I have a string
indent code by 4 spaces and 6 letters and by 5 constants
I want to extract each number separately.
1st number : 4
2nd number : 6
The string may change, so I need a one-line command that does it.
Read all numbers into an array:
readarray -t arr < \
<(grep -Eo '[[:digit:]]+' \
<<< 'indent code by 4 spaces and 6 letters and by 5 constants')
Now you can access all matches with their index:
$ echo "${arr[0]}, ${arr[1]}"
4, 6
Related
I have a .txt file and on each line is some amount of numbers. What I need is to filtrate these which does not contain the same number. So I want the output to be only the lines which have all the numbers different. I have to use command grep!
Example:
File_input:
1 1 2 3 4 5
1 2 3 4 5 6
6 6 6 6 6 6
What I want
File_output:
1 2 3 4 5 6
First and third lines contains same numbers so these has to be filtrated out.
This should work for your example:
grep -v "\([0-9]\).*\1" myfile
Idea is to catch any single digit [0-9] and store it \(\) and search for the existing same pattern \1 on the same line. You can easily extend to any word made of digits.
With the given input you can use
sed -r '/([0-9]+).+\1/d' File_input
You will have problems with suubstrings: 1 matches 12 and 12 matches 1.
ou can add word boundaries \b with
sed -r '/\b([0-9]+)\b.*\b\1\b/d' File_input
suppose I have file containing numbers like:
1 4 7
2 5 8
and I want to add 1 to all these numbers, making the output like:
2 5 8
3 6 9
is there a simple one-line command (e.g. awk) to realize this?
try following once.
awk '{for(i=1;i<=NF;i++){$i=$i+1}} 1' Input_file
EDIT: As per OP's request without loop, here is a solution(written as per shown sample only).
With hardcoding of number of fields.
awk -v RS='[ \n]' '{ORS=NR%3==0?"\n":" ";print $0+1}' Input_file
OR
Without hardcoding number of fields.
awk -v RS='[ \n]' -v col=$(awk 'FNR==1{print NF}' Input_file) '{ORS=NR%col==0?"\n":" ";print $0+1}' Input_file
Explanation: So in EDIT section 1st solution I have hardcoded the number of fields by mentioning 3 there, in OR solution of EDIT, I am creating a variable named col which will read the very first line of Input_file to get the number of fields. Then it will not read all the Input_file, Now coming onto the code I have set Record separator as space or new line to it will add them without using a loop and it will add space each time after incrementing 1 in their values. It will print new line only when number of lines are completely divided by value of col(which is why we have taken number of fields in -v col section).
In native bash (no awk or other external tool needed):
#!/usr/bin/env bash
while read -r -a nums; do # read a line into an array, splitting on spaces
out=( ) # initialize an empty output array for that line
for num in "${nums[#]}"; do # iterate over the input array...
out+=( "$(( num + 1 ))" ) # ...and add n+1 to the output array.
done
printf '%s\n' "${out[*]}" # then print that output array with a newline following
done <in.txt >out.txt # with input from in.txt and output to out.txt
You can do this using gnu awk:
awk -v RS="[[:space:]]+" '{$0++; ORS=RT} 1' file
2 5 8
3 6 9
If you don't mind Perl:
perl -pe 's/(\d+)/$1+1/eg' file
Substitute any number composed of multiple digits (\d+) with that number ($1) plus 1. /e means to execute the replacement calculation, and /g means globally throughout the file.
As mentioned in the comments, the above only works for positive integers - per the OP's original sample file. If you wanted it to work with negative numbers, decimals and still retain text and spacing, you could go for something like this:
perl -pe 's/([-]?[.0-9]+)/$1+1/eg' file
Input file
Some column headers # words
1 4 7 # a comment
2 5 cat dog # spacing and stray words
+5 0 # plus sign
-7 4 # minus sign
+1000.6 # positive decimal
-21.789 # negative decimal
Output
Some column headers # words
2 5 8 # a comment
3 6 cat dog # spacing and stray words
+6 1 # plus sign
-6 5 # minus sign
+1001.6 # positive decimal
-20.789 # negative decimal
For a file that contains entries similar to as follows:
foo 1 6 0
fam 5 11 3
wam 7 23 8
woo 2 8 4
kaz 6 4 9
faz 5 8 8
How would you replace the nth field of every mth line with the same element using bash or awk?
For example, if n = 1 and m = 3 and the element = wot, the output would be:
foo 1 6 0
fam 5 11 3
wot 7 23 8
woo 2 8 4
kaz 6 4 9
wot 5 8 8
I understand you can call / print every mth line using e.g.
awk 'NR%7==0' file
So far I have tried to keep this in memory but to no avail... I need to keep the rest of the file as well.
I would prefer answers using bash or awk, but sed solutions would also be helpful. I'm a beginner in all three. Please explain your solution.
awk -v m=3 -v n=1 -v el='wot' 'NR % m == 0 { $n = el } 1' file
Note, however, that the inter-field whitespace is not guaranteed to be preserved as-is, because awk splits a line into fields by any run of whitespace; as written, the output fields of modified lines will be separated by a single space.
If your input fields are consistently separated by 2 spaces, however, you can effectively preserve the input whitespace by adding -F' ' -v OFS=' ' to the awk invocation.
-v m=3 -v n=1 -v el='wot' defines Awk variables m, n, and el
NR % m == 0 is a pattern (condition) that evaluates to true for every m-th line.
