I'm new in bash and i'd like to know why my script doesn t work the way i'd like it work..
I have this bash script:
#!/bin/bash
read n
var=($(cat))
bim=${var[*]}
toto=$(echo $bim | sed 's/ /+/g' | bc)
echo $toto
bobo=$(($toto/$n | bc -l))
echo $bobo | awk '{printf "%.3f\n", $1}'
This is supposed to add up all the values that "cat" has stored in an array and divide the total by the first value that "read" reads. And the result should return me a decimal value of three decimal places. However, it only returns a round number to me when I use bc -l! And when I use awk '{printf% .3f ", $ 1}' it prints .000!
Do you know why?
Thanks
bash only does integer arithmetics and the arithmetic expansion
bobo=$(($toto/$n | bc -l))
will not do what you think. The pipe sign in | bc -l is not a pipe sign. It's a bitwise OR. Broken down:
$toto is interpreted as a variable (as would toto)
$n is interpreted as a variable (as would n)
| is bitwise OR
bc is interpreted as a variable (with the value 0)
- is interpreted as a minus sign (in an arithmetic expression)
l is interpreted as a variable (with the value 0)
So, it becomes $toto/$n | 0 -0 which is the same as the integer division toto/n ($ is not needed for variables in arithmetic expansions).
You could instead use bc for the division too, but you need to set the scale in bc. Here's an example setting it to 3 before performing the division:
bobo=$(echo "scale=3;$toto/$n" | bc)
Note that you don't need to echo this through awk.
Just echo $bobo and you should get the result you want.
Example input:
3
11.3
9
8
Output after having applied the suggested changes:
28.3
9.433
Related
I have a bash function that is currently set up as:
MB=$(( $(echo $(FUNCTION_THAT_RETURNS_Kb_OR_Mb) | cut -d "K" -f 1 | sed 's/^.*- //') / 1000 ))
where the middle portion echo $(FUNCTION_THAT_RETURNS_Kb_OR_Mb) returns a value that ends in K or M, (for example: 515223 K or 36326 M) for Kilobytes or Megabytes. I currently have designed the function to strip the trailing units indicator for K, and then divide by 1000 to convert to megabytes. However, when the inside part of it ends in M, it fails. How can I write a function that detects if its in kilobytes or megabytes?
Don't reinvent the wheel - there is numfmt:
function_that_returns_Kb_or_Mb() { echo "515223 K"; }
mb=$(function_that_returns_Kb_or_Mb | numfmt -d '' --from=iec --to-unit=Mi)
# mb=504
function_that_returns_Kb_or_Mb() { echo "36326 M"; }
mb=$(function_that_returns_Kb_or_Mb | numfmt -d '' --from=iec --to-unit=Mi)
# mb=36326
Notes:
echo $(FUNCTION_THAT_RETURNS_Kb_OR_Mb) is a useless use of echo. It's like echo $(echo $(echo $(...)))). Just FUNCTION_THAT_RETURNS_Kb_OR_Mb | blabla.
By convention UPPERCASE VARIABLES are used for exported variables, like PATH COLUMNS UID PWD etc. - use lower case identifiers in your scripts.
I assumed input and output is using IEC scale, for SI scale use --from=si --to-unit=M.
I'm trying to convert whatever numbers the user inputs into 2 decimal places.
For instance
What is the total cost in cents? 2345
output: 23.45
this is the code i have so far
percentage=20 #cannot change numerical value must convert into 0.20
echo -n "What is the total cost? ";
read cost_in_cents
echo "scale 1; $cost_in_cents" | bc
I'm also going to be doing some multiplication with percentage, how can i also convert the percentage into a float (0.20)
awk to the rescue!
you can define your own floating point calculator with awk, e.g.
$ calc() { awk "BEGIN{ printf \"%.2f\n\", $* }"; }
now you can call
$ calc 43*20/100
which will return
8.60
Perhaps it's nostalgia for reverse polish notation desk calculators, but I'd use dc rather than bc here:
dc <<<"2 k $cost_in_cents 100 / p"
Output is, properly, a float (with two digits past the decimal point of precision).
The exact same code, with no changes whatsoever, will work to convert 20 to .20.
See BashFAQ #22 for a full discussion on floating-point math in bash.
Bash itself could not process floats.
It can, however, printf them:
$ printf 'value: %06.2f\n' 23.45
value: 023.45
So, you need an external program to do the math:
$ echo "scale=4;2345/100*20/100" | bc
4.6900
Or, equivalent:
$ bc <<<"scale=4;2345*20/10^4"
4.6900
Then, you can format the float with printf:
$ printf 'result: %06.2f\n' $(bc <<<"scale=4;2345*20/10^4")
result: 004.69
Or you can use a program that can process floats; like awk.
How about this:
read -p "What is the total cost? " input
percent=20
echo "scale=2; $input / 100 * $percent / 100" | bc
# input = 2345 , output = 4.69
I am a new bash learner. I want to print the result of an expression given as input having 3 digits after decimal point with rounding if needed.
