What I am trying to do is list all the numbers that are even, between the two numbers the user enters via a KornShell (ksh) script. So if user enters for the first digit 2 then the second digit 25 it would display
2,4,6,8,10,12,14,16,18,20,22,24
first=2 # from user
last=25 # from user
seq $first 2 $last
This should work with ksh93 and bash, doesn't require seq or perl which might not be installed depending on the OS used.
function evens {
for((i=($1+($1%2));i<($2-3);i+=2));do printf "%s," $i;done
echo $((i+2))
}
$ evens 2 25
2,4,6,8,10,12,14,16,18,20,24
$ evens 3 24
4,6,8,10,12,14,16,18,20,24
$ evens 0 9
0,2,4,8
In ksh, assuming you have used variables start and end:
set -A evens # use an array to store the numbers
n=0
i=$start
(( i % 2 == 1 )) && (( i+=1 )) # start at an even number
while (( i <= end )); do
evens[n]=$i
(( n+=1 ))
(( i+=2 ))
done
IFS=,
echo "${evens[*]}" # output comma separated string
outputs
2,4,6,8,10,12,14,16,18,20,22,24
there are many ways to do it in shell, shell script, awk, seq etc...
since you tagged question with vi, I added one with vim:
fun! GetEven(f,t)
let ff=a:f%2?a:f+1:a:f
echom join(range(ff,a:t,2),",")
endf
source that function, and type :call GetEven(2,25) you will see your expected output.
It currently echoes in command area, if you want it to be shown in file, just use put or setline, easy too.
Using perl:
perl -e 'print join q{,}, grep { $_ % 2 == 0 } (shift .. shift)' 2 25
It yields:
2,4,6,8,10,12,14,16,18,20,22,24
EDIT to fix the trailing newline:
perl -e 'print join( q{,}, grep { $_ % 2 == 0 } (shift .. shift) ), "\n"' 2 25
By setting first=$(($1+($1%2))) and using the -s option to format the output you can use seq:
first=$(($1+($1%2)))
last=$2
seq -s, $first 2 $last
Save as a script called evens and call with even values of $first:
$ ./evens 2 25
2,4,6,8,10,12,14,16,18,20,22,24
Or odd values of $first:
$ ./evens 3 25
4,6,8,10,12,14,16,18,20,22,24
Related
I'm would like to substitute a set of edit: single byte characters with a set of literal strings in a stream, without any constraint on the line size.
#!/bin/bash
for (( i = 1; i <= 0x7FFFFFFFFFFFFFFF; i++ ))
do
printf '\a,\b,\t,\v'
done |
chars_to_strings $'\a\b\t\v' '<bell>' '<backspace>' '<horizontal-tab>' '<vertical-tab>'
The expected output would be:
<bell>,<backspace>,<horizontal-tab>,<vertical-tab><bell>,<backspace>,<horizontal-tab>,<vertical-tab><bell>...
I can think of a bash function that would do that, something like:
chars_to_strings() {
local delim buffer
while true
do
delim=''
IFS='' read -r -d '.' -n 4096 buffer && (( ${#buffer} != 4096 )) && delim='.'
if [[ -n "${delim:+_}" ]] || [[ -n "${buffer:+_}" ]]
then
# Do the replacements in "$buffer"
# ...
printf "%s%s" "$buffer" "$delim"
else
break
fi
done
}
But I'm looking for a more efficient way, any thoughts?
Since you seem to be okay with using ANSI C quoting via $'...' strings, then maybe use sed?
sed $'s/\a/<bell>/g; s/\b/<backspace>/g; s/\t/<horizontal-tab>/g; s/\v/<vertical-tab>/g'
Or, via separate commands:
sed -e $'s/\a/<bell>/g' \
-e $'s/\b/<backspace>/g' \
-e $'s/\t/<horizontal-tab>/g' \
-e $'s/\v/<vertical-tab>/g'
Or, using awk, which replaces newline characters too (by customizing the Output Record Separator, i.e., the ORS variable):
$ printf '\a,\b,\t,\v\n' | awk -vORS='<newline>' '
{
gsub(/\a/, "<bell>")
gsub(/\b/, "<backspace>")
gsub(/\t/, "<horizontal-tab>")
gsub(/\v/, "<vertical-tab>")
print $0
}
'
<bell>,<backspace>,<horizontal-tab>,<vertical-tab><newline>
For a simple one-liner with reasonable portability, try Perl.
for (( i = 1; i <= 0x7FFFFFFFFFFFFFFF; i++ ))
do
printf '\a,\b,\t,\v'
done |
perl -pe 's/\a/<bell>/g;
s/\b/<backspace>/g;s/\t/<horizontal-tab>/g;s/\v/<vertical-tab>/g'
Perl internally does some intelligent optimizations so it's not encumbered by lines which are longer than its input buffer or whatever.
