英文:
Merge multiple files based on regex
问题
如何使用for循环合并具有匹配字符/数字的两个带下划线的文件？目录中有许多文件
输入：
SRR9200887_1.fastq
SRR9200887_2.fastq
SRR9200888_1.fastq
SRR9200888_2.fastq
SRR9200889_1.fastq
SRR9200889_2.fastq
期望输出：
SRR9200887.fastq
SRR9200888.fastq
SRR9200889.fastq
我的尝试：
for l in $(ls *.fastq | cut -d_ -f1 | sort | uniq); do cat ${l}*.fastq
英文:
How do I merge two files with matching characters/digits before an underscore using a for loop? I have many files in the directory
Input:
SRR9200887_1.fastq
SRR9200887_2.fastq
SRR9200888_1.fastq
SRR9200888_2.fastq
SRR9200889_1.fastq
SRR9200889_2.fastq
Expected output:
SRR9200887.fastq
SRR9200888.fastq
SRR9200889.fastq
My attempt:
for l in $(ls *.fastq | cut -d_ -f1 | sort |uniq); do cat ${l}*.fastq
答案1
得分: 5
With bash 和它的 Parameter Expansion：
for i in *_1.fastq; do
cat "${i%_*.fastq}_1.fastq" "${i%_*.fastq}_2.fastq" > "${i%_*.fastq}.fastq";
done
${i%_*.fastq} 输出 $i 但不包含 _ 及其后的部分，例如 SRR9200887。
英文:
With bash and its Parameter Expansion:
for i in *_1.fastq; do
cat "${i%_*.fastq}_1.fastq" "${i%_*.fastq}_2.fastq" > "${i%_*.fastq}.fastq";
done
${i%_*.fastq} outputs $i without _ and all following it, e.g. SRR9200887.
答案2
得分: 3
for f in *_*.fastq; do cat "$f" >> "${f%_*}.fastq"; done
英文:
for f in *_*.fastq; do cat "$f" >> "${f%_*}.fastq"; done
答案3
得分: 2
为了将文件合并在一起，假设您有匹配的"_1.fastq"和"_2.fastq"文件，每个"SRR"对应一个，一个潜在的选项是：
SRR_array=(*_1.fastq)
for f in "${SRR_array[@]%%_*}"
do
cat "$f"_1.fastq "$f"_2.fastq > "$f".fastq
done
如果您想在合并后删除"_1.fastq"和"_2.fastq"文件：
SRR_array=(*_1.fastq)
for f in "${SRR_array[@]%%_*}"
do
cat "$f"_1.fastq "$f"_2.fastq > "$f".fastq
rm "$f"_1.fastq "$f"_2.fastq
done
英文:
To cat the files together, assuming you have matching "_1.fastq" and "_2.fastq" for every "SRR", one potential option is:
SRR_array=(*_1.fastq)
for f in "${SRR_array[@]%%_*}"
do
cat "$f"_1.fastq "$f"_2.fastq > "$f".fastq
done
If you wanted to delete the _1.fastq and _2.fastq files after merging them together:
SRR_array=(*_1.fastq)
for f in "${SRR_array[@]%%_*}"
do
cat "$f"_1.fastq "$f"_2.fastq > "$f".fastq
rm "$f"_1.fastq "$f"_2.fastq
done
答案4
得分: 1
One bash idea:
while read -r pfx
do
cat "${pfx}"_*.fastq >> "${pfx}".fastq
done < <(find . -name "*_*.fastq" | cut -d'_' -f1 | sort -u)
Tweaking OP's current code:
for l in $(ls -1 *_*.fastq | cut -d_ -f1 | sort | uniq)
do
cat ${l}_*.fastq >> "${l}".fastq
done
Where:
- we look for files with a
_in the name; if the script is run more than once this will insure we don't pick up the previous concatenated files - make sure
lslists one file per line (hence the-1) - in this case
sort | uniqcould be replaced withsort -u
英文:
One bash idea:
while read -r pfx
do
cat "${pfx}"_*.fastq >> "${pfx}".fastq
done < <(find . -name "*_*.fastq" | cut -d'_' -f1 | sort -u)
Tweaking OP's current code:
for l in $(ls -1 *_*.fastq | cut -d_ -f1 | sort | uniq)
do
cat ${l}_*.fastq >> "${l}".fastq
done
Where:
- we look for files with a
_in the name; if the script is run more than once this will insure we don't pick up the previous concatenated files - make sure
lslists one file per line (hence the-1) - in this case
sort | uniqcould be replaced withsort -u
答案5
得分: 0
使用任何awk（未经测试）：
FNR==1 {
out = FILENAME
sub(/_[0-9]+/,"",out)
if ( out != prev ) {
close(prev)
prev = out
}
}
{ print > out }
英文:
Using any awk (untested):
awk '
FNR==1 {
out = FILENAME
sub(/_[0-9]+/,"",out)
if ( out != prev ) {
close(prev)
prev = out
}
}
{ print > out }
' *_*.fastq
That will concatenate files with the same suffix no matter how many files have the same suffix, not just 2.
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