How do I iterate over a range of numbers in Bash when the range is given by a variable?
I know I can do this (called "sequence expression" in the Bash documentation):
for i in {1..5}; do echo $i; done
Which gives:
1
2
3
4
5
Yet, how can I replace either of the range endpoints with a variable? This doesn't work:
END=5
for i in {1..$END}; do echo $i; done
Which prints:
{1..5}
I know this question is about bash
, but - just for the record - ksh93
is smarter and implements it as expected:
$ ksh -c 'i=5; for x in {1..$i}; do echo "$x"; done'
1
2
3
4
5
$ ksh -c 'echo $KSH_VERSION'
Version JM 93u+ 2012-02-29
$ bash -c 'i=5; for x in {1..$i}; do echo "$x"; done'
{1..5}
Another layer of indirection:
for i in $(eval echo {1..$END}); do
:
This works fine in bash
:
END=5
i=1 ; while [[ $i -le $END ]] ; do
echo $i
((i = i + 1))
done
The seq
method is the simplest, but Bash has built-in arithmetic evaluation.
END=5
for ((i=1;i<=END;i++)); do
echo $i
done
# ==> outputs 1 2 3 4 5 on separate lines
The for ((expr1;expr2;expr3));
construct works just like for (expr1;expr2;expr3)
in C and similar languages, and like other ((expr))
cases, Bash treats them as arithmetic.
There are many ways to do this, however the ones I prefer is given below
seq
Synopsis from
man seq
$ seq [-w] [-f format] [-s string] [-t string] [first [incr]] last
Syntax
Full command
seq first incr last
Example:
$ seq 1 2 10
1 3 5 7 9
Only with first and last:
$ seq 1 5
1 2 3 4 5
Only with last:
$ seq 5
1 2 3 4 5
{first..last..incr}
Here first and last are mandatory and incr is optional
Using just first and last
$ echo {1..5}
1 2 3 4 5
Using incr
$ echo {1..10..2}
1 3 5 7 9
You can use this even for characters like below
$ echo {a..z}
a b c d e f g h i j k l m n o p q r s t u v w x y z
I've combined a few of the ideas here and measured performance.
seq
and {..}
are really fastfor
and while
loops are slow$( )
is slowfor (( ; ; ))
loops are slower$(( ))
is even slowerThese are not conclusions. You would have to look at the C code behind each of these to draw conclusions. This is more about how we tend to use each of these mechanisms for looping over code. Most single operations are close enough to being the same speed that it's not going to matter in most cases. But a mechanism like for (( i=1; i<=1000000; i++ ))
is many operations as you can visually see. It is also many more operations per loop than you get from for i in $(seq 1 1000000)
. And that may not be obvious to you, which is why doing tests like this is valuable.
# show that seq is fast
$ time (seq 1 1000000 | wc)
1000000 1000000 6888894
real 0m0.227s
user 0m0.239s
sys 0m0.008s
# show that {..} is fast
$ time (echo {1..1000000} | wc)
1 1000000 6888896
real 0m1.778s
user 0m1.735s
sys 0m0.072s
# Show that for loops (even with a : noop) are slow
$ time (for i in {1..1000000} ; do :; done | wc)
0 0 0
real 0m3.642s
user 0m3.582s
sys 0m0.057s
# show that echo is slow
$ time (for i in {1..1000000} ; do echo $i; done | wc)
1000000 1000000 6888896
real 0m7.480s
user 0m6.803s
sys 0m2.580s
$ time (for i in $(seq 1 1000000) ; do echo $i; done | wc)
1000000 1000000 6888894
real 0m7.029s
user 0m6.335s
sys 0m2.666s
# show that C-style for loops are slower
$ time (for (( i=1; i<=1000000; i++ )) ; do echo $i; done | wc)
1000000 1000000 6888896
real 0m12.391s
user 0m11.069s
sys 0m3.437s
# show that arithmetic expansion is even slower
$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $i; i=$(($i+1)); done | wc)
1000000 1000000 6888896
real 0m19.696s
user 0m18.017s
sys 0m3.806s
$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $i; ((i=i+1)); done | wc)
1000000 1000000 6888896
real 0m18.629s
user 0m16.843s
sys 0m3.936s
$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $((i++)); done | wc)
1000000 1000000 6888896
real 0m17.012s
user 0m15.319s
sys 0m3.906s
# even a noop is slow
$ time (i=1; e=1000000; while [ $((i++)) -le $e ]; do :; done | wc)
0 0 0
real 0m12.679s
user 0m11.658s
sys 0m1.004s
The seq
method is the simplest, but Bash has built-in arithmetic evaluation.
