I'm getting load average in a bash shell script like so
load=`echo $(cat /proc/loadavg | awk '{print $1}')`
I know piping to bc
load=`echo $(cat /proc/loadavg | awk '{print $1}') \> 3 | bc -l`
is used in almost all examples of how to cast $load as an int but this box does not have bc installed and I am not allowed to add it.
I tried
int=`perl -E "say $load - 0"`
I tried
int=${load%.*}
I tried
int=`printf -v int %.0f "$load"`
What I want to be able to do is
if [ "$int" -gt 3.5 ]; then
How do I get that to evaluate as intended?
You can use awk to produce a success/failure depending on the condition:
# exit 0 (success) when load average greater than 3.5, so take the branch
if awk '{ exit !($1 > 3.5) }' /proc/loadavg; then
# load average was greater than 3.5
fi
Unfortunately, since "success" is 0 in the shell, you have to invert the logic of the condition to make awk exit with the required status. Obviously, you can do this in a number of ways, such as changing > to <=.
You don't need any external tools (like awk) to read this stuff. Load average from /proc/loadavg is always formatted with two decimal places, so you can do this:
read load _ < /proc/loadavg
if [ ${load/./} -gt 350 ]; then
# do something
fi
Greetings!
I uses bс to make some calculations in my script. For example:
bc
scale=6
1/2
.500000
For further usage in my script I need "0.500000" insted of ".500000".
Could you help me please to configure bc output number format for my case?
In one line:
printf "%0.6f\n" $(bc -q <<< scale=6\;1/2)
Just do all your calculations and output in awk:
float_scale=6
result=$(awk -v scale=$floatscale 'BEGIN { printf "%.*f\n", scale, 1/2 }')
As an alternative, if you'd prefer to use bc and not use AWK alone or with 'bc', Bash's printf supports floating point numbers even though the rest of Bash doesn't.
result=$(echo "scale=$float_scale; $*" | bc -q 2>/dev/null)
result=$(printf '%*.*f' 0 "$float_scale" "$result")
The second line above could instead be:
printf -v $result '%*.*f' 0 "$float_scale" "$result"
Which works kind of like sprintf would and doesn't create a subshell.
Quick and dirty, since scale only applies to the decimal digits and bc does not seem to have a sprintf-like function:
$ bc
scale = 6
result = 1 / 2
if (0 <= result && result < 1) {
print "0"
}
print result;
echo "scale=3;12/7" | bc -q | sed 's/^\\./0./;s/0*$//;s/\\.$//'
I believe here is modified version of the function:
float_scale=6
function float_eval()
{
local stat=0
local result=0.0
if [[ $# -gt 0 ]]; then
result=$(echo "scale=$float_scale; $*" | bc -q | awk '{printf "%f\n", $0}' 2>/dev/null)
stat=$?
if [[ $stat -eq 0 && -z "$result" ]]; then stat=1; fi
fi
echo $result
return $stat
}
Can you put the bc usage into a little better context? What are you using the results of bc for?
Given the following in a file called some_math.bc
scale=6
output=1/2
print output
on the command line I can do the following to add a zero:
$ bc -q some_math.bc | awk '{printf "%08f\n", $0}'
0.500000
If I only needed the output string to have a zero for formatting purposes, I'd use awk.
Let's say I have a variable like this:
list='1
1
1.4
1
1
1'
Now I want to add the value from each line to the value from the line before. Like this:
result='1
2
3.4
4.4
5.4'
It must work with floating numbers too, so I guess awk is the best tool for it.
I was stating with a while loop:
while read line; do
add=$( awk 'BEGIN{ print "'"$x"'" + "'"$line"'" }' )
done <<< "$list"
But this doesn't work because I don't know how to save the value from the line before.
This would be my bash snippet:
prev=0
for val in $list; do
val=$(echo "$val + $prev"|bc)
prev=$val
result="${result}${val}\n"
done
echo -e "$result"
I don't know whether using zsh (which has floating point support) is an option, but if it is, you can simply do:
s=0;for i in $list; do s=$(($s+$i)); echo $s; done
Using awk you can do:
awk 'NR==1{print; s=$1; next} {s+=$1; print s}' file
1
2
3.4
4.4
5.4
6.4
Here is a way of doing it using bc:
#!/bin/bash
list='1
1
1.4
1
1
1'
results=( )
total="0.0"
index=0
for line in $list; do
total=$(bc -l <<< "$line + $total")
echo $total
results[$index]=$total
let index++
done
echo ${results[#]}
Also:
total="0.0"; while read v; do total=$(bc -l <<< "$v + $total"); echo $total; done <<< "$list"
I have a bash variable with value something like this:
10:3.0,16:4.0,32:4.0,39:2.0,65:3.0,95:4.0,110:4.0,111:4.0,2312:1.0
There are no spaces within value. This value can be very long or very short. Here pairs such as 65:3.0 exist. I know the value of a number from the first part of pair, say 65. I want to extract the number 3.0 or pair 65:3.0. I am not aware of the position (offset) of 65.
I will be grateful for a bash-script that can do such extraction. Thanks.
