I made a loop to find the results of calculating each number from (1 to 10 mod 5) + 2
for (( i = 0; i <= 10; i++))
do
calculate=$(( i % 5 + 2 ))
echo "($i % 5) + 2 = " $calculate
done
average=$(($calculate/10))
echo $average`
My problem is fixing my code so that I can take all the results of the loop and find the average of them
Its returning back 0 for the average
You have to keep a full total - ($calculate/10) is just the last iteration. Keep a running total initialized to zero before the loop total = 0 ... then add the calculated value to the total in each iteration of the loop total = $( $total + $calculate ) Then the average is total/10 (not calculate/10).
#!/bin/bash
total=0
for (( i = 0; i <= 10; i++))
do
calculate=$(( i % 5 + 2 ))
total=$(( $total + $calculate ))
echo "($i % 5) + 2 = " $calculate
done
#average=$(($calculate/10))
average=$(($total/10))
echo $average
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How to code the program in the Shell Script to calculate the sum of the Fibonacci series with the upper limit input from the keyboard? Input for example: "Enter the limit of the Fibonacci series? 10". and the result: 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 = 88
#!/bin/bash
if [ $# -eq 1 ]
then
Num=$1
else
echo -n "Enter the limits of the Fibonacci sequence : "
read Num
fi
a=0
b=1
echo "Result : "
for (( i=0;i<=Num;i++ ))
do
echo -n “$a + “
n=$((a+b))
a=$b
b=$n
done
echo
Anyone know how to get the results?
my expected output is:
Enter the limits of the Fibonacci sequence : 10
Result : 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 = 88
I'm stuck trying
$ cat fib
#!/bin/bash
Num=${1-5}
a=0
b=1
unset s
printf "Result: "
for (( i=0; i<Num; i++ )); do
printf "${s:+ + }$a"
s=$(($s + $a))
n=$((a+$b))
a=$b
b=$n
done
echo " = $s"
$ ./fib 5
Result: 0 + 1 + 1 + 2 + 3 = 7
$ ./fib 10
Result: 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 = 88
This works:
a=1;
b=0;
sum=$a;
for ((i=1;i<10-1;++i)); do
((tmp=b));
((b=a));
((a+=tmp));
((sum+=a));
done;
echo $sum;
Output from my terminal:
[ichramm#hypernova][Wed, 28 Oct 16:19:24][~]
$ a=1; b=0; sum=$a; for ((i=1;i<10-1;++i)); do ((tmp=b)); ((b=a)); ((a+=tmp)); ((sum+=a)); done; echo $sum;
88
I'm new to Bash and I've been having issues with creating a script. What this script does is take numbers and add them to a total. However, I can't get total to work.It constantly claims that total is a non-variable despite it being assigned earlier in the program.
error message (8 is an example number being entered)
./adder: line 16: 0 = 0 + 8: attempted assignment to non-variable (error token is "= 0 + 8")
#!/bin/bash
clear
total=0
count=0
while [[ $choice != 0 ]]; do
echo Please enter a number or 0 to quit
read choice
if [[ $choice != 0 ]];
then
$(($total = $total + $choice))
$(($count = $count + 1))
echo Total is $total
echo
echo Total is derived from $count numbers
fi
done
exit 0
Get rid of some of the dollar signs in front of the variable names. They're optional inside of an arithmetic context, which is what ((...)) is. On the left-hand side of an assignment they're not just optional, they're forbidden, because = needs the variable name on the left rather than its value.
Also $((...)) should be plain ((...)) without the leading dollar sign. The dollar sign will capture the result of the expression and try to run it as a command. It'll try to run a command named 0 or 5 or whatever the computed value is.
You can write:
((total = $total + $choice))
((count = $count + 1))
or:
((total = total + choice))
((count = count + 1))
or even:
((total += choice))
((count += 1))
I'm trying to write a for loop that goes from 1 to 10, then calculates ( 1 through 10 mod 5) + 2. After that I want to display it like this (1 to 10 mod 5) + 2 = answer. However i'm getting an error at the beginning of the loop which is a syntax error.
for (( i = 0; i <= 10; i++)); do
calculate=(i % 5) + 2
echo ("("i "% 5) + 2" calculate)
done
Try these changes:
calculate=$(( i % 5 + 2 ))
# $(( ... )) is the shell's way to do arithmetic
echo "($i % 5) + 2 = " $calculate
# $x is a way to refer to the value of variable x
# (also inside a double-quoted string)
The for loop header is actually OK.
I want to calculate the average of 15 files:- ifile1.txt, ifile2.txt, ....., ifile15.txt. Number of columns and rows of each file are same. But some of them are missing values. Part of the data looks as
ifile1.txt ifile2.txt ifile3.txt
3 ? ? ? . 1 2 1 3 . 4 ? ? ? .