END=5
for ((i=1;i<=END;i++)); do
echo $i
done
# ==> outputs 1 2 3 4 5 on separate lines
The for ((expr1;expr2;expr3));
construct works just like for (expr1;expr2;expr3)
in C and similar languages, and like other ((expr))
cases, Bash treats them as arithmetic.
You can use
for i in $(seq $END); do echo $i; done
Here is why the original expression didn't work.
From man bash:
Brace expansion is performed before any other expansions, and any characters special to other expansions are preserved in the result. It is strictly textual. Bash does not apply any syntactic interpretation to the context of the expansion or the text between the braces.
So, brace expansion is something done early as a purely textual macro operation, before parameter expansion.
Shells are highly optimized hybrids between macro processors and more formal programming languages. In order to optimize the typical use cases, the language is made rather more complex and some limitations are accepted.
Recommendation
I would suggest sticking with Posix1 features. This means using for i in <list>; do
, if the list is already known, otherwise, use while
or seq
, as in:
#!/bin/sh
limit=4
i=1; while [ $i -le $limit ]; do
echo $i
i=$(($i + 1))
done
# Or -----------------------
for i in $(seq 1 $limit); do
echo $i
done
Replace {}
with (( ))
:
tmpstart=0;
tmpend=4;
for (( i=$tmpstart; i<=$tmpend; i++ )) ; do
echo $i ;
done
Yields:
0
1
2
3
4
This is another way:
end=5
for i in $(bash -c "echo {1..${end}}"); do echo $i; done
If you want to stay as close as possible to the brace-expression syntax, try out the range
function from bash-tricks' range.bash
.
For example, all of the following will do the exact same thing as echo {1..10}
:
source range.bash
one=1
ten=10
range {$one..$ten}
range $one $ten
range {1..$ten}
range {1..10}
It tries to support the native bash syntax with as few "gotchas" as possible: not only are variables supported, but the often-undesirable behavior of invalid ranges being supplied as strings (e.g. for i in {1..a}; do echo $i; done
) is prevented as well.
The other answers will work in most cases, but they all have at least one of the following drawbacks:
seq
is a binary which must be installed to be used, must be loaded by bash, and must contain the program you expect, for it to work in this case. Ubiquitous or not, that's a lot more to rely on than just the Bash language itself.{a..z}
; brace expansion will. The question was about ranges of numbers, though, so this is a quibble.{1..10}
brace-expanded range syntax, so programs that use both may be a tiny bit harder to read.$END
variable is not a valid range "bookend" for the other side of the range. If END=a
, for example, an error will not occur and the verbatim value {1..a}
will be echoed. This is the default behavior of Bash, as well--it is just often unexpected.Disclaimer: I am the author of the linked code.
This works in Bash and Korn, also can go from higher to lower numbers. Probably not fastest or prettiest but works well enough. Handles negatives too.
function num_range {
# Return a range of whole numbers from beginning value to ending value.
# >>> num_range start end
# start: Whole number to start with.
# end: Whole number to end with.
typeset s e v
s=${1}
e=${2}
if (( ${e} >= ${s} )); then
v=${s}
while (( ${v} <= ${e} )); do
echo ${v}
((v=v+1))
done
elif (( ${e} < ${s} )); then
v=${s}
while (( ${v} >= ${e} )); do
echo ${v}
((v=v-1))
done
fi
}
function test_num_range {
num_range 1 3 | egrep "1|2|3" | assert_lc 3
num_range 1 3 | head -1 | assert_eq 1
num_range -1 1 | head -1 | assert_eq "-1"
num_range 3 1 | egrep "1|2|3" | assert_lc 3
num_range 3 1 | head -1 | assert_eq 3
num_range 1 -1 | tail -1 | assert_eq "-1"
}
These are all nice but seq is supposedly deprecated and most only work with numeric ranges.
