I have below input and output needed.
input :
insert xxx to table xx.xxx.
1
insert yyy to table yy.yyy
10000
output:
insert yyy to table yy.yyy
10000
I want to print the a line before the value >= 10000.
tried wasnt wrking:
awk '($1>10000) {print$1)' < log2 > log3 | awk '/[0-9]$/ {print $1)' < log
You can use this awk:
awk '/10000/{print line ORS $0} {line=$0}' file
insert yyy to table yy.yyy
10000
You can use this code:
awk '{if (($1+0==$1) && $1 >= 10000) {print a; print}; a=$0}' < input
Here, ($1+0==$1) makes sure that $1 is a number, and $1 >= 10000 check that it is not smaller than 10000. If both conditions are met, it prints the current and the last line, which was saved using a=$0.
Like it's stated in the question, you need to get the before line and the pattern ('10000' for your example) line right?
I think the easiest way to do it is using grep:
grep -B 1 '^([1-9]\d{4,})$' file
or if you want to print it to an output file:
grep -B 1 '^([1-9]\d{4,})$' file > outputFile
It gets values that start with 1 and have 4 or more digits after. ^ and $ to guarantee that the line only contains the numeric value in it (^ is the line beginning and $ is the line end).
This, for the input you suggested will output:
insert yyy to table yy.yyy
10000
Related
I have a data file (file.txt) contains the below lines:
123 pro=tegs, ETA=12:00, team=xyz,user1=tom,dom=dby.com
345 pro=rbs, team=abc,user1=chan,dom=sbc.int,ETA=23:00
456 team=efg, pro=bvy,ETA=22:00,dom=sss.co.uk,user2=lis
I'm expecting to get the first column ($1) only if the ETA= number is greater than 15, like here I will have 2nd and 3rd line first column only is expected.
345
456
I tried like cat file.txt | awk -F [,TPF=]' '{print $1}' but its print whole line which has ETA at the end.
Using awk
$ awk -F"[=, ]" '{for (i=1;i<NF;i++) if ($i=="ETA") if ($(i+1) > 15) print $1}' input_file
345
456
With your shown samples please try following GNU awk code. Using match function of GNU awk where I am using regex (^[0-9]+).*ETA=([0-9]+):[0-9]+ which creates 2 capturing groups and saves its values into array arr. Then checking condition if 2nd element of arr is greater than 15 then print 1st value of arr array as per requirement.
awk '
match($0,/(^[0-9]+).*\<ETA=([0-9]+):[0-9]+/,arr) && arr[2]+0>15{
print arr[1]
}
' Input_file
I would harness GNU AWK for this task following way, let file.txt content be
123 pro=tegs, ETA=12:00, team=xyz,user1=tom,dom=dby.com
345 pro=rbs, team=abc,user1=chan,dom=sbc.int,ETA=23:00
456 team=efg, pro=bvy,ETA=02:00,dom=sss.co.uk,user2=lis
then
awk 'substr($0,index($0,"ETA=")+4,2)+0>15{print $1}' file.txt
gives output
345
Explanation: I use String functions, index to find where is ETA= then substr to get 2 characters after ETA=, 4 is used as ETA= is 4 characters long and index gives start position, I use +0 to convert to integer then compare it with 15. Disclaimer: this solution assumes every row has ETA= followed by exactly 2 digits.
(tested in GNU Awk 5.0.1)
Whenever input contains tag=value pairs as yours does, it's best to first create an array of those mappings (v[]) below and then you can just access the values by their tags (names):
$ cat tst.awk
BEGIN {
FS = "[, =]+"
OFS = ","
}
{
delete v
for ( i=2; i<NF; i+=2 ) {
v[$i] = $(i+1)
}
}
v["ETA"]+0 > 15 {
print $1
}
$ awk -f tst.awk file
345
456
With that approach you can trivially enhance the script in future to access whatever values you like by their names, test them in whatever combinations you like, output them in whatever order you like, etc. For example:
$ cat tst.awk
BEGIN {
FS = "[, =]+"
OFS = ","
}
{
delete v
for ( i=2; i<NF; i+=2 ) {
v[$i] = $(i+1)
}
}
(v["pro"] ~ /b/) && (v["ETA"]+0 > 15) {
print $1, v["team"], v["dom"]
}
$ awk -f tst.awk file
345,abc,sbc.int
456,efg,sss.co.uk
Think about how you'd enhance any other solution to do the above or anything remotely similar.