Perl by itself is not POSIX, of course; but it can be expected to be installed on any even remotely modern platform (short of perhaps embedded systems etc).
Assuming the overall objective is to provide the ability to process a stream of data in real time without having to wait for a EOL/End-of-buffer occurrence to trigger processing ...
A few items:
continue to use the while/read -n loop to read a chunk of data from the incoming stream and store in buffer variable
push the conversion code into something that's better suited to string manipulation (ie, something other than bash); for sake of discussion we'll choose awk
within the while/read -n loop printf "%s\n" "${buffer}" and pipe the output from the while loop into awk; NOTE: the key item is to introduce an explicit \n into the stream so as to trigger awk processing for each new 'line' of input; OP can decide if this additional \n must be distinguished from a \n occurring in the original stream of data
awk then parses each line of input as per the replacement logic, making sure to append anything leftover to the front of the next line of input (ie, for when the while/read -n breaks an item in the 'middle')
General idea:
chars_to_strings() {
while read -r -n 15 buffer # using '15' for demo purposes otherwise replace with '4096' or whatever OP wants
do
printf "%s\n" "${buffer}"
done | awk '{print NR,FNR,length($0)}' # replace 'print ...' with OP's replacement logic
}
Take for a test drive:
for (( i = 1; i <= 20; i++ ))
do
printf '\a,\b,\t,\v'
sleep 0.1 # add some delay to data being streamed to chars_to_strings()
done | chars_to_strings
1 1 15 # output starts printing right away
2 2 15 # instead of waiting for the 'for'
3 3 15 # loop to complete
4 4 15
5 5 13
6 6 15
7 7 15
8 8 15
9 9 15
A variation on this idea using a named pipe:
mkfifo /tmp/pipeX
sleep infinity > /tmp/pipeX # keep pipe open so awk does not exit
awk '{print NR,FNR,length($0)}' < /tmp/pipeX &
chars_to_strings() {
while read -r -n 15 buffer
do
printf "%s\n" "${buffer}"
done > /tmp/pipeX
}
Take for a test drive:
for (( i = 1; i <= 20; i++ ))
do
printf '\a,\b,\t,\v'
sleep 0.1
done | chars_to_strings
1 1 15 # output starts printing right away
2 2 15 # instead of waiting for the 'for'
3 3 15 # loop to complete
4 4 15
5 5 13
6 6 15
7 7 15
8 8 15
9 9 15
# kill background 'awk' and/or 'sleep infinity' when no longer needed
don't waste FS/OFS - use the built-in variables to take 2 out of the 5 needed :
echo $' \t abc xyz \t \a \n\n ' |
mawk 'gsub(/\7/, "<bell>", $!(NF = NF)) + gsub(/\10/,"<bs>") +\
gsub(/\11/,"<h-tab>")^_' OFS='<v-tab>' FS='\13' ORS='<newline>'
<h-tab> abc xyz <h-tab> <bell> <newline><newline> <newline>
To have NO constraint on the line length you could do something like this with GNU awk:
awk -v RS='.{1,100}' -v ORS= '{
$0 = RT
gsub(foo,bar)
print
}'
That will read and process the input 100 chars at a time no matter which chars are present, whether it has newlines or not, and even if the input was one multi-terabyte line.
Replace gsub(foo,bar) with whatever substitution(s) you have in mind, e.g.:
$ printf '\a,\b,\t,\v' |
awk -v RS='.{1,100}' -v ORS= '{
$0 = RT
gsub(/\a/,"<bell>")
gsub(/\b/,"<backspace>")
gsub(/\t/,"<horizontal-tab>")
gsub(/\v/,"<vertical-tab>")
print
}'
<bell>,<backspace>,<horizontal-tab>,<vertical-tab>
and of course it'd be trivial to pass a list of old and new strings to awk rather than hardcoding them, you'd just have to sanitize any regexp or backreference metachars before calling gsub().
How to generate 9 digit random number in shell?
I am trying something like this but it only gave numbers below 32768.
