linux - Taking average of decimal number using bash -

how can take average of decimal values written in file: testfile as

time write: 0.000118000 sec  time write: 0.000119000 sec  time write: 0.000122000 sec 

wrong soln:

following prints 0 i.e. 0

awk '{sum+=$7}end{print sum/nr}' testfile 

edit

since having trouble, , seem return 0 try using this

grep -op "\d+\.\d+" testfile | awk -vx=0 '{x += $1} end {print x/nr}'

this work if file double spaced or not. prints match of file has decimal number. , sends awk.

the -p flag perl regular expression. same -e, \d+ matches 1 or more digits, \. matches period. . has special meaning in regular expressions , need escaped , \d+ matches 1 or more digits put together, '\d+\.\d+', have decimal.

lastly, if continue scientific notation may consider printf achieve floating point noation

awk -vx=0 '{x += $4} end { printf ("%8.9f", x/nr) } testfile'

you can specifiy smaller "%4.3f" print 4 numbers after decimial, conversely, use %e print in scientific notation

more using printf in awk

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old information, see above <hr>

awk -vx=0 '{x += $4} end {print x/nr}' testfile

which outputs:

0.000119667

for each line append $4, in test file number, x. @ end divide x number of lines.

if file double spaced run following first:

sed -i '/^$/d' testfile remove blank lines. may want consider not editing testfile in place removing -i , doing sed '/^$/d' testfile > newfile

or combine 2 files , pipe stdout sed awk

sed '/^$/d' testfile | awk -vx=0 '{x += $4} end {print x/nr}'

if returns 0, may have problem testfile.