从另一个文件中删除与特定模式匹配的行

Question

我有两个文件（我只显示了这些文件的开头）:

patterns.txt

m64071_201130_104452/13
m64071_201130_104452/26
m64071_201130_104452/46
m64071_201130_104452/49
m64071_201130_104452/113
m64071_201130_104452/147

myfile.txt

>m64071_201130_104452/13/ccs
ACAGTCGAGCG
>m64071_201130_104452/16/ccs
ACAGTCGAGCG
>m64071_201130_104452/20/ccs
CAGTCGAGCGC
>m64071_201130_104452/22/ccs
CACACATCTCG
>m64071_201130_104452/26/ccs
TAGACAATGTA

我应该得到这样的输出：

>m64071_201130_104452/13/ccs
ACAGTCGAGCG
>m64071_201130_104452/26/ccs
TAGACAATGTA

如果 patterns.txt 中的行与 myfile.txt 中的行匹配，我想创建一个新文件。我需要保留与相关模式关联的字母 ACTG。我使用：

for i in $(cat patterns.txt); do 
     grep -A 1 $i myfile.txt; done > my_newfile.txt

它可以工作，但创建新文件的速度非常慢...我处理的文件非常大但不是太多（patterns.txt 为 14M，myfile.txt 为 700M）。

我也尝试使用 grep -v，因为我有另一个文件，其中包含 patterns.txt 中不存在的 myfile.txt 的其他模式。但是还是一样的“速填文件”问题

如果您看到解决方案..

Answer 1

使用您显示的示例，请尝试执行以下操作。在 GNU awk.

中编写和测试

awk '
FNR==NR{
  arr[[=10=]]
  next
}
/^>/{
  found=0
  match([=10=],/.*\//)
  if((substr([=10=],RSTART+1,RLENGTH-2)) in arr){
    print
    found=1
  }
  next
}
found
'  patterns.txt myfile.txt

说明： 为以上添加详细说明。

awk '                         ##Starting awk program from here.
FNR==NR{                      ##Checking condition which will be TRUE when patterns.txt is being read.
  arr[[=11=]]                     ##Creating array with index of current line.
  next                        ##next will skip all further statements from here.
}
/^>/{                         ##Checking condition if line starts from > then do following.
  found=0                     ##Unsetting found here.
  match([=11=],/.*\//)            ##using match to match a regex to till / in current line.
  if((substr([=11=],RSTART+1,RLENGTH-2)) in arr){  ##Checking condition if sub string of matched regex is present in arr then do following.
    print                     ##Printing current line here.
    found=1                   ##Setting found to 1 here.
  }
  next                        ##next will skip all further statements from here.
}
found                         ##Printing the line if found is set.
'  patterns.txt myfile.txt    ##Mentioning Input_file names here.

Answer 2

另一个 awk:

$ awk -F/ '                            # / delimiter
NR==FNR {
    a[,]                           # hash patterns to a
    next
}
{
    if( tf=((substr(,2),) in a) )  # if first part found in hash
        print                          # output and store found result in var tf
    if(getline && tf)                  # read next record and if previous record was found
        print                          # output
}' patterns myfile

输出：

>m64071_201130_104452/13/ccs
ACAGTCGAGCG
>m64071_201130_104452/26/ccs
TAGACAATGTA

编辑：输出不找到的：

$ awk -F/ '                              # / delimiter
NR==FNR {
    a[,]                             # hash patterns to a
    next
}
{
    if( tf=((substr(,2),) in a) ) {  # if first part found in hash
        getline                          # consume the next record too
        next
    }
    print                                # otherwise output
}' patterns myfile

输出：

>m64071_201130_104452/16/ccs
ACAGTCGAGCG
>m64071_201130_104452/20/ccs
CAGTCGAGCGC
>m64071_201130_104452/22/ccs
CACACATCTCG

从另一个文件中删除与特定模式匹配的行

Removing lines which match with specific pattern from another file

awk

grep

fasta