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}
for i in $(seq 1 $END); do echo $i; doneedit: I prefer seq over the other methods because I can actually remember it ;)
seq $END would suffice, as the default is to start from 1. From man seq: "If FIRST or INCREMENT is omitted, it defaults to 1". –
Radio {4..1} gives 4 3 2 1 since the negative "increment" is deduced automatically. Conversely, seq 4 1 gives nothing, because the default increment is 1. In such cases, the full command is to be used: seq 4 -1 1. –
Pesthole $END is just a shell variable which presumed to be containing some number. –
Bowleg 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.
typeset -i i END though. In the pre-bash times (i.e. ksh) it made a difference, but computers were much slower then. –
Ivatts seq. Use it! –
Mutinous #!/bin/bash the first line of your script. wiki.ubuntu.com/… –
Faceharden seq if there's a chance you want the loop to make zero iterations. seq 0 0 results in 1 iteration. seq 0 results in TWO iterations. lol! –
Lais 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).
for ((i=$1;i<=$2;++i)); do echo $i; done in a script works fine for me on bash v.4.1.9, so I don't see a problem with command line arguments. Do you mean something else? –
Ivatts time for i in $(seq 100000); do :; done is a lot quicker! –
Whitson /bin/sh (the dash, not Bash) –
Earthly 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
seq for large ranges. E.g., echo {1..1000000} | wc reveals that the echo produces 1 line, a million words, and 6,888,896 bytes. Trying seq 1 1000000 | wc yields a million lines, a million words, and 6,888,896 bytes and is also more than seven times faster, as measured by the time command. –
Universal while method previously in my answer: https://mcmap.net/q/44968/-how-do-i-iterate-over-a-range-of-numbers-defined-by-variables-in-bash But glad you agree :-) –
Ful printf "%02i\n" $i in the for loop. Try printf "%02i\n" $((i++)) for the while loop method. Format string: "0" indicates pad with zero vs space, "2" is field width, "i" for integer, and don't forget your newline :) –
Letdown 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.
x, not the entire expression. $((x + 1)) is just fine. –
Duong seq (BSD seq allows you to set a sequence termination string with -t), FreeBSD and NetBSD also have seq since 9.0 and 3.0, respectively. –
Fairy $((x+1)) and $((x + 1)) parse exactly the same, as when the parser tokenizes x+1 it will be split into 3 tokens: x, +, and 1. x isn't a valid numerical token, but it is a valid variable name token, yet x+ is not, hence the split. + is a valid arithmetic operator token, yet +1 is not, so the token is again split there. And so forth. –
Fairy You can use
for i in $(seq $END); do echo $i; done
Another layer of indirection:
for i in $(eval echo {1..$END}); do
∶
for n in {001..10}; do echo $n; done –
Broadnax I've combined a few of the ideas here and measured performance.
seq and {..} are really fastfor and while loops are slow$( ) 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
$(seq) is about the same speed as {a..b}. Also, each operation takes about the same time, so adds about 4μs to each iteration of the loop for me. Here an operation is an echo in the body, an arithmetic compare, an increment, etc. Is any of this surprising? Who cares how long it takes the loop paraphernalia to do its job—the runtime is likely to be dominated by the contents of the loop. –
Mutinous 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
printf "%02d\n" $i be easier than printf "%2.0d\n" $i |sed "s/ /0/"? –
Asperse There are many ways to do this, however the ones I prefer is given below
seqSynopsis 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
If you're on BSD / OS X you can use jot instead of seq:
for i in $(jot $END); do echo $i; done
seq –
Rakel The seq command first appeared in Plan 9 from Bell Labs. A seq command appeared in NetBSD 3.0, and ported to FreeBSD 9.0. This command was based on the command of the same name in Plan 9 from Bell Labs and the GNU core utilities. The GNU seq command first appeared in the 1.13 shell utilities release. –
Rakel This works fine in bash:
END=5
i=1 ; while [[ $i -le $END ]] ; do
echo $i
((i = i + 1))
done
echo $((i++)) works and combines it onto one line. –
Andreaandreana bash, we can simply do while [[ i++ -le "$END" ]]; do to do the (post-) increment in the test –
Pandorapandour bash, you could even more simply do while (( i++ < END )); do; there's almost no reason to ever use [[ ... ]] for arithmetic expressions or comparisons. –
Duong 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}
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.
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.
Replace {} with (( )):
tmpstart=0;
tmpend=4;
for (( i=$tmpstart; i<=$tmpend; i++ )) ; do
echo $i ;
done
Yields:
0
1
2
3
4
If you're doing shell commands and you (like I) have a fetish for pipelining, this one is good:
seq 1 $END | xargs -I {} echo {}
if you don't wanna use 'seq' or 'eval' or jot or arithmetic expansion format eg. for ((i=1;i<=END;i++)), or other loops eg. while, and you don't wanna 'printf' and happy to 'echo' only, then this simple workaround might fit your budget:
a=1; b=5; d='for i in {'$a'..'$b'}; do echo -n "$i"; done;' echo "$d" | bash
PS: My bash doesn't have 'seq' command anyway.
Tested on Mac OSX 10.6.8, Bash 3.2.48
This works in Bash and Korn, also can go from higher to lower numbers. Probably not fastest or prettiest but works well enough. Handles negatives too.
function num_range {
# Return a range of whole numbers from beginning value to ending value.
# >>> num_range start end
# start: Whole number to start with.
# end: Whole number to end with.
typeset s e v
s=${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"
}
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for i in {01..10}; do echo $i; donewould give numbers like01, 02, 03, ..., 10. – Planogametemyarray=('a' 'b' 'c'); for i in ${!myarray[@]}; do echo $i; done(note the exclamation mark). It's more specific than the original question, but could help. See bash parameter expansions – Saraband{jpg,png,gif}which isn't directly addressed here, though the answer will be identical. See Brace expansion with variable? [duplicate] which is marked as a duplicate of this one. – Natchseqnow "these days" FWIW... – Debarfor i in {1..${END}}; do echo ${i}; done. This works in terminal command line but not from within a SHELL script, which confuses me. – GironENDends up being equal or larger thanSTART, e.g.END=0andSTART=1,seqwill generate a non-empty sequence, and accordingly your loop will run at least 1 iteration. In most cases, that's not what you want. It's a bug. – Forewent