{ $n = el } is the associated action that replaces the nth field of the input line with variable el, causing the line to be rebuilt, implicitly using OFS, the output-field separator, which defaults to a space.
1 is a common Awk shorthand for printing the (possibly modified) input line at hand.
Great little exercise. While I would probably lean toward an awk solution, in bash you can also rely on parameter expansion with substring replacement to replace the nth field of every mth line. Essentially, you can read every line, preserving whitespace, then check your line count, e.g. if c is your line counter and m your variable for mth line, you could use:
if (( $((c % m )) == 0)) ## test for mth line
If the line is a replacement line, you can read each word into an array after restoring default word-splitting and then use your array element index n-1 to provide the replacement (e.g. ${line/find/replace} with ${line/"${array[$((n-1))]}"/replace}).
If it isn't a replacement line, simply output the line unchanged. A short example could be similar to the following (to which you can add additional validations as required)
#!/bin/bash
[ -n "$1" -a -r "$1" ] || { ## filename given an readable
printf "error: insufficient or unreadable input.\n"
exit 1
}
n=${2:-1} ## variables with default n=1, m=3, e=wot
m=${3:-3}
e=${4:-wot}
c=1 ## line count
while IFS= read -r line; do
if (( $((c % m )) == 0)) ## test for mth line
then
IFS=$' \t\n'
a=( $line ) ## split into array
IFS=
echo "${line/"${a[$((n-1))]}"/$e}" ## nth replaced with e
else
echo "$line" ## otherwise just output line
fi
((c++)) ## advance counter
done <"$1"
Example Use/Output
n=1, m=3, e=wot
$ bash replmn.sh dat/repl.txt
foo 1 6 0
fam 5 11 3
wot 7 23 8
woo 2 8 4
kaz 6 4 9
wot 5 8 8
n=1, m=2, e=baz
$ bash replmn.sh dat/repl.txt 1 2 baz
foo 1 6 0
baz 5 11 3
wam 7 23 8
baz 2 8 4
kaz 6 4 9
baz 5 8 8
n=3, m=2, e=99
$ bash replmn.sh dat/repl.txt 3 2 99
foo 1 6 0
fam 5 99 3
wam 7 23 8
woo 2 99 4
kaz 6 4 9
faz 5 99 8
An awk solution is shorter (and avoids problems with duplicate occurrences of the replacement string in $line), but both would need similar validation of field existence, etc.. Learn from both and let me know if you have any questions.
This question already has answers here:
Length of string in bash
(11 answers)
Closed 6 years ago.
Is it even possible? I currently have a one-liner to count the number of words in a file. If I output what I currently have it looks like this:
3 abcdef
3 abcd
3 fec
2 abc
This is all done in 1 line without loops and I was thinking if I could add a column with length of each word in a column. I was thinking I could use wc -m to count the characters, but I don't know if I can do that without a loop?
As seen in the title, no AWK, sed, perl.. Just good old bash.
What I want:
3 abcdef 6
3 abcd 4
3 fec 3
2 abc 3
Where the last column is length of each word.
while read -r num word; do
printf '%s %s %s\n' "$num" "$word" "${#word}"
done < file
You can do something like this also:
File
> cat test.txt
3 abcdef
3 abcd
3 fec
2 abc
Bash script
> cat test.txt.sh
#!/bin/bash
while read line; do
items=($line) # split the line
strlen=${#items[1]} # get the 2nd item's length
echo $line $strlen # print the line and the length
done < test.txt
Results
> bash test.txt.sh
3 abcdef 6
3 abcd 4
3 fec 3
2 abc 3
I'm having the following issue. I have an array of numbers:
text="\n1\t2\t3\t4\t5\n6\t7\t8\t9\t0"
And I'd like to delete the leading newline.
I've tried
sed 's/.//' <<< "$text"
cut -c 1- <<< "$text"
and some iterations. But the issue is that both of those delete the first character AFTER EVERY newline. Resulting in this:
text="\n\t2\t3\t4\t5\n\t7\t8\t9\t0"
This is not what I want and there doesn't seem to be an answer to this case.
Is there a way to tell either of those commands to treat newlines like characters and the entire string as one entity?
awk to the rescue!
awk 'NR>1'
of course you can do the same with tail -n +2 or sed 1d as well.
You can probably use the substitution modifier (see parameter expansion and ANSI C quoting in the Bash manual):
$ text=$'\n1\t2\t3\t4\t5\n6\t7\t8\t9\t0'
$ echo "$text"
1 2 3 4 5
6 7 8 9 0
$ echo "${text/$'\n'/}"
1 2 3 4 5
6 7 8 9 0
$
It replaces the first newline with nothing, as requested. However, note that it is not anchored to the first character:
$ alt="${text/$'\n'/}"
$ echo "${alt/$'\n'/}"
1 2 3 4 56 7 8 9 0
$
Using a caret ^ before the newline doesn't help — it just means there's no match.
As pointed out by rici in the comments, if you read the manual page I referenced, you can find how to anchor the pattern at the start with a # prefix:
$ echo "${text/#$'\n'/}"
1 2 3 4 5
6 7 8 9 0
$ echo "${alt/#$'\n'/}"
1 2 3 4 5
6 7 8 9 0
$
The notation bears no obvious resemblance to other regex systems; you just have to know it.