It's unclear why you think your attempt would do anything of the sort. Your attempt uses a completely different field separator and does not compare anything against the number 15.
You'll also want to get rid of the useless use of cat.
When you specify a column separator with -F that changes what the first column $1 actually means; it is then everything before the first occurrence of the separator. Probably separately split the line to obtain the first column, space-separated.
awk -F 'ETA=' '$2 > 15 { split($0, n, /[ \t]+/); print n[1] }' file.txt
The value in $2 will be the data after the first separator (and up until the next one) but using it in a numeric comparison simply ignores any non-numeric text after the number at the beginning of the field. So for example, on the first line, we are actually literally checking if 12:00, team=xyz,user1=tom,dom=dby.com is larger than 15 but it effectively checks if 12 is larger than 15 (which is obviously false).
When the condition is true, we split the original line $0 into the array n on sequences of whitespace, and then print the first element of this array.
Using awk you could match ETA= followed by 1 or more digits. Then get the match without the ETA= part and check if the number is greater than 15 and print the first field.
awk '/^[0-9]/ && match($0, /ETA=[0-9]+/) {
if(substr($0, RSTART+4, RLENGTH-4)+0 > 15) print $1
}' file
Output
345
456
If the first field should start with a number:
awk '/^[0-9]/ && match($0, /ETA=[0-9]+/) {
if(substr($0, RSTART+4, RLENGTH-4) > 15)+0 print $1
}' file
How to call the output file as the string in 4th column of output (or according to 4th column of ith row of the input)?
I tried:
for i in {1..321}; do
awk '(FNR==i) {outfile = $4 print $0 >> outfile}' RV1_phase;
done
or
for i in {1..321}; do
awk '(FNR==i) {outfile = $4; print $0}' RV1_phase > "$outfile";
done
input file:
1 2 2 a
4 5 6 f
4 4 5 f
....
....
desired input i=1
name: a
1 2 2 a
The aim: I have data that I plotted in gnuplot and I would like to plot set of figures named after string to know which point come from which file. The point will be coloured. I need to get files for plotting in gnuplot so I would like to create them using the cycle from my question.
Simply
for i in {1..321}; do
awk '(FNR==i) {print $0 >> $4}' RV1_phase;
done
The problem with your first attempt was that you didn't use a ; to separate the assignment to outfile from the print command. The separate variable isn't necessary, though.
You don't need a bash loop, either:
awk '1 <= FNR && FNR <= 321 {print $0 >> $4}' RV1_phase;
I have a file of two columns. The first column is dates and the second contains a corresponding number. The two commas are separated by a column. I want to take the average of the first three numbers and print it to a new file. Then do the same for the 2nd-4th number. Then 3rd-5th and so on. For example:
File1
date1,1
date2,1
date3,4
date4,1
date5,7
Output file
2
2
4
Is there any way to do this using awk or some other tool?
