I have a csv file that contains this kind of values:
vm47,8,32794384Ki,16257320Ki
vm47,8,30223304245,15223080Ki
vm48,8,32794384Ki,16257312Ki
vm48,8,30223304245,15223072Ki
vm49,8,32794384Ki,16257320Ki
vm49,8,30223304245,15223080Ki
The columns 3 and 4 are memoy values expressed either in bytes, or kibibytes. The problem is that the "Ki" string appears randomly through the CSV file, particularly in column3, it's inconsistent.
So to make the file consistent, I need to convert everything in bytes. So basically, any value matching a trailing "Ki" needs to have its numeric value multiplied by 1024, and then replace the corresponding XXXXXKi match.
The reason why I want to do it with awk is because I am already using awk to generate that csv format, but I am happy to do it with sed too.
This is my code so far but obviously it's wrong as it's multiplying any value in columns 3 and 4 by 1024 even though it does not match "Ki". I am not sure at this point how to ask awk "if you see Ki at the end, then multiply by 1024".
kubectl describe node --context=$context| sed -E '/Name:|cpu:|ephemeral-storage:|memory:/!d' | sed 's/\s//g' | awk '
BEGIN {FS = ":"; OFS = ","}
{record[$1] = $2}
$1 == "memory" {print record["Name"], record["cpu"], record["ephemeral-storage"], record["memory"]}
' | awk -F, '{print $1,$2,$3,$3*1024,$4,$4*1024}' >> describe_nodes.csv
Edit: I made a mistake, you need to multiply by 128 to convert KiB in bytes, not 1024.
"if you see Ki at the end, then multiply by 1024
You may use:
awk 'BEGIN{FS=OFS=","} $3 ~ /Ki$/ {$3 *= 1024} $4 ~ /Ki$/ {$4 *= 1024} 1' file
vm47,8,33581449216,16647495680
vm47,8,30223304245,15588433920
vm48,8,33581449216,16647487488
vm48,8,30223304245,15588425728
vm49,8,33581449216,16647495680
vm49,8,30223304245,15588433920
Or a bit shorter:
awk 'BEGIN{FS=OFS=","} {
for (i=3; i<=4; ++i) $i ~ /Ki$/ && $i *= 1024} 1' file
With your shown samples/attempts, please try following awk code. Simple explanation would be, traverse through fields from 3rd field onwards and look for if a value has Ki(ignore cased manner) then multiply it with 128, print all edited/non-edited lines at last.
awk 'BEGIN{FS=OFS=","} {for(i=3;i<=NF;i++){if($i~/[Kk][Ii]$/){$i *= 128}}} 1' Input_file
You could try numfmt:
$ numfmt -d, --field 3,4 --from=auto --to=none <<EOF
vm47,8,32794384Ki,16257320Ki
vm47,8,30223304245,15223080Ki
EOF
vm47,8,33581449216,16647495680
vm47,8,30223304245,15588433920
Related
I have a table of numbers I am printing in awk using printf.
The printf accomplishes some truncation for the numbers.
(cat <<E\OF
Name,Where,Grade
Bob,Sydney,75.12
Sue,Sydney,65.2475
George,Sydney,84.6
Jack,Sydney,35
Amy,Sydney,
EOF
)|gawk 'BEGIN{FS=","}
FNR==1 {print("Name","Where","Grade");next}
{if ($3<50) {$3=0}
printf("%s,%s,%d \n",$1,$2,$3)}'
This produces:
Name Where Grade
Bob,Sydney,75
Sue,Sydney,65
George,Sydney,84
Jack,Sydney,0
Amy,Sydney,0
What I want is to display scores which are less than 50, or missing, as a dash ("-").
Name Where Grade
Bob,Sydney,75
Sue,Sydney,65
George,Sydney,84
Jack,Sydney,-
Amy,Sydney,-
This requires the 3rd string format in printf change from %d to %s.
So in some rows, the third column should be a value, and in some rows, the third column should be a string. How can I tell this to GAWK? Or should I just pipe through another awk to re-format?
$ gawk 'BEGIN{FS=","}
FNR==1 {print("Name","Where","Grade");next}
{if ($3<50) {$3="-"} else {$3=sprintf("%d", $3)}
printf("%s,%s,%s \n",$1,$2,$3)}' ip.txt
Name Where Grade
Bob,Sydney,75
Sue,Sydney,65
George,Sydney,84
Jack,Sydney,-
Amy,Sydney,-
use if-else to assign value to $3 as needed
sprintf allows to assign result of formatting to a variable
for this case, you could use int function as well
now printf will have %s for $3 as well
Assuming you missed the commas for the header and space after third column is not needed, you could do this with a simple one-liner
$ awk -F, -v OFS=, 'NR>1{$3 = $3 < 50 ? "-" : int($3)} 1' ip.txt
Name,Where,Grade
Bob,Sydney,75
Sue,Sydney,65
George,Sydney,84
Jack,Sydney,-
Amy,Sydney,-
?: ternary operator is alternate for if-else
1 is an awk idiom to print contents of $0
I have a single long column and want to reformat it into three comma separated columns, as indicated below, using awk or any Unix tool.
