How to sum reciprocals in bash? [duplicate] - bash

This question already has answers here:
How do I use floating-point arithmetic in bash?
(23 answers)
Closed 4 years ago.
I am attempting to find the sum of a list of numbers' reciprocals. To illustrate what I am trying to do, here's a basic example:
With the file:
1
2
3
4
I would be trying to find the sum of 1/1, 1/2, 1/3 and 1/4. Is there a simple bash one-liner to do this? (I am new to bash, so explanations would be welcome!)

You could do something like this:
sed 's|^|1/|' file | paste -sd+ | bc -l
sed 's|^|1/|' prepends 1/ to every line
paste -sd+ joins all lines with a plus sign creating an arithmetic expression 1/1+1/2+1/3+1/4
bc -l evaluates that arithmetic expression and outputs the result

If you're looking for an arithmetical progression, you can use this bash one-liner using the bc command
d=0; for c in {1..4}; do d=`echo "$d + 1/$c" | bc -l`; done; echo "$d"
Its output is 1 + 0.5 + 0.3333 + 0.25 =
2.08333333333333333333
It works by
Setting a variable named d to 0
Creating a for loop that counts from 1 to 4
In the for loop it sets the d variable to the new value $d + 1/$c passed to the bc -l command executing the arithmetic
And outputs the value with an echo command

Related

addition in bash with decimal using bc

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

Generate uniform random number in range of floats in bash [duplicate]

This question already has answers here:
Random numbers generation with awk in BASH shell
(3 answers)
Closed 2 years ago.
[SOLVED]
I want to generate a uniform random float number in the range of float numbers in the bash script. range e.g. [3.556,6.563]
basically, I am creating LSH(Latin hypercube sampling) function in bash. There I would like to generate an array as one can do with this python command line.
p = np.random.uniform(low=l_lim, high=u_lim, size=[n]).
sample code :
lhs(){
l_lim=($(seq $1 $2 $(echo $3 - $dif | bc)))
h_lim=($(seq $(echo $1 + $dif | bc) $2 $3))
points=()
for ((i=0;i<$n;i++)) ; do
di=${l_lim[i]}
dj=${h_lim[i]}
echo $di, $dj
p=$(awk -v min=6.50 -v max=8.45 -v seed=$RANDOM 'BEGIN{srand(seed);print min+rand()*int(1000*(max-min)+1)/1000}')
points+=("${p}")
done
}
n=5
a=(3 5)
b=(1 3)
dif=$(div $(echo ${a[1]} - ${a[0]} | bc) $n)
lhs ${a[0]} 0.45 ${a[1]}
echo ${points[#]}
I have tried $RANDOM, awk but it did not work for me. I do not want to use python -c.
Most common rand() implementations at least generate a number in the range [0...1), which is really all you need. You can scale a random number in one range to a number in another using the techniques outlined in the answers to this question, eg:
NewValue = (((OldValue - OldMin) * (NewMax - NewMin)) / (OldMax - OldMin)) + NewMin
For bash you have two choices: integer arithmetic or use a different tool.
Some of your choices for tools that support float arithmetic from the command-line include:
a different shell (eg, zsh)
perl: my $x = $minimum + rand($maximum - $minimum);
ruby: x = min + rand * (max-min)
awk: awk -v min=3 -v max=17 'BEGIN{srand(); print min+rand()*int(1000*(max-min)+1)/1000}'
note: The original answer this was copied from is broken; the above is a slight modification to help correct the problem.
bc: printf '%s\n' $(echo "scale=8; $RANDOM/32768" | bc )
... to name a few.

rounding off in bash [duplicate]

This question already has answers here:
How do I use floating-point arithmetic in bash?
(23 answers)
Closed 4 years ago.
I was taking a challenge on hackerrank .
The aim is:
read (int) N; then read N integers and print their avg to three decimal places.
Here's the code:
#!/bin/bash
#file name:rdlp.sh
read N
s=0
i=1
while (($i<=$N))
do
read a
s=$((s+a))
i=$((i+1))
done
s=$s/$N
echo "scale=3;$s"|bc -l
fi
When I run the code for some inputs:
3 #(value of N)
4 #(N = 3 integers)
4
3
Then the output is 3.666, but it should be 3.667.
So the QUESTION is that is there anyway to get it right (correct rounding off), or does it work like that only?
(the question came off when the above code was run for Testcase2 of the challenge at hackerrank)
bc rounds down with scale=x.
You can printf:
$ printf "%.3f\n" $(echo "scale=1000; 11/3"|bc -l)
3.667
or some tricky bc by adding 0.0005:
$ echo "scale=1000; v=11/3; v=v+0.0005; scale=3; v/1" | bc -l
3.667

Generating random number between 1 and 10 in Bash Shell Script [duplicate]

This question already has answers here:
Random number from a range in a Bash Script
(19 answers)
Closed 7 years ago.
How would I generate an inclusive random number between 1 to 10 in Bash Shell Script?
Would it be $(RANDOM 1+10)?
$(( ( RANDOM % 10 ) + 1 ))
EDIT. Changed brackets into parenthesis according to the comment.
http://web.archive.org/web/20150206070451/http://islandlinux.org/howto/generate-random-numbers-bash-scripting
Simplest solution would be to use tool which allows you to directly specify ranges, like gnu shuf
shuf -i1-10 -n1
If you want to use $RANDOM, it would be more precise to throw out the last 8 numbers in 0...32767, and just treat it as 0...32759, since taking 0...32767 mod 10 you get the following distribution
0-8 each: 3277
8-9 each: 3276
So, slightly slower but more precise would be
while :; do ran=$RANDOM; ((ran < 32760)) && echo $(((ran%10)+1)) && break; done
To generate random numbers with bash use the $RANDOM internal Bash function. Note that $RANDOM should not be used to generate an encryption key. $RANDOM is generated by using your current process ID (PID) and the current time/date as defined by the number of seconds elapsed since 1970.
echo $RANDOM % 10 + 1 | bc
You can also use /dev/urandom:
grep -m1 -ao '[0-9]' /dev/urandom | sed s/0/10/ | head -n1
To generate in the range: {0,..,9}
r=$(( $RANDOM % 10 )); echo $r
To generate in the range: {40,..,49}
r=$(( $RANDOM % 10 + 40 )); echo $r
Here is example of pseudo-random generator when neither $RANDOM nor /dev/urandom is available
echo $(date +%S) | grep -o .$ | sed s/0/10/

Shell Script: How do I add the digits of a number?

I am making a shell script that takes a single number (length is unimportant) from the command line and adds the digits of it together. I thought I had it, but it won't work and either displays "0+3+4+5" if the command input is 345 or it displays the variables when I use expr to add them.
#!/bin/bash
sum=0
i="$(expr length $1)"
s=$1
for i in $(seq 0 $((${#s} - 1))); do
value=${s:$i:1}
typeset -i value
sum=$sum+$value
done
echo $sum
Also doesn't work when I replace it with sum='expr $sum + $value'
any ideas?
What you are looking for is sum=$(($sum+$value)).
#!/bin/bash
expr $(echo $1| sed 's/./& + /g;s/..$//')
For example, if the argument is 12345, this translates it to the string 1 + 2 + 3 + 4 + 5 and uses expr to evaluate it.

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