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how to sum variables from file in bash
How to name a file in the deepest level of a directory treeUse & (ampersand) in single line bash loopScript with multiple arithmetic conditions failingbash division of numbers in text fileProblem with data comparison in if loopArithmetic syntax error with a number variablePrint file line-by-line, but handle the file changing mid-executionPutting string read from file inside double quotebash script to find files thinks the file name should be an integerString to integer in Shell
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
Hello I need to sum numbers from a file line by line.
The file:
1.0
0.46
0.67
I want to sum, then divide 3.
I currently have:
while IFS= read -r var
do
x=$(($var + $x)) | bc -l
done < "file.txt"
echo "$x / 3"
My error:
-bash: 1.0 + 0: syntax error: invalid arithmetic operator (error token is ".0 + 0")
bash arithmetic
add a comment |
Hello I need to sum numbers from a file line by line.
The file:
1.0
0.46
0.67
I want to sum, then divide 3.
I currently have:
while IFS= read -r var
do
x=$(($var + $x)) | bc -l
done < "file.txt"
echo "$x / 3"
My error:
-bash: 1.0 + 0: syntax error: invalid arithmetic operator (error token is ".0 + 0")
bash arithmetic
add a comment |
Hello I need to sum numbers from a file line by line.
The file:
1.0
0.46
0.67
I want to sum, then divide 3.
I currently have:
while IFS= read -r var
do
x=$(($var + $x)) | bc -l
done < "file.txt"
echo "$x / 3"
My error:
-bash: 1.0 + 0: syntax error: invalid arithmetic operator (error token is ".0 + 0")
bash arithmetic
Hello I need to sum numbers from a file line by line.
The file:
1.0
0.46
0.67
I want to sum, then divide 3.
I currently have:
while IFS= read -r var
do
x=$(($var + $x)) | bc -l
done < "file.txt"
echo "$x / 3"
My error:
-bash: 1.0 + 0: syntax error: invalid arithmetic operator (error token is ".0 + 0")
bash arithmetic
bash arithmetic
edited 2 hours ago
Jesse_b
14.7k33574
14.7k33574
asked 2 hours ago
AhmetMelihBaşbuğAhmetMelihBaşbuğ
1175
1175
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
bash/shell arithmetic can't handle floating point arithmetic. You can accomplish your task using awk:
awk 'sum= sum+$1 END print sum/3' file
This will read through your file and add each line to sum then after it completes reading the file it will print sum divided by 3.
add a comment |
Just for fun
$ paste -sd+ file | dc -e2k0 -f- -e+3/p
.71
add a comment |
Try this:
x=0
while IFS= read -r var; do
x="$(echo "$var + $x" | bc -l)"
done < "file.txt"
echo "$x / 3" | bc -l
xwas not defined.- You need an
echoforbcto work (not mixed bash arithmetic).
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
bash/shell arithmetic can't handle floating point arithmetic. You can accomplish your task using awk:
awk 'sum= sum+$1 END print sum/3' file
This will read through your file and add each line to sum then after it completes reading the file it will print sum divided by 3.
add a comment |
bash/shell arithmetic can't handle floating point arithmetic. You can accomplish your task using awk:
awk 'sum= sum+$1 END print sum/3' file
This will read through your file and add each line to sum then after it completes reading the file it will print sum divided by 3.
add a comment |
bash/shell arithmetic can't handle floating point arithmetic. You can accomplish your task using awk:
awk 'sum= sum+$1 END print sum/3' file
This will read through your file and add each line to sum then after it completes reading the file it will print sum divided by 3.
bash/shell arithmetic can't handle floating point arithmetic. You can accomplish your task using awk:
awk 'sum= sum+$1 END print sum/3' file
This will read through your file and add each line to sum then after it completes reading the file it will print sum divided by 3.
answered 2 hours ago
Jesse_bJesse_b
14.7k33574
14.7k33574
add a comment |
add a comment |
Just for fun
$ paste -sd+ file | dc -e2k0 -f- -e+3/p
.71
add a comment |
Just for fun
$ paste -sd+ file | dc -e2k0 -f- -e+3/p
.71
add a comment |
Just for fun
$ paste -sd+ file | dc -e2k0 -f- -e+3/p
.71
Just for fun
$ paste -sd+ file | dc -e2k0 -f- -e+3/p
.71
answered 1 hour ago
steeldriversteeldriver
38.2k45489
38.2k45489
add a comment |
add a comment |
Try this:
x=0
while IFS= read -r var; do
x="$(echo "$var + $x" | bc -l)"
done < "file.txt"
echo "$x / 3" | bc -l
xwas not defined.- You need an
echoforbcto work (not mixed bash arithmetic).
add a comment |
Try this:
x=0
while IFS= read -r var; do
x="$(echo "$var + $x" | bc -l)"
done < "file.txt"
echo "$x / 3" | bc -l
xwas not defined.- You need an
echoforbcto work (not mixed bash arithmetic).
add a comment |
Try this:
x=0
while IFS= read -r var; do
x="$(echo "$var + $x" | bc -l)"
done < "file.txt"
echo "$x / 3" | bc -l
xwas not defined.- You need an
echoforbcto work (not mixed bash arithmetic).
Try this:
x=0
while IFS= read -r var; do
x="$(echo "$var + $x" | bc -l)"
done < "file.txt"
echo "$x / 3" | bc -l
xwas not defined.- You need an
echoforbcto work (not mixed bash arithmetic).
answered 2 hours ago
FreddyFreddy
2,241210
2,241210
add a comment |
add a comment |
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