I would like to compare a number, which is the output of a command, with a constant and do some manipulation. That is, if $id < 10, I want to see 590$id and if it is above 10, I want to see 59$id.
I found that expr doesn't working here:
ID=3
NUM=59$ID
if [ `expr $ID` -lt 10]; then
NUM=590$ID
fi
echo $NUM
The output of the code is 593 and not 5903. Even, $(($ID + 5900)) -lt 5910 writes 593.
How can I fix that?
Could you please try following.
cat script.sh
#!/bin/bash
ID=$(printf "%02d" 3 )
##NUM=59$ID ##Commented this to check if, condition is getting satisfied or not. Doesn't seem to be fit here.
(( $ID < 10 )) && NUM="59$ID"
echo "$NUM"
Output will be 5903 after running above code.
Don't use expr. It's old and tricky.
Don't use backticks `. They are discouraged and $( ... ) is preferred.
For arithmetic comparisons use arithmetic expansions. Just
if (( ID < 10 )); then
Note that bash is space aware and your script has a syntax error, it is missing a space - the 10]; should be 10 ];.
Note that by convention uppercase variables should be used for exported variables.
Looking at your code I think you just want:
NUM=$((5900 + ID))
Related
echo 3+3
How can I evaluate such expressions in Bash, in this case to 6?
echo $(( 3+3 ))
expr is the standard way, but it only handles integers.
bash has a couple of extensions, which only handle integers as well:
$((3+3)) returns 6
((3+3)) used in conditionals, returns 0 for true (non-zero) and 1 for false
let 3+3 same as (( ))
let and (( )) can be used to assign values, e.g.
let a=3+3
((a=3+3))
for floating point you can use bc
echo 3+3 | bc
in shells such as zsh/ksh, you can use floats for maths. If you need more maths power, use tools like bc/awk/dc
eg
var=$(echo "scale=2;3.4+43.1" | bc)
var=$(awk 'BEGIN{print 3.4*43.1}')
looking at what you are trying to do
awk '{printf "%.2f\n",$0/59.5}' ball_dropping_times >bull_velocities
You can make use of the expr command as:
expr 3 + 3
To store the result into a variable you can do:
sum=$(expr 3 + 3)
or
sum=`expr 3 + 3`
Lots of ways - most portable is to use the expr command:
expr 3 + 3
I believe the ((3+3)) method is the most rapid as it's interpreted by the shell rather than an external binary.
time a large loop using all suggested methods for the most efficient.
Solved thanks to Dennis, an example of BC-use:
$ cat calc_velo.sh
#!/bin/bash
for i in `cat ball_dropping_times`
do
echo "scale=20; $i / 59.5" | bc
done > ball_velocities
My understanding of math processing involves floating point processing.
Using bashj (https://sourceforge.net/projects/bashj/) you can call a java method (with floating point processing, cos(), sin(), log(), exp()...) using simply
bashj +eval "3+3"
bashj +eval "3.5*5.5"
or in a bashj script, java calls of this kind:
#!/usr/bin/bashj
EXPR="3.0*6.0"
echo $EXPR "=" u.doubleEval($EXPR)
FUNCTIONX="3*x*x+cos(x)+1"
X=3.0
FX=u.doubleEval($FUNCTIONX,$X)
echo "x="$X " => f(x)=" $FUNCTIONX "=" $FX
Note the interesting speed : ~ 10 msec per call (the answer is provided by a JVM server).
Note also that u.doubleEval(1/2) will provide 0.5 (floating point) instead of 0 (integer)
One use case that might be useful in this regard is, if one of your operand itself is a bash command then try this.
echo $(( `date +%s\`+10 )) or even echo $(( `date +%s\`+(60*60) ))
In my case I was trying to get Unixtime 10 seconds and hour later than current time respectively.
Taking count from file, say if count = 5, I want to print 5 variables. i.e. A B C D E.
If count = 2, Print 2 variables A B, etc.
I have tried using the ASCII values but couldn't go through it.
for i in {1..5}; do
count=5; a=0;
printf "\x$(printf %x '65+$a')";
count=count+1;
done
if count = 5, I want to print 5 variables. i.e. A B C D E. If count = 2, Print 2 variables A B, etc.
Here's a program that matches your style that does what you are looking for:
a=0
for i in {1..5}; do
printf "\x$(printf %x $(( 65 + a )) )";
a=$((a+1));
done
The first thing to note is that in order to do math in bash, you'll need to use the $(( )) operation. Above, you can see I replaced you '65+$a' with $(( 65 + a )) . That's the big news that you need to get math done.