1 ? ? ? . 1 ? ? ? . 5 ? ? ? .
4 6 5 2 . 2 5 5 1 . 3 4 3 1 .
5 5 7 1 . 0 0 1 1 . 4 3 4 0 .
. . . . . . . . . . . . . . .
I would like to find a new file which will show the average of these 15 fils without considering the missing values.
ofile.txt
2.66 2 1 3 . (i.e. average of 3 1 4, average of ? 2 ? and so on)
2.33 ? ? ? .
3 5 4.33 1.33 .
3 2.67 4 0.66 .
. . . . .
This question is similar to my earlier question Average of multiple files in shell where the script was
awk 'FNR == 1 { nfiles++; ncols = NF }
{ for (i = 1; i < NF; i++) sum[FNR,i] += $i
if (FNR > maxnr) maxnr = FNR
}
END {
for (line = 1; line <= maxnr; line++)
{
for (col = 1; col < ncols; col++)
printf " %f", sum[line,col]/nfiles;
printf "\n"
}
}' ifile*.txt
But I can't able to modify it.
Use this:
paste ifile*.txt | awk '{n=f=0; for(i=1;i<=NF;i++){if($i*1){f++;n+=$i}}; print n/f}'
paste will show all files side by side
awk calculates the averages per line:
n=f=0; set the variables to 0.
for(i=1;i<=NF;i++) loop trough all the fields.
if($i*1) if the field contains a digit (multiplication by 1 will succeed).
f++;n+=$i increment f (number of fields with digits) and sum up n.
print n/f calculate n/f.
awk '
{
for (i = 1;i <= NF;i++) {
Sum[FNR,i]+=$i
Count[FNR,i]+=$i!="?"
}
}
END {
for( i = 1; i <= FNR; i++){
for( j = 1; j <= NF; j++) printf "%s ", Count[i,j] != 0 ? Sum[i,j]/Count[i,j] : "?"
print ""
}
}
' ifile*
assuming file are correctly feeded (no trailing empty space line, ...)
awk 'FNR == 1 { nfiles++; ncols = NF }
{ for (i = 1; i < NF; i++)
if ( $i != "?" ) { sum[FNR,i] += $i ; count[FNR,i]++ ;}
if (FNR > maxnr) maxnr = FNR
}
END {
for (line = 1; line <= maxnr; line++)
{
for (col = 1; col < ncols; col++)
if ( count[line,col] > 0 ) printf " %f", sum[line,col]/count[line,col];
else printf " ? " ;
printf "\n" ;
}
}' ifile*.txt
I just check the '?' ...
I have a matrix,
A(i,j), i=1,m and j=1,n
I can read it in C and FORTRAN, but I can't read it in shell script. I know this is a very simple question, but I am very new to shell script. I want to read all entries and do some calculation e.g. I have a matrix:
A= 1 0 1 1
2 1 0 2
1 0 0 3
1 2 3 0
Now I want to compare each 0 with its above, below, left and right values. Finally I want to do some computation (lets say sum) with these four values around each zero. In the above example the result will be- for five zeros
1st zero: 3
2nd zero: 4
3rd zero: 4
4th zero: 6
5th zero: 6
So in FORTRAN, I can do it by reading all the values as
do j=1,n
do i=1,m
if (A(i,j) .eq. 0) then
B(i,j)=A(i-1,j)+A(i+1,j)+A(i,j+1)+A(i,j-1)
enddo
enddo
But I want to do it in shell script. How to do?
Assuming that data are given in "test.dat" (with no "A = "), I tried it anyway...
#!/bin/bash
inpfile="test.dat"
L=100 # some large value
for (( i = 0; i < L; i++ )) {
for (( j = 0; j < L; j++ )) {
A[ L * i + j ]=0
}
}
i=1
while read buf; do
inp=( $buf ); n=${#inp[#]}
if (( L <= n+1 )); then echo "L is too small"; exit -1; fi
for (( j = 1; j <= n; j++ )) {
A[ L * i + j ]=${inp[j-1]}
}
(( i++ ))
done < $inpfile
nzeros=0
for (( i = 1; i <= n; i++ )) {
for (( j = 1; j <= n; j++ )) {
if (( ${A[ L * i + j ]} == 0 )); then
(( nzeros++ ))
B[ nzeros ]=$(( \
${A[ L * (i-1) + j ]} + \
${A[ L * (i+1) + j ]} + \
${A[ L * i + j+1 ]} + \
${A[ L * i + j-1 ]} ))
fi
}
}
for (( k = 1; k <= nzeros; k++ )) {
printf "%dst zero: %d\n" $k ${B[k]}
}
Conclusion: Very painful. Fortran is recommended...(as expected)