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grep out load average from uptime
(8 answers)
Closed 3 years ago.
When I issue a command on the terminal uptime, it returns:
13:21:52 up 13:02, 3 users, load average: 1.10, 1.09, 0.96
Then, I want to print the load averages which are at field 8,9,10 by using:
uptime | awk '{print $8}' | tr -d ","
Then the result will be: 1.10
But, sometimes, when the pc is up for more than a day, it returns:
13:22:12 up 6 days, 19:50, 3 users, load average: 1.10, 1.09, 0.96
As you can see, the number of field columns is not fixed. I don't want to manually edit the field number I want to print. So, I thought of reading the line starting from the right side. With that, 0.96 will be the field 0, 1.09 will be field 1, and so on. I don't care about the values near the left side.
You can do it with a single call to awk. The basic call would be:
uptime | awk '{ print $(NF - 2) }'
To get rid of the comma at the same time instead of creating another subshell with tr, you could do:
uptime | awk '{ gsub(/,/,""); print $(NF - 2) }'
Let me know if you have any further issues.
In this way, you do not to worry about uptime issue.
cat /proc/loadavg|awk '{print $1,$2,$3}'
Another command with rev because rev is fun :
uptime | rev | cut -d , -f 3 | rev
If you just split by comma, the number of fields may vary, but if you split by colon first, and then by comma, the figure you are interested in is the first field in the last group:
uptime | cut -d : -f 5 | cut -d , -f 1
I have the following script:
curl -s 'https://someonepage=5m' | jq '.[]|.[0],.[1],.[2],.[3],.[4],.[5],.[6],.[7],.[8],.[9],.[10],.[11],.[12]' | perl -p -e 's/\"//g' |awk '/^[0-9]/{print; if (++onr%12 == 0) print ""; }'
This is part of result:
1517773500000
0.10250100
0.10275700
0.10243500
0.10256600
257.26700000
1517773799999
26.38912220
1229
104.32200000
10.70579910
0
1517773800000
0.10256600
0.10268000
0.10231600
0.10243400
310.64600000
1517774099999
31.83806883
1452
129.70500000
13.29758266
0
1517774100000
0.10243400
0.10257500
0.10211800
0.10230000
359.06300000
1517774399999
36.73708621
1296
154.78500000
15.84041910
0
I want to insert this data in a MySQL database. I want for each line this result:
(1517773800000,0.10256600,0.10268000,0.10231600,0.10243400,310.64600000,1517774099999,31.83806883,1452,129.70500000,13.29758266,0)
(1517774100000,0.10243400,0.10257500,0.10211800,0.10230000,359.06300000,151774399999,36.73708621,1296,154.78500000,15.84041910,0)
I need merge lines each 12 lines, any can help me for get this result.
Here's an all-jq solution:
.[] | .[0:12] | #tsv | gsub("\t";",") | "(\(.))"
In the sample, all the subarrays have length 12, so you might be able to drop the .[0:12] part of the pipeline. If using jq 1.5 or later, you could use join(“,”) instead of the #tsv|gsub portion of the pipeline. You might, for example, want to consider:
.[] | join(“,”) | “(\(.))”. # jq 1.5 or later
Invocation: use the -r command-line option
Sample output:
(1517627400000,0.10452300,0.10499000,0.10418200,0.10449400,819.50400000,1517627699999,85.57150693,2340,452.63400000,47.27213035,0)
(1517627700000,0.10435700,0.10449200,0.10366000,0.10370000,717.37000000,1517627999999,74.60582079,1996,321.25500000,33.42273846,0)
(1517628000000,0.10376600,0.10390000,0.10366000,0.10370400,519.59400000,1517628299999,53.88836170,1258,239.89300000,24.88613854,0)
$ awk 'BEGIN {RS=""; OFS=","} {$1=$1; $0="("$0")"}1' file
(1517773500000,0.10250100,0.10275700,0.10243500,0.10256600,257.26700000,1517773799999,26.38912220,1229,104.32200000,10.70579910,0)
(1517773800000,0.10256600,0.10268000,0.10231600,0.10243400,310.64600000,1517774099999,31.83806883,1452,129.70500000,13.29758266,0)
(1517774100000,0.10243400,0.10257500,0.10211800,0.10230000,359.06300000,1517774399999,36.73708621,1296,154.78500000,15.84041910,0)
RS="":
Treat groups of lines separated one or more blank lines as a record
OFS=","
Set the output separator to be a ","
$1=$1
Reconstitute the line, replacing the input separators with the output separator
$0="("$0")"
Surround the record with parens
1
Print the record
Good day all,
I am running below command:
netstat -an | awk '/:25/{ print $4 }' | sed 's/:25//' | paste -sd ',' -
which produces
192.168.2.22,127.0.0.1
I would like to amend the result to something like below (to be parsed as a csv by an application)
Manuallyaddedtext 192.168.2.22,127.0.0.1
Many thanks
echo -n "Mytext " ; netstat...
