7200804 R   ..
>FC305JN_20080525:1:15:1799:100 TTCAGCTTATTGATAAAGAAGCACNNNN    U0  1   0   0   chr6.fa 20979453    R   ..
>FC305JN_20080525:1:15:771:1076 GAGTTCACTAAACAAAAGAGTGTCNNNN    U2  0   0   1   chr6.fa 136877852   R   ..  7A  13G

期望的输出（示例）：

>FC305JN_20080525:1:15:944:72   GATGACTTCCTTAATTTTCTTTATNNNN    chr6.fa 7200804 R
>FC305JN_20080525:1:15:1799:100 TTCAGCTTATTGATAAAGAAGCACNNNN    chr6.fa 20979453    R
>FC305JN_20080525:1:15:771:1076 GAGTTCACTAAACAAAAGAGTGTCNNNN    chr6.fa 136877852   R

Answer 1

简化您的代码（使用从借用的代码）

grep -E "chr6.fa" FC305JN_s_1_eland_result.txt > out.txt
awk '{print , "\t", , "\t", , "\t", , "\t", }' out.txt > outfile.txt

产生输出：

FC305JN_20080525:1:15:944:72     GATGACTTCCTTAATTTTCTTTATNNNN    chr6.fa     7200804     R
FC305JN_20080525:1:15:1799:100   TTCAGCTTATTGATAAAGAAGCACNNNN    chr6.fa     20979453    R
FC305JN_20080525:1:15:771:1076   GAGTTCACTAAACAAAAGAGTGTCNNNN    chr6.fa     136877852   R

Answer 2

完整的 awk:

$ awk 'BEGIN {
    FS=OFS="\t"               # set correct delimiters
}
~/chr6\.fa/ {               # replaces the grep part
    print , , , ,   # output
}'  file                      # your file goes here

输出：

>FC305JN_20080525:1:15:944:72   GATGACTTCCTTAATTTTCTTTATNNNN    chr6.fa 7200804 R
>FC305JN_20080525:1:15:1799:100 TTCAGCTTATTGATAAAGAAGCACNNNN    chr6.fa 20979453        R
>FC305JN_20080525:1:15:771:1076 GAGTTCACTAAACAAAAGAGTGTCNNNN    chr6.fa 136877852       R

Ubuntu：如果制表符分隔文件包含特定字符串，我如何仅提取特定列？

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