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Processing a file in BASH

Here is a sample file which provides counts of how many times a record is duplicated in a file

      2 D1111111111     AAAAAA
      2 D2222222222     BB2222
      2 D3333333333     CC3333
     39 D4444444444     EE4444
      2 T5555555555     DD5555
      2 D6666666666     FF6666
      2 T7777777777     GG7777
      2 D8888888888     HH8888


The first field in the row is a count of how many times the record appeared in a file.  Everything after that is the duplicate row

I would like to do the following:
for every row where the 2nd field starts with D, sum the 1st field.

So what I am expecting is the sum= 2+2+2+39+2+2 = 49 (the 2 rows that commenced with T is excluded)

I don't suspect this file of duplicates to be long, however I have been trying to accomplish this using awk and grep as a one liner,  rather than a line by line reading of the file in a for loop

Any help is appreciated.
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klyles95
Asked:
klyles95
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1 Solution
 
ozoCommented:
awk '$2~/^D/{sum+=$1}END{print sum}'   sample.file
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klyles95Author Commented:
Perfect!!
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klyles95Author Commented:
Hi ozo...one more question.  I am trying to count how many rows where the S2 started with D
awk '$2~/^D/{print NR}'   sample.file

Open in new window

is returning the line number where it appears rather than the count of rows.

Any ideas

Dont worry.  I figured it out

awk '$2~/^D/ {count++} END {print count}'   sample.file

Open in new window

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