how to remove decimal place in shell script.i am multiplying MB with bytes to get value in bytes .I need to remove decimal place.
ex:-
196.3*1024*1024
205835468.8
expected output
205835468
(You did not mention what shell you're using; this answer assumes Bash).
You can remove the decimal values using ${VAR%.*}. For example:
[me#home]$ X=$(echo "196.3 * 1024 * 1024" | bc)
[me#home]$ echo $X
205835468.8
[me#home]$ echo ${X%.*}
205835468
Note that this truncates the value rather than rounds it. If you wish to round it, use printf as shown in Roman's answer.
The ${variable%pattern} syntax deletes the shortest match of pattern starting from tbe back of variable. For more information, read http://tldp.org/LDP/abs/html/string-manipulation.html
Use printf:
printf %.0f $float
This will perform rounding. So if float is 1.8, it'll give you 2.
Related
Here is my script:
d1=0.003
d2=0.0008
d1d2=$((d1 + d2))
mean1=7
mean2=5
meandiff=$((mean1 - mean2))
echo $meandiff
echo $d1d2
But instead of getting my intended output of:
0.0038
2
I am getting the error Invalid Arithmetic Operator, (error token is ".003")?
bash does not support floating-point arithmetic. You need to use an external utility like bc.
# Like everything else in shell, these are strings, not
# floating-point values
d1=0.003
d2=0.0008
# bc parses its input to perform math
d1d2=$(echo "$d1 + $d2" | bc)
# These, too, are strings (not integers)
mean1=7
mean2=5
# $((...)) is a built-in construct that can parse
# its contents as integers; valid identifiers
# are recursively resolved as variables.
meandiff=$((mean1 - mean2))
Another way to calculate floating numbers, is by using AWK rounding capability, for example:
a=502.709672592
b=501.627497268
echo "$a $b" | awk '{print $1 - $2}'
1.08218
In case you do not need floating point precision, you may simply strip off the decimal part.
echo $var | cut -d "." -f 1 | cut -d "," -f 1
cuts the integer part of the value. The reason to use cut twice is to parse integer part in case a regional setting may use dots to separate decimals and some others may use commas.
Edit:
Or, to automate the regional settings one may use locale.
echo $var | cut -d $(locale decimal_point) -f 1
You can change the shell which you are using. If you are executing your script with bash shell bash scriptname.sh try using ksh for your script execution. Bash doesn't support arithmetic operations that involve floating point numbers.
Big shout-out to the bc command - it totally saved my day! It's a simple answer, but it worked like a charm.
a=1.1
b=1.1
echo $a + $b | bc -l
# Output:
2.2
#SUM
sum=$(echo $a + $b | bc -l)
echo $sum
# Output
2.2
bc is a command-line calculator, which allows users to perform mathematical calculations on the terminal.
I have a txt file and i want to assign a specific word from inside the text file and its many lines. e.g Let's say that the text file is this sentence and the value 0.78 is the one that i want to get assigned. How do i do that?
I'm guessing you are looking for a number, so I'll extract anything that looks like a number:
grep -Ewo "[+-]?[0-9]*\.?[0-9]+" file
echo $n
0.78
The regex says it can have an optional +/- at the start, some optional digits, an optional decimal point and some more digits.
If you only want the number with a decimal point in it, and you want that in a bash variable, do this
var=$(grep -Ewo "[+-]?[0-9]*\.[0-9]*" file)
echo $var
I have no idea what your context is, but let's assume you want to match the number immediately following the word "value" in your text file, and you want to do it in pure Bash:
$ [[ $(<yourfile.txt) =~ value\ ([0-9.]*) ]] && echo ${BASH_REMATCH[1]}
0.78
See Conditional Constructs in the Bash manual for more information.
I need some help with a bash mathematical expression script. I need the expression below to be a negative 2 to the power of 63 but I have tried all sorts of combinations from "", '', (()) and even keeping the positive value but multiplying it by -1 but I can't seem to get the syntax right.
Original expression:
kw=`expr 2^63 | bc`
gives me 9223372036854775808 (which is correct) but I need to get a -9223372036854775808.
