如何在bash中计算字符串中的连续（重复）字符？

Question

I am wondering if there is a simple bash or AWK oneliner to get the number of repeated characters, per repeat.

For example considering this string:

AATGATGGAANNNNNGATAGAACGATNNNNNNNNGATAATGANNNNNNNTAGACTGA

Is it possible to get the number of Ns in the first repeat, the number of Ns in the second repeat, etc.?

Thanks!

Expected results, the length of each repeat on a new line.

Answer 1

You can use awk to split fields on each character that not N and print each field and it's length:

s='AATGATGGAANNNNNGATAGAACGATNNNNNNNNGATAATGANNNNNNNTAGACTGA'

awk -F '[^N]+' '{for (i=1; i<=NF; i++) if ($i != "") print $i, length($i)}' <<< "$s"

NNNNN 5
NNNNNNNN 8
NNNNNNN 7

Another option is to use grep + awk :

grep -Eo 'N+' <<< "$s" | awk '{print $1, length($1)}'

And here is pure BASH solution :

shopt -s extglob
while read -r line; do
    [[ -n $line ]] && echo "$line ${#line}"
done <<< "${s//+([!N])/$'\n'}"

NNNNN 5
NNNNNNNN 8
NNNNNNN 7

BASH solution details:

1. It uses extended glob pattern to match 1 or more non-N characters and replace them with line break in +([!N])/$'\\n'}"
2. Using a while loop we iterate through each substring of N characters
3. Inside the loop we print each string and length of that string.

Answer 2

A simple solution:

echo "$string" | grep -oE "N+" | awk '{ print $0, length}'

NNNNN 5
NNNNNNNN 8
NNNNNNN 7

EDIT:
As per suggestion of @Ed-Morton: Changing -P to -E.
Man page of grep says -P is "highly experimental" functionality.
We don't need PCREs to use +, just EREs are sufficient.

Answer 3

With GNU awk for multi-char RS:

$ awk -v RS='N+' 'RT{print length(RT)}' file
5
8
7

$ awk -v RS='N+' 'RT{print RT, length(RT)}' file
NNNNN 5
NNNNNNNN 8
NNNNNNN 7

Answer 4

Here's a Perl one-liner:

perl -ne 'while (m/(.)(\1*)/g) { printf "%5i %s\n", length($2)+1, $1 }' <<<AATGATGGAANNNNNGATAGAACGATNNNNNNNNGATAATGANNNNNNNTAGACTGA
2 A
1 T
1 G
1 A
1 T
2 G
2 A
5 N
1 G
1 A
1 T
1 A
1 G
2 A
1 C
1 G
1 A
1 T
8 N
1 G
1 A
1 T
2 A
1 T
1 G
1 A
7 N
1 T
1 A
1 G
1 A
1 C
1 T
1 G
1 A

The m/(.)(\\1*)/ successively matches as many identical characters as possible, with the /g causing the matching to pick up again on the next iteration for as long as the string still contains something which we have not yet matched. So we are looping over the string in chunks of identical characters, and on each iteration, printing the first character as well as the length of the entire matched string.

The first pair of parentheses capture a character at the beginning of the (remaining unmatched) line, and \\1 says to repeat this character. The * quantifier matches this as many times as possible.

If you are interested in just the N:s, you could change the first parenthesis to (N) , or you could add a conditional like printf("%7i %s\\n", length($2), $1) if ($1 == "N") . Similarly, if you want only hits where there are repeats (more than one occurrence), you can say \\1+ instead of \\1* or add a conditional like ... if length($2) >= 1 .

Answer 5

当您要求 sed 解决方案时，如果您的重复字符链不超过 9 个字符并且您的字符串不包含任何分号，则可以使用此解决方案：

sed 's/$/;NNNNNNNNN0123456789/;:a;s/\\(N\\+\\)\\([^;]*;\\1.\\{9\\}\\)\\(.\\)\\(.*\\)/\\2\\3\\4\\n\\3/;ta;s/[^\\n]*\\n//'

Answer 6

try these two:

First one

sed 's/[^N]/ /g' file | awk '{for(i=1;i<=NF;i++){print $i":"length($i)}}'

Second One

cat file | tr -c 'N' ' ' | awk '{for(i=1;i<=NF;i++){print $i":"length($i)}}'

Answer 7

Short GNU awk approach:

str='AATGATGGAANNNNNGATAGAACGATNNNNNNNNGATAATGANNNNNNNTAGACTGA'

awk -v FPAT='N+' '{for(i=1;i<=NF;i++) print $i,length($i)}' <<< $str

The output:

NNNNN 5
NNNNNNNN 8
NNNNNNN 7

Answer 8

You could take help of the regular expression method.

This is a solution code I get from the following link

Count occurrences of a char in a string using Bash

needle=","
var="text,text,text,text"

number_of_occurrences=$(grep -o "$needle" <<< "$var" | wc -l)

as you can see we get the number of occurrences of "$needle" pretty easily with the help of WC(word count).

You can loop it to satisfy your demand.