Input
akshay#db-3325:/tmp$ cat file.txt
date1,1
date2,1
date3,4
date4,1
date5,7
akshay#db-3325:/tmp$ awk -v n=3 -v FS=, '{
x = $2;
i = NR % n;
ma += (x - q[i]) / n;
q[i] = x;
if(NR>=n)print ma;
}' file.txt
2
2
4
OR below one useful for plotting and keeping reference axis (in your case date) at center of average point
Script
akshay#db-3325:/tmp$ cat avg.awk
BEGIN {
m=int((n+1)/2)
}
{L[NR]=$2; sum+=$2}
NR>=m {d[++i]=$1}
NR>n {sum-=L[NR-n]}
NR>=n{
a[++k]=sum/n
}
END {
for (j=1; j<=k; j++)
print d[j],a[j] # remove d[j], if you just want values only
}
Output
akshay#db-3325:/tmp$ awk -v n=3 -v FS=, -v OFS=, -f avg.awk file.txt
date2,2
date3,2
date4,4
$ awk -F, '{a[NR%3]=$2} (NR>=3){print (a[0]+a[1]+a[2])/3}' file
2
2
4
Add a little bit math tricks here, set $2 to a[NR%3] for each record. So the value in each element would be updated cyclically. And the sum of a[0], a[1], a[2] would be the sum of past 3 numbers.
updated based on the changes made due to the helpful feedback from Ed Morton
here's a quick and dirty script to do what you've asked for. It doesn't have much flexibility in it but you can easily figure out how to extend it.
To run save it into a file and execute it as an awk script either with a shebang line or by calling awk -f
// {
Numbers[NR]=$2;
if ( NR >= 3 ) {
printf("%i\n", (Numbers[NR] + Numbers[NR-1] + Numbers[NR-2])/3)
}
}
BEGIN {
FS=","
}
Explanation:
Line 1: Match all lines, "/" is the match operator and in this case we have an empty match which means "do this thing on every line". Line 3: Use the Record Number (NR) as the key and store the value from column 2 Line 4: If we have 3 or more values read from the file Line 5: Do the maths and print as an integer BEGIN block: Change the Field Separator to a comma ",".
Hi I have a text file with values
A VAL|1|2|3|
C VAL|2|2|3|
D VAL|1|2|3|
[No space between lines]
I want to replace the values in the above as per the first col i.e A VAL,C
VAL,D VAL,
so I want to
1. replace 3 from A VAL row
2. replace 2 value from C VAL row.
3. replace 1 value from D VAL row.
Basically I want to modify the above values by using AWK as AWK helps
treating csv , pipe delimited files
So I tried by using AWK command as
enter code here
`awk 'BEGIN {OFS=FS="|"} {if ($1="A") sub($4,"A1") ;elseif ($1="C") sub
($2,"B1"); print }' myval.txt`
*But I am getting wrong results *
C|B1|2|A1|B1C
C|B1|2|A1|B1C
C|B1|2|3|B1C
>The fisrt column itself is geting replace and the substitution is at wrong
>position.
**Expected output is **
A VAL|1|2|A1|
C VAL|2|2|B1|
D VAL|1|2|3|
You can try this awk:
awk 'BEGIN{OFS=FS="|"} $1 ~ /^A/{$(NF-1)="A1"} $1 ~ /^C/{$(NF-1)="B1"} 1' file.csv
A VAL|1|2|A1|
C VAL|2|2|B1|
D VAL|1|2|3|
awk 'BEGIN{OFS=FS="|"}{if(substr($1,0,1)=="A")sub($3,"A1",$3);else if(substr($1,0,1)=="C")sub($3,"B1",$3);else if(substr($1,0,1)=="D")sub($3,"3",$3);print }' inputtext.txt > outtext.txt
This is working fine
I'm trying to know which are the lines that are repeated X times in a text file, and I'm using awk but I see that awk in my command, not work with lines that begin with the same characters or words. That is, does not recognize the full line individually.
Using this command I try to get the lines that are repeated 3 times:
awk '++A[$1]==3' ./textfile > ./log
This is what you need hopefully:
awk '{a[$0]++}END{for(i in a){if(a[i]==3)print i}}' File
Increment array a with the line($0) as index for each line. In the end, for each index ($0), check if the count(a[i] which is the original a[$0]) equals 3. If so, print the line (i which is the original $0 / line). Hope it's clear.
This returns lines repeated 3 times but adds a space at the beginning of each 3x-repeated line:
sort ./textfile | uniq -c | awk '$1 == 3 {$1 = ""; print}' > ./log