Identifying genes from a list of genes

Question

I have a gene list file. Some thing like this

    SWT21
    SSA1
    NRP1
    EFB1
    TFC3
    MDM10

I have another file which also contains the names of these genes in my list along with other essential information about them. The second file looks like this:

chrI    147593  151166  YAL001C -   TFC3
chrI    143706  147531  YAL002W +   VPS8
chrI    142173  143160  YAL003W +   EFB1
chrI    140759  141407  YAL004W +   YAL004W
chrI    139502  141431  YAL005C -   SSA1
chrI    137697  138345  YAL007C -   ERP2
chrI    136913  137510  YAL008W +   FUN14
chrI    135853  136633  YAL009W +   SPO7
chrI    134183  135665  YAL010C -   MDM10

I want to extract those lines in the 2nd file which have gene names as are present in first file.

Answer 1

All you need is a simple grep:

grep -Fwf gene_list.txt gene_info.txt

The options used are:

• -w : Search for whole words, this ensures that the gene name ERK1 will not match the gene ERK12 (-w is not a standard option but is fairly common)
• -f : Read the patterns to be searched for from a file. In this case gene_list.txt.
• -F : Treat the patterns as strings, not as regular expressions. This ensures that a gene name like TOR* (if such a thing existed) would not match TORRRRRR.

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NOTE: This assumes that there are no spaces around the gene names in your list. If there are, you will need to remove them first (here with GNU sed):

sed -i 's/ //g' gene_list.txt

Answer 2

From an example on the awk wiki:

awk 'FNR==NR {arr[$0];next} $6 in arr' gene_list info_list

Explanation copied from the wiki entry:

• FNR == NR: This test is true when the number of records is equal to the number of records in the file. This is only true for the first file, for the second file NR will be equal to the number of lines of file1 + FNR.
• arr[$0]: This is a classic technique to create an array element index by the whole line. This will create an array with the first names of file1.
• next: this will skip to the next record so no more processing is done on file1.
• $6 in arr: This will only happen on the records of file2 because of the next. If $1 is present in arr, ie in file1, the default action will be executed and the line will be printed.

Answer 3

For more robustness you could alter the gene list into regular expressions to match at the end of the line. This should work (and be POSIX compatible):

sed 's/[[:space:]]*$/[[:space:]]*$/' gene_list | grep -wf - gene_info

Or with leading space in the gene_list (as per the example in the question):

sed 's/$/[[:space:]]*$/; s/[[:space:]]//' gene_list |
  grep -wf - gene_info

Of course this won't work if there are characters included in the gene names which are interpreted by regular expressions. Atlhough, if this is the case then you need to add extra replace expressions to sed to escape them, eg s:\[:\[:.

Answer 4

Assuming the gene list is list.txt and the info file is info.txt, as well as that the gene names never contain whitespace:

for item in $(cat list.txt); do grep "\s$item$" info.txt ; done

Output:

chrI    139502  141431  YAL005C -   SSA1
chrI    142173  143160  YAL003W +   EFB1
chrI    147593  151166  YAL001C -   TFC3
chrI    134183  135665  YAL010C -   MDM10

Answer 5

You can also use grep command in linux,by giving command like egrep -wi "complete genelist (1) with inducing pipe among genes" "file name containg all genes(2)".Here genelist(1) is your genelist says its 20 in number and geneliust (2) contains file with whole gene (1000 in number )