使用 bash 取十进制数的平均值

Question

How can I take average of decimal values written in a file: TestFile as

Time to write: 0.000118000 sec

Time to write: 0.000119000 sec

Time to write: 0.000122000 sec

Wrong Soln:

Following prints just zero ie 0

awk '{sum+=$7}END{print sum/NR}' TestFile

Answer 1

This solution will target calculation at the correct lines:

awk '/Time to write/ {sum+=$4; count++} END {print "avg:", sum/count}' data.txt

Answer 2

since you are having trouble, and seem to return zero try using this

grep -oP "\d+\.\d+" testFile | awk -vx=0 '{x += $1} END {print x/NR}'

this will work if the file is double spaced or not. it only prints the match of a file that has as decimal number. and sends it to awk.

the -P flag is a perl regular expression. you could do the same with -E, but \d+ matches one or more digits, \. matches a period. a. has special meaning in regular expressions and need to be escaped and \d+ matches one or more digits so put that together, '\d+\.\d+' , you have a decimal.

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

awk -vx=0 '{x += $4} END { printf ("%8.9f", x/NR) } testFile'

you can specifiy something smaller like "%4.3f" to print only 4 numbers after a decimial, Conversely, use %e to 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, which in your test file is the number, to x. At the end divide x for number of lines.

if your file is really double spaced run the following first:

sed -i '/^$/d' testFile to remove blank lines. you may want to consider not editing testFile in place by removing -i and doing something like this sed '/^$/d' testFile > newFile

or even combine the two files and pipe stdout from sed to awk

sed '/^$/d' testFile | awk -vx=0 '{x += $4} END {print x/NR}'

if this returns 0, you may have a problem with your testFile.