Is there a one liner that shifts all characters in a string by some i number. The input string can contain any ascii characters. It would be for a cypher.
For example, if b comes after a then command 1 "ab" returns "bc", command 3 "ab" returns "de". It should work with any ascii character not just with letters.
The command you want is called caesar.
this gawk command gives you new sequence with ascii code+1:
awk 'BEGIN{FS=OFS="";s=2;for(n=0;n<=127;n++)ord[sprintf("%c",n)]=n}
{for(i=1;i<=NF;i++)$i=sprintf("%c",(ord[$i]+s)%127)}7'
test with string shifted with step 2:
kent$ echo "xyab+123"|awk 'BEGIN{FS=OFS="";s=2;for(n=0;n<=127;n++)ord[sprintf("%c",n)]=n}{for(i=1;i<=NF;i++)$i=sprintf("%c",(ord[$i]+s)%127)}7'
z{cd-345
you just need pass the s as variable, to define shift step.
Use Perl:
echo -n bbb | perl -F'' -ane 'foreach(#F){$_++; printf "$_"}END{print "\n"}'
ccc
If you need shift a N chars (in the case below 4):
echo -n bbb | perl -F'' -ane 'foreach(#F){ $a=ord($_); $a+=4; print chr($a)} END{print "\n"}'
fff
Shifting to negative value:
echo -n bbb | perl -F'' -ane 'foreach(#F){ $a=ord($_); $a-=1; print chr($a)} END{print "\n"}'
aaa
Related
my file contains lines like this
any1 aaa bbb ccc
The delimiter is space. the number of words in the line is unknown
I want to put the first word into a var1. It's simple with
awk '{print $1}'
Now I want to put the rest of the line into a var2 with awk.
How I can print the rest of the line with awk ?
Better to use read here:
s="any1 aaa bbb ccc"
read var1 var2 <<< "$s"
echo "$var1"
any1
echo "$var2"
aaa bbb ccc
For awk only solution use:
echo "$s" | awk '{print $1; print substr($0, index($0, " ")+1)}'
any1
aaa bbb ccc
$ var=$(awk '{sub(/^[^[:space:]]+[[:space:]]+/,"")}1' file)
$ echo "$var"
aaa bbb ccc
or in general to skip some number of fields use a RE interval:
$ awk '{sub(/^[[:space:]]*([^[:space:]]+[[:space:]]+){1}/,"")}1' file
aaa bbb ccc
$ awk '{sub(/^[[:space:]]*([^[:space:]]+[[:space:]]+){2}/,"")}1' file
bbb ccc
$ awk '{sub(/^[[:space:]]*([^[:space:]]+[[:space:]]+){3}/,"")}1' file
ccc
Note that doing this gets much more complicated if you have a FS that's more than a single char, and the above is just for the default FS since it additionally skips any leading blanks if present (remove the first [[:space:]]* if you have a non-default but still single-char FS).
awk solution:
awk '{$1 = ""; print $0;}'`
Given input such as:
1
1a
1.1b
2.0c
How to extract the integer/decimal number at beginning of each input line, using only Linux/Unix command line utilities?
Using awk, you could say:
awk '{print $0+0}'
Awk is available in Linux, BSD, and many other Unix-like operating systems. It helps in this way:
echo "1" | awk '{a+=$0; print a}' # output 1
echo "1a" | awk '{a+=$0; print a}' # output 1
echo "1.1b" | awk '{a+=$0; print a}' # output 1.1
echo "2.0c" | awk '{a+=$0; print a}' # output 2
Some more awk
For extracting only digits
$ awk 'gsub(/[[:alpha:]].*/,x,$1) + 1' << EOF
1
1a
1.1b
2.0c
EOF
1
1
1.1
2.0
For integer
$ awk '{print int($0)}' << EOF
1
1a
1.1b
2.0c
EOF
1
1
1
2
---edit---
If there is any blank line in file, you can avoid printing zero from following
$ awk 'NF{$0+=0}1' << EOF
1
1a
1.1b
2foot4c
2
EOF
1
1
1.1
2
2
Here is a way to do this with sed:
echo "12.3abc" | sed -n 's/^\([0-9.][0-9.]*\).*/\1/p'
Output:
12.3
The block in parentheses matches all numbers or periods '.' that occur at the beginning of the line. Everything after that is match by the '.*'.
The \1 says to replace the entire line with just the portion that was matched in the parentheses.
Assuming your version of grep supports -o:
grep -o '^[0-9.]\+' data.in
NB: This will match any sequence of digits and decimal points at the start of the line.
I have a string
ABCDEFGHIJ
I would like it to print.
A
B
C
D
E
F
G
H
I
J
ie horizontal, no editing between characters to vertical. Bonus points for how to put a number next to each one with a single line. It'd be nice if this were an awk or shell script, but I am open to learning new things. :) Thanks!
If you just want to convert a string to one-char-per-line, you just need to tell awk that each input character is a separate field and that each output field should be separated by a newline and then recompile each record by assigning a field to itself:
awk -v FS= -v OFS='\n' '{$1=$1}1'
e.g.:
$ echo "ABCDEFGHIJ" | awk -v FS= -v OFS='\n' '{$1=$1}1'
A
B
C
D
E
F
G
H
I
J
and if you want field numbers next to each character, see #Kent's solution or pipe to cat -n.
