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Display how array elements are found in data file?

Display how array elements are found in data file?

Display how array elements are found in data file?

Hello and thanks for reading my post, I really appreciate it.

Need a way to read from the arrayfile and search the datafile (with arrayfile elements) in a way that prints out how the arrayfile elements are found. An awk one liner is preferred.


CODE --> bash


CODE --> bash


CODE --> bash

haha 1110111
hahh 0001000 
So its basically remove the last character from the arrayfile elements, search the datafile until you find the match, and if the next character after match is the same as was removed then write a 1 and if not then write a 0, It also needs to search in the most squeezed way possible (from beginning of each character in datafile, not by blocks like normal search in text editor, as portrayed in my previous post https://www.tek-tips.com/viewthread.cfm?qid=178648....

Thank you.

RE: Display how array elements are found in data file?

Got it by slightly modifying the script from previous post:


# Run:
#   awk -f starrysky1_02.awk datafile.txt arrayfile.txt

  # remove spaces
  gsub(/[ ]+/, "", $0)

  data_len = length(data)

  pattern = $0
  pattern_len = length(pattern)
  printf("%s\t", pattern)
  for (j=1; j+pattern_len-1 <= data_len; j++) {
    if (substr(data, j, pattern_len-1) ~ substr(pattern, 1, pattern_len-1)) {
      if (substr(data, j + pattern_len - 1, 1) ~ substr(pattern, pattern_len, 1)) {
        result = 1
      else {
        result = 0 
      printf("%s", result) 
  print ""



$ awk -f starrysky1_02.awk datafile.txt arrayfile.txt
haha	1110111
hahh	0001000 

Try to do from it an one liner.

RE: Display how array elements are found in data file?

I am eternally grateful!!! Thank you soo much !!

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