在bash的每一行中找到连续的以空格分隔的单个字符

Question

Let say I have the following file

Y M C A
cambridge m a
d m v office
t mobile

and want to convert it to

YMCA
cambridge ma
dmv office
t mobile

that is to detect all consecutive single characters followed by single space of different lengths ( greater than two). For example, the item 'dmv office', we should detect 'dm v' and convert it to 'dmv' but would leave 't mobile store' intact (only one single character).
Is it possible to do this in bash or I have to use a program like python to do this?

Answer 1

Perl one-liner:

echo 'Y M C A' | perl -ple's/\b\w\K\s(?=\w\b)//g'
==> YMCA

echo 't mobile' | perl -ple's/\b\w\K\s(?=\w\b)//g'
==> t mobile

This replaces a space when surrounded by a single word character. You can replace \\w by [a-zA-Z] if it's more convenient for you.

Answer 2

This sed one-liner works for given example:

sed -r 's/ (\S\S)/_\1/g;s/(\S\S) /\1_/g;s/ //g;s/_/ /g' file

Test with your data:

kent$  sed -r 's/ (\S\S)/_\1/g;s/(\S\S) /\1_/g;s/ //g;s/_/ /g' f   
YMCA
cambridge ma
dmv office
t mobile

I used a place holder in above line, the _ , if your text has already _ , you can use \\x99 , in visible char.

Answer 3

With any awk in any shell on any UNIX system:

$ awk '{out=$1; for (i=2;i<=NF;i++) {out = out (length($(i-1)$i)==2 ? "" : OFS) $i} print out}' file
YMCA
cambridge ma
dmv office
t mobile