The sed solution you posted is non-portable and will fail with some seds on some OSs, and it will add an undesirable blank line to the end of your sed output which will then become a trailing line number after your pipe to cat -n so it's not a good alternative. You should accept #Kent's answer.
awk one-liner:
awk 'BEGIN{FS=""}{for(i=1;i<=NF;i++)print i,$i}'
test :
kent$ echo "ABCDEF"|awk 'BEGIN{FS=""}{for(i=1;i<=NF;i++)print i,$i}'
1 A
2 B
3 C
4 D
5 E
6 F
So I figured this one out on my own with sed.
sed 's/./&\n/g' horiz.txt > vert.txt
One more awk
echo "ABCDEFGHIJ" | awk '{gsub(/./,"&\n")}1'
A
B
C
D
E
F
G
H
I
J
This might work for you (GNU sed):
sed 's/\B/\n/g' <<<ABCDEFGHIJ
for line numbers:
sed 's/\B/\n/g' <<<ABCDEFGHIJ | sed = | sed 'N;y/\n/ /'
or:
sed 's/\B/\n/g' <<<ABCDEFGHIJ | cat -n
I want to find the last appearance of a string in a text file with linux commands. For example
1 a 1
2 a 2
3 a 3
1 b 1
2 b 2
3 b 3
1 c 1
2 c 2
3 c 3
In such a text file, i want to find the line number of the last appearance of b which is 6.
I can find the first appearance with
awk '/ b / {print NR;exit}' textFile.txt
but I have no idea how to do it for the last occurrence.
cat -n textfile.txt | grep " b " | tail -1 | cut -f 1
cat -n prints the file to STDOUT prepending line numbers.
grep greps out all lines containing "b" (you can use egrep for more advanced patterns or fgrep for faster grep of fixed strings)
tail -1 prints last line of those lines containing "b"
cut -f 1 prints first column, which is line # from cat -n
Or you can use Perl if you wish (It's very similar to what you'd do in awk, but frankly, I personally don't ever use awk if I have Perl handy - Perl supports 100% of what awk can do, by design, as 1-liners - YMMV):
perl -ne '{$n=$. if / b /} END {print "$n\n"}' textfile.txt
This can work:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
We check every second file being "b" and we record the number of line. It is appended, so by the time we finish reading the file, it will be the last one.
Test:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
6
Update based on sudo_O advise:
$ awk '{if ($2=="b") a=NR} END{print a}' your_file
to avoid having some abc in 2nd field.
It is also valid this one (shorter, I keep the one above because it is the one I thought :D):
$ awk '$2=="b" {a=NR} END{print a}' your_file
Another approach if $2 is always grouped (may be more efficient then waiting until the end):
awk 'NR==1||$2=="b",$2=="b"{next} {print NR-1; exit}' file
or
awk '$2=="b"{f=1} f==1 && $2!="b" {print NR-1; exit}' file
Is there a Linux utility or a Bash command I can use to sort a space delimited string of numbers?
Here's a simple example to get you going:
echo "81 4 6 12 3 0" | tr " " "\n" | sort -g
tr translates the spaces delimiting the numbers, into carriage returns, because sort uses carriage returns as delimiters (ie it is for sorting lines of text). The -g option tells sort to sort by "general numerical value".
man sort for further details about sort.
This is a variation from #JamesMorris answer:
echo "81 4 6 12 3 0" | xargs -n1 | sort -g | xargs
Instead of tr, I use xargs -n1 to convert to new lines. The final xargs is to convert back, to a space separated sequence of numbers.
This is a variation on ghostdog74's answer that's too big to fit in a comment. It shows digits instead of names of numbers and both the original string and the result are in space-delimited strings (instead of an array which becomes a newline-delimited string).
$ s="3 2 11 15 8"
$ sorted=$(echo $(printf "%s\n" $s | sort -n))
$ echo $sorted
2 3 8 11 15
$ echo "$sorted"
2 3 8 11 15
If you didn't use the echo when setting the value of sorted, then the string has newlines in it. In that case echoing it without quotes puts it all on one line, but, as echoing it with quotes would show, each number would appear on its own line. This is the case whether the original is an array or a string.
# demo
$ s="3 2 11 15 8"
$ sorted=$(printf "%s\n" $s | sort -n)
$ echo $sorted
2 3 8 11 15
$ echo "$sorted"
2
3
8
11
15
$ s=(one two three four)
$ sorted=$(printf "%s\n" ${s[#]}|sort)
$ echo $sorted
four one three two
Using Bash parameter expansion (to replace spaces with newlines) we can do:
str="3 2 11 15 8"
sort -n <<< "${str// /$'\n'}"
# alternative
NL=$'\n'
str="3 2 11 15 8"
sort -n <<< "${str// /${NL}}"
If you actually have a space-delimited string of numbers, then one of the other answers provided would work fine. If your list is a bash array, then:
oldIFS="$IFS"
IFS=$'\n'
array=($(sort -g <<< "${array[*]}"))
IFS="$oldIFS"
might be a better solution. The newline delimiter would help if you want to generalize to sorting an array of strings instead of numbers.
Improving on Evan Krall's nice Bash "array sort" by limiting the scope of IFS to a single command:
printf "%q\n" "${IFS}"
array=(3 2 11 15 8)
array=($(IFS=$'\n' sort -n <<< "${array[*]}"))
echo "${array[#]}"
printf "%q\n" "${IFS}"
$ awk 'BEGIN{split(ARGV[1], numbers);for(i in numbers) {print numbers[i]} }' \
"6 7 4 1 2 3" | sort -n
I added this to my .zshrc (or .bashrc) file:
#sort a space-separated list of words (e.g. a list of HTML classes)
sortwords() {
echo $1 | xargs -n1 | sort -g | xargs
}
Call it from the terminal like this:
sortwords "banana date apple cherry"
# apple banana cherry date
Thanks to #FranMowinckel and others for inspiration.