Probably awk is the most straight-forward approach:
awk -F: -v RS=',' '$1==65{print $2}' <<< "$var"
3.0
Or to get the pair:
$ awk -F: -v RS=',' '$1==65' <<< "$var"
65:3.0
Here's a pure Bash solution:
var=10:3.0,16:4.0,32:4.0,39:2.0,65:3.0,95:4.0,110:4.0,111:4.0,2312:1.0
while read -r -d, i; do
[[ $i = 65:* ]] || continue
echo "$i"
done <<< "$var,"
You may use break after echo "$i" if there's only one 65:... in var, or if you only want the first one.
To get the value 3.0: echo "${i#*:}".
Other (pure Bash) approach, without parsing the string explicitly. I'm assuming you're only looking for the first 65 in the string, and that it is present in the string:
var=10:3.0,16:4.0,32:4.0,39:2.0,65:3.0,95:4.0,110:4.0,111:4.0,2312:1.0
value=${var#*,65:}
value=${value%%,*}
echo "$value"
This will be very slow for long strings!
Same as above, but will output all the values corresponding to 65 (or none if there are none):
var=10:3.0,16:4.0,32:4.0,39:2.0,65:3.0,95:4.0,110:4.0,111:4.0,2312:1.0
tmpvar=,$var
while [[ $tmpvar = *,65:* ]]; do
tmpvar=${tmpvar#*,65:}
echo "${tmpvar%%,*}"
done
Same thing, this will be slow for long strings!
The fastest I can obtain in pure Bash is my original answer (and it's fine with 10000 fields):
var=10:3.0,16:4.0,32:4.0,39:2.0,65:3.0,95:4.0,110:4.0,111:4.0,2312:1.0
IFS=, read -ra ary <<< "$var"
for i in "${ary[#]}"; do
[[ $i = 65:* ]] || continue
echo "$i"
done
In fact, no, the fastest I can obtain in pure Bash is with this regex:
var=10:3.0,16:4.0,32:4.0,39:2.0,65:3.0,95:4.0,110:4.0,111:4.0,2312:1.0
[[ ,$var, =~ ,65:([^,]+), ]] && echo "${BASH_REMATCH[1]}"
Test of this vs awk,
where the 65:3.0 is at the end:
printf -v var '%s:3.0,' {100..11000}
var+=65:42.0
time awk -F: -v RS=',' '$1==65{print $2}' <<< "$var"
shows 0m0.020s (rough average) whereas:
time { [[ ,$var, =~ ,65:([^,]+), ]] && echo "${BASH_REMATCH[1]}"; }
shows 0m0.008s (rough average too).
where the 65:3.0 is not at the end:
printf -v var '%s:3.0,' {1..10000}
time awk -F: -v RS=',' '$1==65{print $2}' <<< "$var"
shows 0m0.020s (rough average) and with early exit:
time awk -F: -v RS=',' '$1==65{print $2;exit}' <<< "$var"
shows 0m0.010s (rough average) whereas:
time { [[ ,$var, =~ ,65:([^,]+), ]] && echo "${BASH_REMATCH[1]}"; }
shows 0m0.002s (rough average).
With grep:
grep -o '\b65\b[^,]*' <<<"$var"
65:3.0
Or
grep -oP '\b65\b:\K[^,]*' <<<"$var"
3.0
\K option ignores everything before matched pattern and ignore pattern itself. It's Perl-compatibility(-P) for grep command .
Here is an gnu awk
awk -vRS="(^|,)65:" -F, 'NR>1{print $1}' <<< "$var"
3.0
try
echo $var | tr , '\n' | awk '/65/'
where
tr , '\n' turn comma to new line
awk '/65/' pick the line with 65
or
echo $var | tr , '\n' | awk -F: '$1 == 65 {print $2}'
where
-F: use : as separator
$1 == 65 pick line with 65 as first field
{ print $2} print second field
Using sed
sed -e 's/^.*,\(65:[0-9.]*\),.*$/\1/' <<<",$var,"
output:
65:3.0
There are two different ways to protect against 65:3.0 being the first-in-line or last-in-line. Above, commas are added to surround the variable providing for an occurrence regardless. Below, the Gnu extension \? is used to specify zero-or-one occurrence.
sed -e 's/^.*,\?\(65:[0-9.]*\),\?.*$/\1/' <<<$var
Both handle 65:3.0 regardless of where it appears in the string.
Try egrep like below:
echo $myvar | egrep -o '\b65:[0-9]+.[0-9]+' |
I have two numbers:
value1=686
value2=228.35
I am not able to add an integer and a float. Please help me out to get the result.
I am running it in bash.
echo 1 + 3.5 | bc
awk "BEGIN {print 1+3.5; exit}"
python -c "print 1+3.5"
perl -e "print 1+3.5"
Just replace the numbers with your variables, eg: echo $n1 + $n2 | bc
If you have the bc language installed, you can do the following:
#!bin/bash
numone=1.234
numtwo=0.124
total=`echo $numone + $numtwo | bc`
echo $total
If you don't have bc, then you can try with awk. Just in one single line:
echo 1.234 2.345 | awk '{print $1 + $2}'
There are plenty of other options, also. Like python, perl, php....
Bash doesn't have floating-point types, but you can use a calculator such as bc:
a=686
b=228.35
c=`echo $a + $b | bc`
echo "$c"
#!/bin/Bash
echo "Enter the two numbers to be added:"
read n1
read n2
answer=$(($n1+$n2))
echo $answer