There were a couple of other little issues, but you were stuck on the $(()) stuff so they weren't clear yet. Incidentally, the 'a' variable can be completely removed from the program to just use the 'i' variable like this:
for i in {1..5}; do
printf "\x$(printf %x $(( 64 + i )) )";
done
I had to change the constant to 64 since we are now counting starting at 1.
The {1..5} expression is a good short cut for 1 2 3 4 5, but you won't be able to put a variable into it. So, if you need to add a count variable back in, consider using the seq program instead like this:
count=$1
for i in $(seq 1 $count); do
printf "\x$(printf %x $(( 64 + i )) )";
done
Note that $() is different than the math operator $(()). $() runs a subcommand returning the results.
method 1: simple brace expansion
#!/bin/bash
# generate a lookup table
vars=( - $(echo {A..Z}) )
# use the elements
for i in {1..5}; do
echo ${vars[$i]}
done
{A..Z} generates 26 strings: A, B, ..., Z
which get stored in an array variable by vars=(...)
we prepend a - that we'll ignore
we can then do 1-based indexing into the array
limited to 26 variables (or whatever range we choose)
method 2: multiple brace expansion to generate arbitrary long variables
#!/bin/bash
if [[ ! $1 =~ ^[0-9]+$ ]]; then
echo "Usage: $0 count"
exit
fi
cmd='{A..Z}'
for (( i=$1; i>26; i=i/26 )); do
cmd="${A..Z}$cmd"
done
vars=( $(eval echo $cmd) )
for (( i=0; i<$1; i++ )); do
echo ${vars[$i]}
done
i/26 does integer division (throws away the remainder)
I'm lazy and generate "more than enough" variables rather than attempting to calculate how many is "exactly enough"
{a..b}{a..b}{a..b} becomes aaa aab aba abb baa bab bba bbb
using eval lets us do the brace expansion without knowing in advance how many sets are needed
Sample output:
$ mkvar.sh 10000 |fmt -64 | tail -5
ORY ORZ OSA OSB OSC OSD OSE OSF OSG OSH OSI OSJ OSK OSL OSM
OSN OSO OSP OSQ OSR OSS OST OSU OSV OSW OSX OSY OSZ OTA OTB
OTC OTD OTE OTF OTG OTH OTI OTJ OTK OTL OTM OTN OTO OTP OTQ
OTR OTS OTT OTU OTV OTW OTX OTY OTZ OUA OUB OUC OUD OUE OUF
OUG OUH OUI OUJ OUK OUL OUM OUN OUO OUP
I'm a bit stuck with something. I have a for loop like this:
#!/bin/bash
for i in {10..15}
do
I want to obtain the last digit of the number, so if i is 12, I want to get 2. I'm having difficulties with the syntax though. I've read that I should convert it into a character array, but when I do something like:
j=${i[#]}
echo $j
I don't get 1 0 1 1 1 2 and so on...I get 10, 11, 12...How do I get the numbers to be split up so I can get the last one of i, when I don't always know how many digits will make up i (ex. it may be 1, or 10, or a 100, etc.)?
Trick is to treat $i like a string.
for i in {10..15}; do j="${i: -1}"; echo $j; done
Of course, you do not need to assign to a variable if you don't want to:
for i in {10..15}; do echo "${i: -1}"; done
This answer which uses GNU shell parameter expansion is the most sensible method, I guess.
However, you can also use the double parenthesis construct which allows C-style manipulation of variables in Bash.
for i in {10..15}
do
(( j = i % 10 )) # modulo 10 always gives the ones' digit
echo $j
done
This awk command could solve your problem:
awk '{print substr($0,length,1)}' test_file
I'm assuming that the numbers are saved in a file test_file
If you want to use for loop:
for i in `cat test_1`
do
echo $i |tail -c 2
done
I am trying to read a file in bash forloop. But I do not know how to put write the script for this.
for i in $( seq 0 $step 10 )
do
echo "Rendering: "$(( i + j ))
python auto_fine.py density000000.vtu velocity000000.vtu $(( i + j ))
done
each and every loop I need to call
i -> 0 python auto_fine.py density000000.vtu velocity000000.vtu
i -> 1 python auto_fine.py density000010.vtu velocity000010.vtu
i -> 2 python auto_fine.py density000020.vtu velocity000020.vtu
It seems to me that you need to zero pad the numbers sed provides to you:
As seen in How to zero pad a sequence of integers in bash so that all have the same width?, you need to do something like
$ seq -f "%06g" 0 10 100
Which returns:
000000
000010
000020
...