#!/bin/bash
mo=$((RANDOM%999999999))
echo "********Random"$mo
Please help
output should be ********Random453351111
In Linux with /dev/urandom:
$ rnd=$(tr -cd "[:digit:]" < /dev/urandom | head -c 9) && echo $rnd
463559879
I think this should make it
shuf -i 99999999-999999999 -n 1
As a work around, we could just simply ask for 1 random integer, for n times:
rand=''
for i in {1..9}; do
rand="${rand}$(( $RANDOM % 10 ))"
done
echo $rand
Try it online!
Note [1]: Since RANDOM's upper limit has a final digit of 7, there's a slightly lesser change for the 'generated' number to contain 8 or 9's.
Because of RANDOM's limited range, it can only be used to retrieve four base-10 digits at a time. Thus, to retrieve 9 digits, you need to call it three times.
If we don't care much about performance (are willing to pay process substitution costs), this may look like:
#!/usr/bin/env bash
get4() {
local newVal=32768
while (( newVal > 29999 )); do # avoid bias because of remainder
newVal=$RANDOM
done
printf '%04d' "$((newVal % 10000))"
}
result="$(get4)$(get4)$(get4)"
result=$(( result % 1000000000 ))
printf '%09d\n' "$result"
If we do care about performance, it may instead look like:
#!/usr/bin/env bash
get4() {
local newVal=32768 outVar=$1
while (( newVal > 29999 )); do # avoid bias because of remainder
newVal=$RANDOM
done
printf -v "$outVar" '%04d' "$((newVal % 10000))"
}
get4 out1; get4 out2; get4 out3
result="${out1}${out2}${out3}"
result=$(( result % 1000000000 ))
printf '%09d\n' "$result"
Use perl, as follows :
perl -e print\ rand | cut -c 3-11
Or
perl -MPOSIX -e 'print floor rand 10**9'
Usually when we wish to concatenate several files column wise and the filenames of the files are just consecutive increasing integers we can do the following:
#Imagine I have 10 files
paste {1..10} > out
However, I'm currently working on a script in which the ranges are variables, so I want to be able to do something like this
first=1
last=10
paste {"${first}".."${last}"} > out
This doesn't work as variables can't be correctly expanded within the curly braces. Is there an alternative syntax I can use to achieve the same result?
If you don't want to use eval, you can use seq(1):
seq -s ' ' "$first" "$last"
Like so:
paste $(seq -s ' ' "$first" "$last") > out
Once upon a time a needed a seq like function but a way faster, so i made this
# Create sequence like {0..X}
cnt () { printf -v N %$1s; N=(${N// / 1}); printf "${!N[*]}"; }
$ cnt 5
0 1 2 3 4
And if we modify it a bit
# Create sequence like {X..Y}
cnt () { printf -v N %$2s; N=(${N// / 1}); N=(${!N[#]}); printf "${N[*]:$1} ${#N[#]}"; }
$ cnt 7 11
7 8 9 10 11
I want to print a number with a certain field width for the digits, have the digits right-aligned, and print a sign indicator - not right before the digits, but rather before the spacing. Thus
$ magic -123 7
- 123
rather than
$ magic -123 7
-123
Can I do that with the GNU coreutils version of the printf utility? Other versions of it perhaps?
Note: To be clear, the solution should work for any field spacing and any value, e.g.:
There might be zero, one or many spaces
The number might "overflow" the specified width
Simply transform the output:
printf %+d 12 | sed 's/[+-]/\0 /'
+ 12
To directly answer your question, I do not believe that you can, with the GNU coreutils version of the printf, have space padding be inserted between the sign character and the nonzero digits of the number. printf seems to always group the sign with the unpadded digits, placing any additional space padding to the left of the sign.
You can use a function called magic like this using pure shell utilities:
magic() {
# some sanity checks to make sure you get $1 and $2
[[ $2 -lt 0 ]] && printf "-" || printf "+"
printf "%${1}s\n" "${2#[+-]}"
}
Now use it as:
$> magic 5 120
+ 120
$> magic 5 120234
+120234
$> magic 5 -120234
-120234
$> magic 5 -120
- 120
$> magic 5 1
+ 1
$> magic 5 +120
+ 120
Based on #KarolyHorvath's suggestion, I suppose this should work:
printf "%+7d" 123 | sed -r 's/^( *)([+-])/\2\1/'
magic () {
local sign="+" number=$1 width=$2
if ((number < 0)); then
sign="-"
((number *= -1))
fi
printf '%s%*d\n' "$sign" "$((width - 1))" "$number"
}
or
magic () {
printf '%+*d\n' "$2" "$1" | sed -r 's/^( *)([+-])/\2\1/'
}
Uses the * in the format specification to take the field width from the arguments.
