foo bar 12,300.50
foo bar 2,300.50
abc xyz 1,22,300.50
How do I replace all , from 3rd field using awk and pass output to bc -l in the following format to get sum of all numbers:
12300.50+2300.50+1,22,300.50
You can use gsub() funcion as follows. The syntax is:
gsub("find", "replace")
gsub("find-regex", "replace")
gsub("find-regex", "replace", t)
gsub(r, s [, t])
From the awk man page:
For each substring matching the regular expression r in the string t, substitute the string s, and return the number of substitutions. If t is not supplied, use $0. An & in the replacement text is replaced with the text that was actually matched. Use \& to get a literal &.
You can also use the following syntax:
gensub(r, s, h [, t])
From the awk man page:
Search the target string t for matches of the regular expression r. If h is a string beginning with g or G, then replace all matches of r with s. Otherwise, h is a number indicating which match of r to replace. If t is not supplied, $0 is used instead. Within the replacement text s, the sequence \n, where n is a digit from 1 to 9, may be used to indicate just the text that matched the n’th parenthesized subexpression. The sequence \0 represents the entire matched text, as does the character &. Unlike sub() and gsub(), the modified string is returned as the result of the function, and the original target string is not changed.
Example
Create a data file cat /tmp/data.txt
foo bar 12,300.50 foo bar 2,300.50 abc xyz 1,22,300.50
Type the following awk command:
awk '{ gsub(",","",$3); print $3 }' /tmp/data.txt |
Sample outputs:
12300.50 2300.50 122300.50
You can pass the output to any command or calculate sum of the fields:
awk 'BEGIN{ sum=0} { gsub(",","",$3); sum += $3 } END{ printf "%.2f\n", sum}' /tmp/data.txt |
OR build the list and pass to the bc -l:
awk '{ x=gensub(",","","G",$3); printf x "+" } END{ print "0" }' /tmp/data.txt | bc -l |
Nice tutorial :-)
gsub doesn’t work for me…
data.txt
user1 123,433,345.55
user3 12.50
awk ‘{gsub(“.”,”,”,$2); print $2}’ ./data.txt
result:
,,,,,,,,,,,,,,
,,,,,
the “.” is being evaluated as a wildcard:
In regular expressions, the period (“.”, also called “dot”) is the wildcard pattern character that matches a single character. Combined with the asterisk operator (.*) it will match any number of characters.
Add the “\” escape character in front of it so that it will be interpreted literally.
Too much commands.
Awk can substitute and delete fields in a string with a //// syntax.
Can you please explain what does it mean by //// syntax
i am trying to open a virus data file and replace dna strings with number patterns generated from formula in hopes to innoculate the h.i.v. virus.
awk > virus-data ‘{ gsub(“a”, “84”, $1); gsub(“b”, “177”, $2); gsub(“c”, “16984”, $3)}’ >> pattern
liquiddb.net
Code from the H.I.V. virus can be computed with a formula simplifying strains into segments that would fit into immunity code.
1 2 3 4 5 6 7 8(n)
-4 -4 -4 -4 -4 -4 -4 -4
+1 +1 +1 +1 +1 +1 +1 +1
==========================
-3 3 -1 5 1 7 3 n
2 -2 4 0 6 2 8 4
| | | | | | | |
-1__1__3__5__7__n__n+3_n+4
| | | |
0 + n + 7n + n^2+7
— -1 —
| |- 0
2-| – 1 —
| |
–| 3 —
n-| |- n
–| 5 —
| |
| – 7 —
n^2-| |- 7n
| –n —
| |
–| 3+n–
n^2+4-| |- n^2 +7
–n+4–
0 + n + 7n +n^2 + 7 (virus sequence:01)
01
716
1121691
161627169116016
161627169116016
055545653805615
050011112585154
055010001333441
050511001200103
055540101120113
050014111012102
055013300111112
050512030100001—+
055541233110001—|—+
050013110201001—|—|–+
055012201221101—|—|–|–+
050511021101011—|—|–|–|–+
055540121011110 | | | | |
050014111110001 | | | | |
055013300001001 | | | | |
050512030001101 | | | | |
055541233001011 | | | | |
050013110301110 | | | | |
055012201331001 | | | | |
050511021202101 | | | | |
055540121122111 | | | | |
050014111010100 | | | | |
055013300111110 | | | | |
050512030100001—+ | | | |
055541233110001——-+ | | |
050013110201001———-+ | |
055012201221101————-+ |
050511021101011—————-+
2 + n + n^2 + n^2 +4 (virus sequence:01)
01
69
84177
142429111111
032227800000
031005180000
032105478000
031115131800
032004422780
031204020518
032124422547
031112020313
032001222322
031201100110
032101010101
031111111111
032000000000
031200000000
032120000000
031112000000
031112000000
032001200000
031201120000
032121012000
031111111200
032000000120
031200000112+
032120000101
031112000111
032001200100
031201120110
032121012101
With Python, C and Bash under my belt an extra language made no sense so I’m now using piep a Python pipeline processor for this sort of thing. With piep you do it like this (p is lines, pp is stream):
cat data.txt | piep ‘p.split()[2]’
Pipe the output of cat to piep, split each line into lists then take the third item (Python lists count from 0). So far it seems a good replacement for awk and sed although I can’t seem to give up on grep and pgrep.
any one can solve this problem,
i have one folder, in that many file names are there. i need to print the path of the files by using awk script.
please tell me this is possible or not.
Hi Anil,
it is possible..
You need to use readlink -f
You can use this one line script to get what you want, i hope this one helps
for line in `ls -ltrh|awk -F” ” ‘{print $9}’`; do readlink -f $line; done
Didn’t work for me on Mac OS X. I had to use…
awk '{ gsub("find","replace"); print }' ./file.txtinstead of the suggested…
awk '{ gsub("find","replace",$3); print $3 }' ./file.txtHi,
I have a file with a keyword ‘FIX(number number) ‘ in it. This keyword occurs around 1000+ times.
I want to carry out a multiplication operation on the numbers inside ‘FIX’.
I could get all the keywords and could carry out the operation using awk, but I need to replace the new result in my original file.
Is there a way I can awk, process the data and replace it into an original file ?
Very nice! Thank you for the post. Helped me solve a several problems.