000100
All together,
for i in $(seq -f "%06g" 0 10 100)
do
# echo "Rendering: "$(( i + j )) -- not sure what this does
python auto_fine.py density$i.vtu velocity$i.vtu
done
Bash can do this without requiring external tools like seq.
for i in {0..100}; do
[[ $i = *0 ]] || continue
python auto_fine.py density$(printf '%06d' $i).vtu velocity$(printf '%06d' $i).vtu
done
This uses pattern matching (*0) to limit your list to every 10 numbers, which is a bit of a hack, but will work against your sample data.
You could alternately loop against your zero-padded numeric strings directly:
for i in $(printf '%05d0 ' {0..10}); do
python auto_fine.py density$i.vtu velocity$i.vtu
done
This option shows you every 10 items by placing a zero in the printf format after the incrementing number, which becomes the tens digit. If you want more arbitrary sequencing, you might use multipliers, still without spawning external processes:
low=0
high=100
mult=10
for i in $(eval echo {$low..$((high/mult))}); do
n=$(printf '%06d' $((i*mult)))
python auto_fine.py density$n.vtu velocity$n.vtu
done
Note the eval, which lets you expand variables for use in your sequence expression. (If you are getting these numbers from an external source, have your script validate them before using them!)
If you're using bash version 4 (i.e. not the native version on OSX), you also have increments available in sequence expressions. From the man page:
A sequence expression takes the form {x..y[..incr]}, where x and y are
either integers or single characters, and incr, an optional increment,
is an integer.
So perhaps:
low=0
high=100
mult=10
for i in $(eval "printf '%06d ' {$low..$high..$mult}"); do
python auto_fine.py density$i.vtu velocity$i.vtu
done
Note that in sequence expressions, the first member of the sequence is the first number provided, rather than merely a product of a multiplier. We have quotes around the printf to ensure that the sequence expression is expanded by eval, and not interpreted by the command substitution ($(..)).
looping for all the files in the current dir is trivial:
for i in $( ls -1 )
do
# your code here, variable is referenced with $i
done
what's the j variable you are using?
echo 3+3
How can I evaluate such expressions in Bash, in this case to 6?
echo $(( 3+3 ))
expr is the standard way, but it only handles integers.
bash has a couple of extensions, which only handle integers as well:
$((3+3)) returns 6
((3+3)) used in conditionals, returns 0 for true (non-zero) and 1 for false
let 3+3 same as (( ))
let and (( )) can be used to assign values, e.g.
let a=3+3
((a=3+3))
for floating point you can use bc
echo 3+3 | bc
in shells such as zsh/ksh, you can use floats for maths. If you need more maths power, use tools like bc/awk/dc
eg
var=$(echo "scale=2;3.4+43.1" | bc)
var=$(awk 'BEGIN{print 3.4*43.1}')
looking at what you are trying to do
awk '{printf "%.2f\n",$0/59.5}' ball_dropping_times >bull_velocities
You can make use of the expr command as:
expr 3 + 3
To store the result into a variable you can do:
sum=$(expr 3 + 3)
or
sum=`expr 3 + 3`
Lots of ways - most portable is to use the expr command:
expr 3 + 3
I believe the ((3+3)) method is the most rapid as it's interpreted by the shell rather than an external binary.
time a large loop using all suggested methods for the most efficient.
Solved thanks to Dennis, an example of BC-use:
$ cat calc_velo.sh
#!/bin/bash
for i in `cat ball_dropping_times`
do
echo "scale=20; $i / 59.5" | bc
done > ball_velocities
My understanding of math processing involves floating point processing.
Using bashj (https://sourceforge.net/projects/bashj/) you can call a java method (with floating point processing, cos(), sin(), log(), exp()...) using simply
bashj +eval "3+3"
bashj +eval "3.5*5.5"
or in a bashj script, java calls of this kind:
#!/usr/bin/bashj
EXPR="3.0*6.0"
echo $EXPR "=" u.doubleEval($EXPR)
FUNCTIONX="3*x*x+cos(x)+1"
X=3.0
FX=u.doubleEval($FUNCTIONX,$X)
echo "x="$X " => f(x)=" $FUNCTIONX "=" $FX
Note the interesting speed : ~ 10 msec per call (the answer is provided by a JVM server).
Note also that u.doubleEval(1/2) will provide 0.5 (floating point) instead of 0 (integer)
One use case that might be useful in this regard is, if one of your operand itself is a bash command then try this.
echo $(( `date +%s\`+10 )) or even echo $(( `date +%s\`+(60*60) ))
In my case I was trying to get Unixtime 10 seconds and hour later than current time respectively.