Problem with grep from pattern file with if


 
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# 1  
Data Problem with grep from pattern file with if

I have two sets of data

accept.txt is the list of 50 words
Code:
Tom
Anne
James
...

and I have a file01.txt which has the frequency of words which I got it using
Code:
cat main.file | tr -d '[:punct:]' | tr ' ' '\n' | tr 'A-Z' 'a-z' | sort | uniq -c | sort -n -r > file01.txt

so my file now looks like this
Code:
155 the
100 a
55  this
34  Anne
5   James

I need the output to look like this
Code:
0
34
5

I tried using grep but dont know how to replace things with 0 and just print one column
Code:
for name in `cat accept.txt ` 
do 
grep $name file01.txt;
done

I have to do this for 20 files in the same folder and would like to eventually print it as

file01 file 02 .... file20
Tom 0 54 .... 0
Anne 34 0 ... 43
James 0 43 .... 12

can someone please help me?
Thank you in advance for your time
# 2  
Try this script:

Code:
#!/bin/ksh

ls file* | sort | tr '\n' '\t' | sed 's/^/\t/g'
echo ""

for name in `cat accept.txt`
do
printf "${name}"
  for i_file in `ls file* | sort`
  do  
    count=`grep -c "${name}" "${i_file}"`
    printf "\t${count}"
  done
echo""
done

# 3  
Thanks for the help... I find a solution for if but yours look much more pro Smilie
 

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