Memory 在shell脚本文件中多次Awking

我正在向一个slurm集群发送一个批处理文件，该集群将awks相同填充的字段并附加到输出中。前几行行行得通，但突然间就消失了。它在第二次回声之前就消失了。如果有人能调试或向我展示一个更好的脚本，那就太好了，但我想知道为什么或者当bash文件一遍又一遍地执行相同的工作时，它是否会崩溃

#!/bin/bash
#SBATCH -A b2014091
#SBATCH -p core
#SBATCH -n 1
#SBATCH -t 23:00:00
#SBATCH -J sbatch

#input was MK...vcf

input=$1

output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="1/1") && (substr($15,1,3)=="1/1") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt
output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="0/0") && (substr($15,1,3)=="0/0") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt
output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="1/1") && (substr($15,1,3)=="./.") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt
output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="./.") && (substr($15,1,3)=="1/1") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt
output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="0/0") && (substr($15,1,3)=="./.") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt
output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="./.") && (substr($15,1,3)=="0/0") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt

错误消息：

slurmstepd:获取\u退出\u代码任务0被信号终止

字段12-15的示例：

1/1:0,38:38:99:1149,114,0       1/1:0,39:39:99:1161,117,0       1/1:0,46:46:99:1383,138,0       0/1:21,18:39:99:413,0,524 

---

也许你可以改变这一切：

input=$1

output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="1/1") && (substr($15,1,3)=="1/1") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt
output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="0/0") && (substr($15,1,3)=="0/0") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt
output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="1/1") && (substr($15,1,3)=="./.") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt
output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="./.") && (substr($15,1,3)=="1/1") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt
output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="0/0") && (substr($15,1,3)=="./.") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt
output=$(grep -v '^#' ${input} | awk '(substr($12,1,3)=="./.") && (substr($15,1,3)=="0/0") {print $0}')
echo "$output\n" >> /proj/b2014091/nobackup/potHomoF1.txt

这个

awk '!/^#/ && substr($12,1,3)~/[01.]\/[01.]/ && substr($15,1,3)~/[01.]\/[01.]/' <<< $1 > /proj/b2014091/nobackup/potHomoF1.txt

或

---

当它失败时，错误消息是什么？您能给出一个从中获取子字符串的示例吗？这不包括我正在避免的字符串之一的0/1吗？可能类似于/\[01.]\/\1/是的。我可能认为你必须把它分开，但你仍然可以在一个awk中完成所有的工作。试试这个awk’！/^/&&substr$12,1,3~/1\/1\/0\/0\/\./&&substr$15,1,3~/1\/1\/0\/0\/\./'什么都没有发生，什么都没有。一分钟后完成，输出为空，没有错误消息。我不希望$12==1/1&&&$15==0/0，因为我认为这会给我。

awk '{a=substr($12,1,3);b=substr($15,1,3)} !/^#/ && ((a=="1/1" && b=="1/1") || (a=="0/0" && b=="0/0") || (a=="1/1" && b=="./.") || (a=="./." && b=="1/1") || (a=="0/0" && b=="./.") || (a=="./." && b=="0/0"))' $1 > /proj/b2014091/nobackup/potHomoF1.txt

input=$1
awk '{a=substr($12,1,3);b=substr($15,1,3)} !/^#/ && ((a=="1/1" && b=="1/1") || (a=="0/0" && b=="0/0") || (a=="1/1" && b=="./.") || (a=="./." && b=="1/1") || (a=="0/0" && b=="./.") || (a=="./." && b=="0/0"))' ${input} > /proj/b2014091/nobackup/potHomoF1.txt