结果哪儿不一样了,是大小吗?如果大小不一样通常是由于排序导致的压缩率不同。排完序去重之后应该都是一样的。
for i in Z102A; do
> echo "RUN CMD: bwa mem $BWA_INDEX $workdir/2.data_qc/${i}_1.clean.fq.gz \
> $workdir/2.data_qc/${i}_2.clean.fq.gz -t 16 -M \
> -R '@RG\tID:${i}\tLB:${i}\tPL:ILLUMINA\tSM:${i}' \
> |samtools view -bS -h - > $workdir/3.map/bwa/${i}.bam"
>
> bwa mem $BWA_INDEX $workdir/2.data_qc/${i}_1.clean.fq.gz \
> $workdir/2.data_qc/${i}_2.clean.fq.gz -t 16 -M \
> -R "@RG\tID:${i}\tLB:${i}\tPL:ILLUMINA\tSM:${i}" \
> |samtools view -bS -h - > $workdir/3.map/bwa/${i}.bam
> done
RUN CMD: bwa mem /work/my_reseq/ref/ZS11.v0.fa /work/my_reseq/2.data_qc/Z102A_1.clean.fq.gz /work/my_reseq/2.data_qc/Z102A_2.clean.fq.gz -t 16 -M -R '@RG\tID:Z102A\tLB:Z102A\tPL:ILLUMINA\tSM:Z102A' |samtools view -bS -h - > /work/my_reseq/3.map/bwa/Z102A.bam
[M::bwa_idx_load_from_disk] read 0 ALT contigs
[M::process] read 1072224 sequences (160000246 bp)...
[M::process] read 1072130 sequences (160000268 bp)...
for i in Z102A; do
> echo "RUN CMD: bwa mem $BWA_INDEX $workdir/2.data_qc/${i}_1.clean.fq.gz \
> $workdir/2.data_qc/${i}_2.clean.fq.gz -t 56 -M \
> -R '@RG\tID:${i}\tLB:${i}\tPL:ILLUMINA\tSM:${i}' \
> |samtools view -bS -h - > $workdir/3.map/bwa/${i}.bam"
>
> bwa mem $BWA_INDEX $workdir/2.data_qc/${i}_1.clean.fq.gz \
> $workdir/2.data_qc/${i}_2.clean.fq.gz -t 56 -M \
> -R "@RG\tID:${i}\tLB:${i}\tPL:ILLUMINA\tSM:${i}" \
> |samtools view -bS -h - > $workdir/3.map/bwa/${i}.bam
> done
RUN CMD: bwa mem /work/my_reseq/ref/ZS11.v0.fa /work/my_reseq/2.data_qc/Z102A_1.clean.fq.gz /work/my_reseq/2.data_qc/Z102A_2.clean.fq.gz -t 56 -M -R '@RG\tID:Z102A\tLB:Z102A\tPL:ILLUMINA\tSM:Z102A' |samtools view -bS -h - > /work/my_reseq/3.map/bwa/Z102A.bam
[M::bwa_idx_load_from_disk] read 0 ALT contigs
[M::process] read 3752560 sequences (560000074 bp)...
[M::process] read 3752124 sequences (560000292 bp)...