Input:
Xaa
Ybb
Mdd
Tmmn
UUnx
THM
THSS
THEY
DDe
Output:
Xaa,Ybb,Mdd
Tmmn,UUnx,THM
THSS,THEY,DDe
$ awk '{printf "%s%s",$0,NR%3?",":"\n";}' file
Xaa,Ybb,Mdd
Tmmn,UUnx,THM
THSS,THEY,DDe
How it works
For every line of input, this prints the line followed by, depending on the line number, either a comma or a newline.
The key part is this ternary statement:
NR%3?",":"\n"
This takes the line number modulo 3. If that is non-zero, then it returns a comma. If it is zero, it returns a newline character.
Handling files that end before the final line is complete
The assumes that the number of lines in the file is an integer multiple of three. If it isn't, then we probably want to assure that the last line has a newline. This can be done, as Jonathan Leffler suggests, using:
awk '{printf "%s%s",$0,NR%3?",":"\n";} END { if (NR%3 != 0) print ""}' file
If the final line is short of three columns, the above code will leave a trailing comma on the line. This may or may not be a problem. If we do not want the final comma, then use:
awk 'NR==1{printf "%s",$0; next} {printf "%s%s",(NR-1)%3?",":"\n",$0;} END {print ""}' file
Jonathan Leffler offers this slightly simpler alternative to achieve the same goal:
awk '{ printf("%s%s", pad, $1); pad = (NR%3 == 0) ? "\n" : "," } END { print "" }'
Improved portability
To support platforms which don't use \n as the line terminator, Ed Morton suggests:
awk -v OFS=, '{ printf("%s%s", pad, $1); pad = (NR%3?OFS:ORS)} END { print "" }' file
There is a tool for this. Use pr
pr -3ats,
3 columns width, across, suppress header, comma as separator.
xargs -n3 < file | awk -v OFS="," '{$1=$1} 1'
xargs uses echo as default action, $1=$1 forces rebuild of $0.
Using only awk I would go with this (which is similar to what proposed by #jonathan-leffler and #John1024)
{
sep = NR == 1 ? "" : \
(NR-1)%3 ? "," : \
"\n"
printf sep $0
}
END {
printf "\n"
}
eg, each row of the file is like :
1, 2, 3, 4,..., 1000
How can print out
1 2 3 4 ... 1000
?
If you just want to delete the commas, you can use tr:
$ tr -d ',' <file
1 2 3 4 1000
If it is something more general, you can set FS and OFS (read about FS and OFS) in your begin block:
awk 'BEGIN{FS=","; OFS=""} ...' file
You need to set OFS (the output field separator). Unfortunately, this has no effect unless you also modify the string, leading the rather cryptic:
awk '{$1=$1}1' FS=, OFS=
Although, if you are happy with some additional space being added, you can leave OFS at its default value (a single space), and do:
awk -F, '{$1=$1}1'
and if you don't mind omitting blank lines in the output, you can simplify further to:
awk -F, '$1=$1'
You could also remove the field separators:
awk -F, '{gsub(FS,"")} 1'
Set FS to the input field separators. Assigning to $1 will then reformat the field using the output field separator, which defaults to space:
awk -F',\s*' '{$1 = $1; print}'
See the GNU Awk Manual for an explanation of $1 = $1
I have a rather big file with 255 coma separated columns and I need to print out every third column only.
I was trying something like this
awk '{ for (i=0;i<=NF;i+=3) print $i }' file
but that doesn't seem to be the solution, since it prints to only one long column. Anybody can help? Thanks
Here is one way to do this.
The script prog.awk:
BEGIN {FS = ","} # field separator
{for (i = 1; i <= NF; i += 3) printf ("%s%c", $i, i + 3 <= NF ? "," : "\n");}
Invocation:
awk -f prog.awk <input.csv >output.csv
Example input.csv:
1,2,3,4,5,6,7,8,9,10
11,12,13,14,15,16,17,18,19,20
Example output.csv:
1,4,7,10
11,14,17,20
It behaves like that because by default awk splits fields in spaces. You have to tell it to split them with commas, and it's done using the FS variable or the -F switch. Besides that, first field is number one. The zero is the whole line, so also change the initial value of the for loop:
awk -F',' '{ for (i=1;i<=NF;i+=3) print $i }' file
I have a tab delimited file that looks something like this:
foo 0 4
boo 3 2
blah 4 0
flah 1 1
I am trying to calculate log2 for between the two columns for each row. my problem is with the division by zero
What I have tried is this:
cat file.txt | awk -v OFS='\t' '{print $1, log($3/$2)log(2)}'
when there is a zero as the denominator, the awk will crash. What I would want to do is some sort of conditional statement that would print an "inf" as the result when the denominator is equal to 0.
I am really not sure how to go about this?
Any help would be appreciated
Thanks
You can implement that as follows (with a few additional tweaks):
awk 'BEGIN{OFS="\t"} {if ($2==0) {print $1, "inf"} else {print $1, log($3/$2)log(2)}} file.txt
Explanation:
if ($2==0) {print $1, "inf"} else {...} - First check to see if the 2nd field ($2) is zero. If so, print $1 and inf and move on to the next line; otherwise proceed as usual.
BEGIN{OFS="\t"} - Set OFS inside the awk script; mostly a preference thing.
... file.txt - awk can read from files when you specify it as an argument; this saves the use of a cat process. (See UUCA)
awk -F'\t' '{print $1,($2 ? log($3/$2)log(2) : "inf")}' file.txt