I want to be able to do a kw=expr -2^63 | bc to get a negative results but it's not as straight-forward as I'd hoped. Hence my numerous attempts of different permutations to the expression.
Any help will be very appreciated.
Here you go:
$ kw=$(echo -2^63 | bc)
$ echo $kw
-9223372036854775808
UPDATE
#DigitalTrauma is right, if you're in bash, then using a bash here string is better (one less process, more efficient):
kw=$(bc <<< -2^63)
Since this is bash, you don't even need the echo; you can use a bash here string instead:
$ kw=$(bc <<< -2^63)
$ echo $kw
-9223372036854775808
What might be the most concise way in bash to convert a number into a bitfield character string like 1101?
In effect I am trying to do the opposite of
echo $[2#1101]
Why: I need to send a parameter to a program that takes bitfields in the form of a full string like "0011010110" but often only need to enable one or few bits as in:
SUPPRESSbits=$[1<<16] runscript.sh # OR
SUPPRESSbits=$[1<<3 + 1<<9] runscript.sh # much more readable when I know what bits 3 and 9 toggle in the program
Then runscript.sh then sees in its env a SUPPRESSbits=65536 rather than SUPPRESSbits="1000000000000000" and ends in parse error.
The easy way:
$ dc <<<2o123p
1111011
$ bc <<<'obase=2; 123'
1111011
I doubt about bash but you always can use perl:
a=123; b=$(perl -e 'printf "%b", "'$a'"'); echo $b
1111011
I wrote a .sh file to compile and run a few programs for a homework assignment. I have a "for" loop in the script, but it won't work unless I use only integers:
#!/bin/bash
for (( i=10; i<=100000; i+=100))
do
./hw3_2_2 $i
done
The variable $i is an input for the program hw3_2_2, and I have non-integer values I'd like to use. How could I loop through running the code with a list of decimal numbers?
I find it surprising that in five years no one ever mentioned the utility created just for generating ranges, but, then again, it comes from BSD around 2005, and perhaps it wasn't even generally available on Linux at the time the question was made.
But here it is:
for i in $(seq 0 0.1 1)
Or, to print all numbers with the same width (by prepending or appending zeroes), use -w. That helps prevent numbers being sent as "integers", if that would cause issues.
The syntax is seq [first [incr]] last, with first defaulting to 1, and incr defaulting to either 1 or -1, depending on whether last is greater than or less than first. For other parameters, see seq(1).
you can use awk to generate your decimals eg steps of0.1
num=$(awk 'BEGIN{for(i=1;i<=10;i+=0.1)print i}')
for n in $num
do
./hw3_2_2 $n
done
or you can do it entirely in awk
awk 'BEGIN{cmd="hw3_2_2";for(i=1;i<=10;i+=0.1){c=cmd" "i;system(cmd) } }'
The easiest way is to just list them:
for a in 1.2 3.4 3.11 402.12 4.2 2342.40
do
./hw3_2_2 $a
done
If the list is huge, so you can't have it as a literal list, consider dumping it in a file and then using something like
for a in $(< my-numbers.txt)
do
./hw3_2_2 $a
done
The $(< my-numbers.txt) part is an efficient way (in Bash) to substitute the contents of the names file in that location of the script. Thanks to Dennis Williamson for pointing out that there is no need to use the external cat command for this.
Here's another way. You can use a here doc to include your data in the script:
read -r -d '' data <<EOF
1.1
2.12
3.14159
4
5.05
EOF
for i in "$data"
do
./hw3_2_2 "$i"
done
Similarly:
array=(
1.1
2.12
3.14159
4
5.05
)
for i in "${array[#]}"
do
./hw3_2_2 "$i"
done
I usually also use "seq" as per the second answer, but just to give an answer in terms of a precision-robust integer loop and then bc conversion to a float:
#!/bin/bash
for i in {2..10..2} ; do
x=`echo "scale=2 ; ${i}/10" | bc`
echo $x
done
gives:
.2
.4
.6
.8
1.0
bash doesn't do decimal numbers. Either use something like bc that can, or move to a more complete programming language. Beware of accuracy problems though.