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How to calculate difference:?

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# 1  
Old 07-23-2013
How to calculate difference:?
Experts,

file1 : Want to find the difference of $3 field from next line's 3rd field,


The difference to be calculated from next lines 3rd field, to current lines lines 3rd field.


file1 :
Code:
Jun24_2013.06242013     3301244928 3133059904 167370640   95%
Jun25_1124.06252013     3301244928 3145497448 154988472   95%
Jun26_1016.06262013     3301244928 3157360376 143177152   96%
Jun27_1112.06272013     3301244928 3144037776 156356096   95%
Jun27_1714.06272013     3301244928 3144066576 156327520   95%
Jun28_1036.06282013     3301244928 3155272808 145171224   96%
Jun28_1114.06282013     3301244928 3155276936 145167128   96%
Jul01_1105.07012013     3301244928 3100088440 200823992   94%
Jul02_1033.07022013     3301244928 3112573560 188395312   94%
Jul03_1015.07032013     3301244928 3124571632 176449880   95%
Jul04_1404.07042013     3301244928 3136850640 164224552   95%
Jul05_1036.07052013     3301244928 3136862344 164212944   95%
Jul08_1031.07082013     3301244928 3149128144 152001000   95%
Jul09_1055.07092013     3301244928 3161187472 139993752   96%
Jul10_1006.07102013     3301244928 3172878672 128354128   96%


The desired output should be :
<diff-value>/1024 "MB"

Jun24_2013.06242013 xxMB 95% # Where xx is the calculated value


please advise,
Thanks.
# 2  
Old 07-23-2013
Like this?
Code:
awk '{printf "%s %10.0f MB %s\n", $1, ($3-p)/1024, $NF}{p=$3}' file

This User Gave Thanks to Scrutinizer For This Post:
rveri
# 3  
Old 07-23-2013
try:
Code:
awk '{a[NR]=$1; b[NR]=$3; c[NR]=$NF} END {for (i=1; i<NR; i++) printf("%s %8d MB %s\n", a[i], (b[i+1]-b[i])/1024, c[i])} {p=$3}' infile

This User Gave Thanks to rdrtx1 For This Post:
rveri
# 4  
Old 07-23-2013
Thanks both,
I got different output with Scrutinizer's code , and first entry showing way too much difference in the output ,

Code:
#  =>awk '{printf "%s %10.0f MB %s\n", $1, ($3-p)/1024, $NF}{p=$3}' file
Jun24_2013.06242013    3059629 MB 95%
Jun25_1124.06252013      12146 MB 95%
Jun26_1016.06262013      11585 MB 96%
Jun27_1112.06272013     -13010 MB 95%
Jun27_1714.06272013         28 MB 95%
Jun28_1036.06282013      10944 MB 96%
Jun28_1114.06282013          4 MB 96%
Jul01_1105.07012013     -53895 MB 94%
Jul02_1033.07022013      12192 MB 94%
Jul03_1015.07032013      11717 MB 95%
Jul04_1404.07042013      11991 MB 95%
Jul05_1036.07052013         11 MB 95%
Jul08_1031.07082013      11978 MB 95%
Jul09_1055.07092013      11777 MB 96%
Jul10_1006.07102013      11417 MB 96%
#


- rdrtx1
Code:
#  =>awk '{a[NR]=$1; b[NR]=$3; c[NR]=$NF} END {for (i=1; i<NR; i++) printf("%s %8d MB %s\n", a[i], (b[i+1]-b[i])/1024, c[i])} {p=$3}' file
Jun24_2013.06242013    12146 MB 95%
Jun25_1124.06252013    11584 MB 95%
Jun26_1016.06262013   -13010 MB 96%
Jun27_1112.06272013       28 MB 95%
Jun27_1714.06272013    10943 MB 95%
Jun28_1036.06282013        4 MB 96%
Jun28_1114.06282013   -53895 MB 96%
Jul01_1105.07012013    12192 MB 94%
Jul02_1033.07022013    11716 MB 94%
Jul03_1015.07032013    11991 MB 95%
Jul04_1404.07042013       11 MB 95%
Jul05_1036.07052013    11978 MB 95%
Jul08_1031.07082013    11776 MB 95%
Jul09_1055.07092013    11417 MB 96%
#  =>

# 5  
Old 07-23-2013
Yes that is the original value. I left the first line in so it would be clear in relation to what value the differences in next lines would be... It can be easily left out:
Code:
awk 'NR>1{printf "%s %10.0f MB %s\n", $1, ($3-p)/1024, $NF}{p=$3}' file

or did you mean the other way around?
Code:
awk 'NR>1{printf "%s %10.0f MB %s\n", f1, ($3-f2)/1024, f3}{f1=$1; f2=$3; f3=$NF}' file

This User Gave Thanks to Scrutinizer For This Post:
rveri
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