I have a homework assignment that is asking to shift a decimal number by a specified amount of digits. More clearly this bash script will take two input arguments, the first is the number(maximum 9 digits) that the shift will be performed on and the second is the number(-9 to 9) of digits to shift. Another requirement is that when a digit is shifted off the end, it should be attached to the other end of the number. One headache of a requirement is that we cannot use control statements of any kind: no loops, no if, and switch cases.
Example: 12345 3 should come out to 345000012 and 12345 -3 should be 12345000
I know that if I mod 12345 by 10^3 I get 345 and then if I divide 12345 by 10^3 I get 12 and then I can just concatenate those two variables together to get 34512. I am not quite sure if that is exactly correct but that is the closest I can get as of now. As far as the -3 shift, I know that 10^-3 is .001 and would work however when I try using 10^-3 in bash I get an error.
I am just lost at this point, any tips would be greatly appreciated.
EDIT: After several hours of bashing (pun intended) my head against this problem, I finally came up with a script that for the most part works. I would post the code right now but I fear another student hopelessly lost might stumble upon it. I will check back and post what I came up with in a week or two. I was able to do it with mods and division. Thank you all for the responses, it really helped me to open up and think about the problem from different angles.
Here's a hint:
echo ${string:0:3}
echo ${#string}
Edit (2011-02-11):
Here's my solution. I added some additional parameters with defaults.
rotate-string ()
{
local s=${1:-1} p=${2:--1} w=${3:-8} c=${4:-0} r l
printf -vr '%0*d' $w 0 # save $w zeros in $r
r=${r//0/$c}$s # change the zeros to the character in $c, append the string
r=${r: -w} # save the last $w characters of $r
l=${r: -p%w} # get the last part of $r ($p mod %w characters)
echo "$l${r::w-${#l}}" # output the Last part on the Left and the Right part which starts at the beginning and goes for ($w minus the_length_of_the_Left_part) characters
}
usage: rotate-string string positions-to-rotate width fill-character
example: rotate-string abc -4 9 =
result: ==abc====
Arguments can be omitted starting from the end and these defaults will be used:
fill-character: "0"
width: 8
positions-to-rotate: -1
string: "1"
More examples:
$ rotate-string
00000010
$ rotate-string 123 4
01230000
Fun stuff:
$ for i in {126..6}; do printf '%s\r' "$(rotate-string Dennis $i 20 .)"; sleep .05; done; printf '\n'
$ while true; do for i in {10..1} {1..10}; do printf '%s\r' "$(rotate-string : $i 10 .)"; sleep .1; done; done
$ while true; do for i in {40..2} {2..40}; do printf '%s\r' "$(rotate-string '/\' $i 40 '_')"; sleep .02; done; done
$ d=0; while true; do for i in {1..10} {10..1}; do printf '%s\r' "$(rotate-string $d $i 10 '_')"; sleep .02; done; ((d=++d%10)); done
$ d=0; while true; do for i in {1..10}; do printf '%s\r' "$(rotate-string $d $i 10 '_')"; sleep .2; ((d=++d%10)); done; done
$ shape='▁▂▃▄▅▆▇█▇▆▅▄▃▂▁'; while true; do for ((i=1; i<=COLUMNS; i++)); do printf '%s\r' "$(rotate-string "$shape" $i $COLUMNS ' ')"; done; done
In the absence of control structures, you need to use recursion, with index values as "choice selections", which is how functional programming often works.
#!/bin/sh
#
# cshift NUMBER N
cshift() {
let num=10#$1
num=`printf '%09d' $num`
lshift="${num:1:8}${num:0:1}"
rshift="${num:8:1}${num:0:8}"
next=( "cshift $lshift $(($2 + 1))" "echo $num" "cshift $rshift $(( $2 - 1 ))" )
x=$(( $2 == 0 ? 1 : $2 < 0 ? 0 : 2 ))
eval "${next[x]}"
}
cshift $1 $2
and, the testing:
$ for ((i=-9;i<=9;i++)); do cshift 12345 $i ; done
000012345
500001234
450000123
345000012
234500001
123450000
012345000
001234500
000123450
000012345
500001234
450000123
345000012
234500001
123450000
012345000
001234500
000123450
000012345
You can also do some math on the indexes and avoid the recursion, but I don't mind making the computer work harder so I don't have to. It's easy to think of how to do the shift by one in either direction, and then I use an evaluated choice that is selected by the signum of the shift value, outputting a value and stopping when the shift value is zero.