Bash Script- adding numbers to each line in file

Question

I have a .txt file containing a column of values, something like:

2453882.157576
2453882.157947
2453882.159972
2453882.160354
2453882.160724

And I want to add a single value to each line in the code, let's say 5. So the output would be:

2453887.157576
2453887.157947
2453887.159972
2453887.160354
2453887.160724

I'm fairly new to bash commands and I am a bit unsure what commands to use. I thought it might be useful to try to iterate through the file and simply add through it:

while read i; do
  shift=$(($i + $5)) 
done <filename.txt > shifted_by_5_filename.txt

but this outputs an error "invalid arithmetic operator". Are there any suggestions on how to accomplish this task?

UPDATE: With some of the suggestions below, I've looked into using awk to try to add floats (rather than interger addition). I've found another question: how to add Integer number and a float number in unix shell script

that has an answer to add two floats:

echo 1.234 2.345 | awk '{print $1 + $2}'

To try to extend this to my problem, I've tried the following:

while read i
do echo $i 5 | awk '{print sprintf("%.9f", ($1+ $2))}'
done < filename.txt > shifted_by_5_filename.txt

The sprintf command is to try to produce an output not in scientific notation. However, as with the suggestions below, this attempt does not add by 5 rather it adds by 2.771413. I'm a bit perplexed by this. Any ideas?

Answer 1

Bash doesn't handle floating point arithmetics. You need an external command, eg bc :

sed 's/$/+5/' file.txt | bc

The sed command just adds +5 to the end of each line, it's faster and easier than reading the input line by line by the shell.

Answer 2

Bash can only do integer arithmetic. The simplest way is probably to use bc :

while read i; do
   echo $(bc <<< "$i + 5")                                                  
done <filename.txt > shifted_by_5_filename.txt

Alternatively, you could use some general purpose interpreter, eg Perl:

perl -ple '$_ += 5' filename.txt > shifted_by_5_filename.txt

Answer 3

Arithmetic expansion in bash works only for integers, so for floats you'll need bc , awk , python , perl , ruby , or similar. For example with bc :

while read i; do
   echo "$i + 5" | bc -l
done <filename.txt > shifted_by_5_filename.txt

Or with this simple awk script:

awk '{print $0+5}' filename.txt >shifted_by_5_filename.txt

Answer 4

I believe that there are two pretty clear issues here. First of all it looks like you are missing a parenthesis when you are performing the addition. Secondly, you do not need a dollar on the number 5.

With a shift variable:

while read i; do
    shift=5
    sum=$(($i+$shift))
    echo $sum
done < input.txt > output.txt

Or if you want to use your constant:

while read i; do
    sum=$(($i+5))
    echo $sum
done < input.txt > output.txt