遍历字符串并在shell中查找某些字符

Question

lets say I have the following string stored in a variable:

string="1245aaa./ ssasaaa* kjdsaaa"

Is there a way to somehow loop through this string and find out that it contains 3 "words" so to speak separated by blank spaces and that the word with the most "a" is the second one and there are total of 4 "a" in the second word?

I've been trying to google something like this but with no luck.

Answer 1

Another method is grepping for the line with at least n (in your example 4) a 's.
First you must find the number you need to grep for.
In steps (requested in comment):
Split the words in the string into lines by replacing ( tr , translate) spaces with newlines.

echo "${string}" | tr " " "\n"

With sed 's/old/new/g' you can s (substitute) the old string (pattern) with the new string g (globally). So you can echo "Have all characters a banned" | sed 's/a//g' echo "Have all characters a banned" | sed 's/a//g' . You want to replace all characters except for the character a. The ^ in [^a] stands for not , the [] for a class of characters.

echo "${string}" | tr " " "\n" | sed 's/[^a]//g'

You can find the longest string of a's by sorting them. After sorting, the last line will have most. With tail -1 you get the last line:

echo "${string}" | tr " " "\n" | sed 's/[^a]//g'|sort | tail -1

Now put the result in a variable. You can assign the output of another (set of) unix command(s) to a variable with var=$(command) , be aware that you do not add spaces around the = sign ( var = $(xxx) will fail).

most_a=$(echo "${string}" | tr " " "\n" | sed 's/[^a]//g'|sort | tail -1)

When you want to see the contents of a variable, use $var or prefer ${var} . With {} everybody knows that the other_chars in ${var}other_chars are not part of the variable name. With an # in ${#var} you ask for a number of chars. And always use double quotes when using echo until you understa

echo "The word with the highest number of a's has ${#most_a} of those"

Now you can grep the word with this number of a's out of a list of words. When you want to grep strings with at least 4 a's you will need .* (any character repeated 0 or more times), so grep for a.*a.*a.*a or a.*a.*a.*a.* . You can tell grep that the pattern (a.*) is repeated {4} or {${#most_a}} times. Now you need some backslashes to activate the special meaning of the (){} characters and start splitting the original string in words:

echo "${string}" | tr " " "\n" | grep "\(a.*\)\{${#most_a}\}"

To print the string and number, use something like

printf "%s %s\n" ${#most_a} $(echo "${string}" | tr " " "\n" | grep "\(a.*\)\{${#most_a}\}" )

Answer 2

awk can handle this:

string="1245aaa./ ssasaaa* kjdsaaa"

awk -v k='a' -v RS=' ' '{n = split($0, a, k)-1} 
     n > max{max=n; maxw=$0} END{print maxw, max}' OFS=, <<< "$string"

Output:

ssasaaa*,4

Answer 3

You can do this in Bash alone.

Given:

$ string="1245aaa./ ssasaaa* kjdsaaa"

You can break that string into 'words' by breaking on the current IFS into an array:

$ words=( $string )

Then loop over each word and count the regex matches:

$ for word in "${words[@]}"
> do
> printf "%i %s\n" $(egrep -o 'a' <<<$word | wc -l) $word 
> done
3 1245aaa./
4 ssasaaa*
3 kjdsaaa

And pipeline the result of that into sort to sort by match count and head to get the top one:

for word in "${words[@]}"
do
    printf "%i %s\n" $(egrep -o 'a' <<<$word | wc -l) $word 
done | sort -n -r | head -1
4 ssasaaa*

awk is more efficient, but you can do this way too.

Answer 4

 string="1245aaa./ ssasaaa* kjdsaaa"

 echo $string | tr ' ' '\n' | while read s
 do  
 echo "`echo $s | tr -dc 'a' | wc -c` $s"
 done | sort -nr

or

echo $string | xargs -n 1 bash -c 'for s; do echo "`echo $s | tr -dc 'a' | wc -c` $s"; done' bash | sort -nr