If you enclose your for loop in double quotes, the start and end variables will be dereferenced when you echo the string, and you can ship the string right back to BASH for execution. $i
needs to be escaped with \'s so it is NOT evaluated before being sent to the subshell.
RANGE_START=a
RANGE_END=z
echo -e "for i in {$RANGE_START..$RANGE_END}; do echo \\${i}; done" | bash
This output can also be assigned to a variable:
VAR=`echo -e "for i in {$RANGE_START..$RANGE_END}; do echo \\${i}; done" | bash`
The only "overhead" this should generate should be the second instance of bash so it should be suitable for intensive operations.
if you don't wanna use 'seq
' or 'eval
' or jot
or arithmetic expansion format eg. for ((i=1;i<=END;i++))
, or other loops eg. while
, and you don't wanna 'printf
' and happy to 'echo
' only, then this simple workaround might fit your budget:
a=1; b=5; d='for i in {'$a'..'$b'}; do echo -n "$i"; done;' echo "$d" | bash
PS: My bash doesn't have 'seq
' command anyway.
Tested on Mac OSX 10.6.8, Bash 3.2.48
The POSIX way
If you care about portability, use the example from the POSIX standard:
i=2
end=5
while [ $i -le $end ]; do
echo $i
i=$(($i+1))
done
Output:
2
3
4
5
Things which are not POSIX:
(( ))
without dollar, although it is a common extension as mentioned by POSIX itself.[[
. [
is enough here. See also: What is the difference between single and double square brackets in Bash?for ((;;))
seq
(GNU Coreutils){start..end}
, and that cannot work with variables as mentioned by the Bash manual.let i=i+1
: POSIX 7 2. Shell Command Language does not contain the word let
, and it fails on bash --posix
4.3.42the dollar at i=$i+1
might be required, but I'm not sure. POSIX 7 2.6.4 Arithmetic Expansion says:
If the shell variable x contains a value that forms a valid integer constant, optionally including a leading plus or minus sign, then the arithmetic expansions "$((x))" and "$(($x))" shall return the same value.
but reading it literally that does not imply that $((x+1))
expands since x+1
is not a variable.
If you're on BSD / OS X you can use jot instead of seq:
for i in $(jot $END); do echo $i; done
Using seq
is fine, as Jiaaro suggested. Pax Diablo suggested a Bash loop to avoid calling a subprocess, with the additional advantage of being more memory friendly if $END is too large. Zathrus spotted a typical bug in the loop implementation, and also hinted that since i
is a text variable, continuous conversions to-and-fro numbers are performed with an associated slow-down.
This is an improved version of the Bash loop:
typeset -i i END
let END=5 i=1
while ((i<=END)); do
echo $i
…
let i++
done
If the only thing that we want is the echo
, then we could write echo $((i++))
.
ephemient taught me something: Bash allows for ((expr;expr;expr))
constructs. Since I've never read the whole man page for Bash (like I've done with the Korn shell (ksh
) man page, and that was a long time ago), I missed that.
So,
typeset -i i END # Let's be explicit
for ((i=1;i<=END;++i)); do echo $i; done
seems to be the most memory-efficient way (it won't be necessary to allocate memory to consume seq
's output, which could be a problem if END is very large), although probably not the “fastest”.
eschercycle noted that the {a..b} Bash notation works only with literals; true, accordingly to the Bash manual. One can overcome this obstacle with a single (internal) fork()
without an exec()
(as is the case with calling seq
, which being another image requires a fork+exec):
for i in $(eval echo "{1..$END}"); do
Both eval
and echo
are Bash builtins, but a fork()
is required for the command substitution (the $(…)
construct).
If you need it prefix than you might like this
for ((i=7;i<=12;i++)); do echo `printf "%2.0d\n" $i |sed "s/ /0/"`;done
that will yield
07
08
09
10
11
12
If you're doing shell commands and you (like I) have a fetish for pipelining, this one is good:
seq 1 $END | xargs -I {} echo {}
Source: Stackoverflow.com