从file2中给出的特定单词开始提取file1中的行

Question

I have one file "file1" that contains lines like:

643 2   3   4   5
6433    2   3   4   5
64  2   3   4   5
1234    2   3   4   5
1240    2   3   4   5
12  2   3   4   5

and I would like to extract from it all lines whose first word contained in file 2, which is like:

12
64

Thus, the final result should be:

12  2   3   4   5
64  2   3   4   5

In bash I think I have to use a loop for examining each word in file 2, but I do not know the command to extract the line in file1 containing the exact word.

For example, using:

sed -n '/^64/p' file1

I get:

643 2 3 4 5 6433 2 3 4 5 64 2 3 4 5

which is not correct, because I would like only line: 64 2 3 4 5

Do you know a bash method (sed, grep, awk, or python if you prefer) to do it?

Answer 1

I'd say:

awk 'NR == FNR { a[$1] = 1; next } a[$1]' file2 file1

That is:

NR == FNR {    # while processing the first file (file2)
  a[$1] = 1    # remember what values you saw
  next         # do nothing else
}
a[$1]          # after that (while processing file1): print those whose first
               # field was seen in the pass over file2.

Answer 2

You can use:

awk 'NR==FNR{a[$1]; next} $1 in a' file2 file1
64  2   3   4   5
12  2   3   4   5

Answer 3

I think you can try with grep -w and more exactly:

 -w    Searches for the expression as a word as if surrounded
       by \< and \>.

so yu can try:

grep -w 64 file1

Running on Solaris 10

Answer 4

要使用grep，还要在搜索文件中的模式中添加锚点：

grep -wf <(sed 's/^/^/' file2) file1