I can use the following code, but it does not round. Say if I give 5+50*3/20 + (19*2)/7 as input for the following code, the given output is 17.928. Actual result is 17.92857.... So, it is truncating instead of rounding. I want to round it, that means the output should be 17.929. My code:
read a
echo "scale = 3; $a" | bc -l
Equivalent C++ code can be(in main function):
float a = 5+50*3.0/20.0 + (19*2.0)/7.0;
cout<<setprecision(3)<<fixed<<a<<endl;
What about
a=`echo "5+50*3/20 + (19*2)/7" | bc -l`
a_rounded=`printf "%.3f" $a`
echo "a = $a"
echo "a_rounded = $a_rounded"
which outputs
a = 17.92857142857142857142
a_rounded = 17.929
?
You can use awk:
awk 'BEGIN{printf "%.3f\n", (5+50*3/20 + (19*2)/7)}'
17.929
%.3f output format will round up the number to 3 decimal points.
Try using this:
Here bc will provide the bash the functionality of caluculator and -l will read every single one in string and finally we are printing only three decimals at end
read num
echo $num | bc -l | xargs printf "%.3f"
In a text file, test.txt, I have the next information:
sl-gs5 desconnected Wed Oct 10 08:00:01 EDT 2012 1001
I want to extract the hour of the event by the next command line:
hour=$(grep -n sl-gs5 test.txt | tail -1 | cut -d' ' -f6 | awk -F ":" '{print $1}')
and I got "08". When I try to add 1,
14 echo $((hour+1))
I receive the next error message:
./test2.sh: line 14: 08: value too great for base (error token is "08")
If variables in Bash are untyped, why?
See ARITHMETIC EVALUATION in man bash:
Constants with a leading 0 are interpreted as octal numbers.
You can remove the leading zero by parameter expansion:
hour=${hour#0}
or force base-10 interpretation:
$((10#$hour + 1))
what I'd call a hack, but given that you're only processing hour values, you can do
hour=08
echo $(( ${hour#0} +1 ))
9
hour=10
echo $(( ${hour#0} +1))
11
with little risk.
IHTH.
You could also use bc
hour=8
result=$(echo "$hour + 1" | bc)
echo $result
9
Here's an easy way, albeit not the prettiest way to get an int value for a string.
hour=`expr $hour + 0`
Example
bash-3.2$ hour="08"
bash-3.2$ hour=`expr $hour + 0`
bash-3.2$ echo $hour
8
In Short: In order to deal with "Leading Zero" numbers (any 0 digit that comes before the first non-zero) in bash
- Use bc An arbitrary precision calculator language
Example:
a="000001"
b=$(echo $a | bc)
echo $b
Output: 1
From Bash manual:
"bc is a language that supports arbitrary precision numbers with interactive execution
of statements. There are some similarities in the syntax to the C programming lan-
guage. A standard math library is available by command line option. If requested, the
math library is defined before processing any files. bc starts by processing code from
all the files listed on the command line in the order listed. After all files have
been processed, bc reads from the standard input. All code is executed as it is read.
(If a file contains a command to halt the processor, bc will never read from the standard input.)"
Since hours are always positive, and always 2 digits, you can set a 1 in front of it and subtract 100:
echo $((1$hour+1-100))
which is equivalent to
echo $((1$hour-99))
Be sure to comment such gymnastics. :)
The leading 0 is leading to bash trying to interpret your number as an octal number, but octal numbers are 0-7, and 8 is thus an invalid token.
If I were you, I would add some logic to remove a leading 0, add one, and re-add the leading 0 if the result is < 10.
How about sed?
hour=`echo $hour|sed -e "s/^0*//g"`
I read price from user input. When i multiply the input with int like this
T="$((PRICE*QTY))"|bc; gives
line 272: 12.00: syntax error: invalid arithmetic operator (error token is ".00")
or .50
depending on user input. How do i multiply these two variables and get a total with 2 decimal points?
this works:
PRICE=1.1
QTY=21
RES=$(echo "scale=4; $PRICE*$QTY" | bc)
echo $RES
var=$(echo "scale=2;$PRICE*$QTY" |bc)
You can also use awk
awk -vp=$PRICE -vq=$QTY 'BEGIN{printf "%.2f" ,p * q}'
T="$(echo "$PRICE*$QTY" | bc)"
You can use
mul=0.8
exp=200
texp=awk -vp=$mul -vq=$exp 'BEGIN{printf "%.2f" ,p * q}'
Hope this is going to work.
First, trying to do floating-point arithmetic with bc(1) without using the -l flag is bound to give you some funny answers:
sarnold#haig:~$ bc -q
3.5 * 3.5
12.2
sarnold#haig:~$ bc -q -l
3.5 * 3.5
12.25
Second, the $((...)) is an attempt to do arithmetic in your shell; neither my bash nor dash can handle floating point numbers.
If you want to do the arithmetic in your shell, note printf(1) as well as (probably) your shell's built-in printf function. If you want to do the arithmetic in bc, note the special variable scale.