I'm new in awk scripting and would like to have some help in calculating 95th percentile value for a file that consist of this data:
0.0001357
0.000112
0.000062
0.000054
0.000127
0.000114
0.000136
I tried:
cat filename.txt | sort -n |
awk 'BEGIN{c=0} {total[c]=$1; c++;} END{print total[int(NR*0.95-0.5)]}'
but I dont seem to get the correct value when I compare it to excel.
I am not sure if Excel does some kind of weighted percentile, but if you actually want one of the numbers that was in your original set, then your method should work correctly for rounding.
You can simplify a little bit like this, but it's the same thing.
sort -n input.txt | awk '{all[NR] = $0} END{print all[int(NR*0.95 - 0.5)]}'
Following the calculation suggested here, you can do this:
sort file -n | awk 'BEGIN{c=0} length($0){a[c]=$0;c++}END{p5=(c/100*5); p5=p5%1?int(p5)+1:p5; print a[c-p5-1]}'
Output for given input:
sort file -n | awk 'BEGIN{c=0} length($0){a[c]=$0;c++}END{p5=(c/100*5); p5=p5%1?int(p5)+1:p5; print a[c-p5-1]}'
0.0001357
Explanation:
Sort the file numerically
drop the top 5%
pick the next value
PS. The statement p5=p5%1?int(p5)+1:p5 is doing a ceil operation available in many languages.
Just for the record, there is also solution, inspired by merlin2011 answer, that prints several desired percentiles:
# get amount of values
num="$(wc -l input.txt | cut -f1 -d' ')";
# sort values
sort -n input.txt > temp && mv temp input.txt
# print the desired percentiles
for p in 50 70 80 90 92 95 99 100; do
printf "%3s%%: %-5.5sms\n" "$p" "$(head input.txt -n "$((num / 100 * $p))" | tail -n1)";
done
Update: I messed it up. Bash math can't handle floating numbers, even not if used during a "single expression". That only works for files with 100*(N>0) values. So either bc or awk is required to do the math.
In case you have an "odd" amount of values, you should replace "$((num / 100 * $p))" with "$(awk "BEGIN {print int($num/100*$p)}")" in the code above.
Finally awk is part of that answer. ;)
I have set of files that looks like the following. I'm looking for a good way to count all files that have unique prefixes, where "prefix" is defined by all characters before the second hyphen.
0406-0357-9.jpg 0591-0349-9.jpg 0603-3887-27.jpg 59762-1540-40.jpg 68180-517-6.jpg
0406-0357-90.jpg 0591-0349-90.jpg 0603-3887-28.jpg 59762-1540-41.jpg 68180-517-7.jpg
0406-0357-91.jpg 0591-0349-91.jpg 0603-3887-29.jpg 59762-1540-42.jpg 68180-517-8.jpg
0406-0357-92.jpg 0591-0349-92.jpg 0603-3887-3.jpg 59762-1540-5.jpg 68180-517-9.jpg
0406-0357-93.jpg 0591-0349-93.jpg 0603-3887-30.jpg 59762-1540-6.jpg
Depending on what you actually want output, either of these might be what you want:
ls | awk -F'-' '{c[$1"-"$2]++} END{for (p in c) print p, c[p]}'
or
ls | awk -F'-' '!seen[$1,$2]++{count++} END{print count+0}'
If it's something else, update your question to show the output you're looking for.
This should do it:
ls *.jpg | cut -d- -s -f1,2 | uniq | wc -l
Or if your prefixes are always 4 digits, one dash, 4 digits, you don't need cut:
ls *.jpg | uniq -w9 | wc -l
Parses ls (bad, but it doesn't look like it will cause a problem with these filenames),
uses awk to set the field separator as -.
!seen[$1,$2]++) uses an associative array with $1,$2 as the key and increments, then checks if the value equals 0 to ensure it is only printed once (based on $1 and $2).
print prints on screen :)
ls | awk 'BEGIN{FS="-" ; printf("%-20s%-10s\n","Prefix","Count")} {seen[$1"-"$2]++} END{ for (k in seen){printf("%-20s%-10i\n",k,seen[k])}}'
Will now count based on prefix with headers :)