bash loop through list of numbers except given number - bash

Question

to loop through a continous list of numbers in bash I can do
for s in $(seq 1 5);do
echo ${s}
done
to loop through a continous list of numbers leaving a given number out in python I can do:
list = [s2 for s2 in range(6)[1:] if s2 != s1]
for s1 in list:
print s1
where list contains all numbers in range except s1
How do I do the same in bash?

Answer 1

Just use continue to skip this step:
for s in {1..5} # note there is no need to use $(seq...)
do
[ "$s" -eq 3 ] && continue # if var is for example 3, jump to next loop
echo "$s"
done
This returns:
1
2
4 # <--- 3 is skipped
5
From Bash Reference Manual → 4.1 Bourne Shell Builtins:
continue
continue [n]
Resume the next iteration of an enclosing for, while, until, or select
loop. If n is supplied, the execution of the nth enclosing loop is
resumed. n must be greater than or equal to 1. The return status is
zero unless n is not greater than or equal to 1.

Answer 2

Add a short circuit evaluation, || (logical OR) :
for s in $(seq 1 5); do
(( s == 3 )) || echo "$s"
done
(( s == 3 )) checks if $s is equal to 3, if not (||) echo the number.
With the reverse check ($s not equal to 3) and logical AND (&&):
for s in $(seq 1 5); do
(( s != 3 )) && echo "$s"
done
The classic way, if with test ([), non-equity test:
for s in $(seq 1 5); do
if [ "$s" -ne 3 ]; then
echo "$s"
fi
done
Reverse test, equity check:
for s in $(seq 1 5); do
if [ "$s" -eq 3 ]; then
continue
fi
echo "$s"
done
continue will make the loop control to go at the top rather than evaluating the following commands.
There is also a bash keyword [[ which behaves similarly in most cases but more robust.

Answer 3

You can use BASH arithmetic construct ((...)) like this:
s1=3 # skip this
s2=6 # upper count
for ((i=1; i<s2; i+=(i==s1-1?2:1) )); do echo $i; done
1
2
4
5
About: i+=(i==s1-1?2:1)
In the for loop instead of always incrementing i by 1 here we are incrementing i by 2 when i is 1 less then the number to be skipped.
Alternatively solution using BASH array:
arr=({1..5}) # populate 1 to 5 in an array
unset arr[s1-1] # delete element s1-1
# print the array
printf "%s\n" "${arr[#]}"
1
2
4
5