How can i use sed to exclude patterns when joining lines together? - sed

Question

I am trying to use sed to look for lines that start with '1' and join them with the following line, while ignoring lines that start with '1.'
my source file looks like this:
name cat
1
7.75
2
1.27
X
5.10
The desired output is:
name cat
1 7.75
2
1.27
X
5.10
I have a command that looks for lines that start with 1 and joins the following line, however because I also have lines with 1.* which i want to ignore. I have tried the following sed command and used to try and ignore decimals, however it does not work.
The command i am using is:
sed '/^\<1\>/N;s/\n/ /'
but it gives this output:
name cat
1 7.75
2
1.27 X
5.10
How can I join lines starting with '1' with the following line, while ignoring lines that start with 1.* ?
Edit:
I only want to join lines that contain '1' (nothing else on the line) with the following line
Some lines start with a float , eg 1.2 , i want to ignore these so the next line is not appended to this.

Answer 1

sed '/^1/{/^1\./!N;s/\n/ /}'
If a line does start with a 1, then if it does not start with 1. then append next line. Then replace the newline for a space.
Or just:
sed '/^1\([^\.]\|$\)/N;s/\n/ /'
# same without `\(\|\)`
sed '/^1[^\.]/N;/^1$/N;s/\n/ /'
If a line start with a 1 and then has anything else then a comma or it's the end of line, then append next line. Replace the newline for a space
I only want to join lines that contain '1' (nothing else on the line) with the following line
So just match the 1.
sed '/^1$/N;s/\n/ /'
Maybe you want to just match 1 followed by any whitespace?
sed '/^1[[:space:]]*$/N;s/\n/ /'
Or by spaces only?
sed '/^1 *$/N;s/\n/ /'
The Sed - An Introduction and Tutorial by Bruce Barnett is a great place to learn how to use sed. To learn regexes, I recommend playing with regex crosswords, they let you learn regexes fast and with fun.

Answer 2

You can begin your sed script by starting a new cycle when a line begins with 1.:
#!/bin/sed -f
/^1\./n # don't change 1.x
/^1\b/N # \b is GNU sed word-boundary
s/\n/ /
Thus, only lines not beginning 1. get the following line appended.
Example output:
name cat
1 7.75
2
1.27
X
5.10
According to later comments on the question, it seems you only want to join lines containing 1 and optional trailing spaces, which makes the script much simpler:
#!/bin/sed -f
/^1[[:space:]]*$/N # match the whole line
y/\n/ /

Answer 3

Could you please try following.
awk '
$0==1{
prev=$0
next
}
prev{
$0=prev OFS $0
prev=""
}
1
END{
if(prev){
print prev
}
}
' Input_file
Explanation: Adding detailed explanation for above code.
awk ' ##Starting awk program from here.
$0==1{ ##Checking condition if line is having 1 value then do following.
prev=$0 ##Creating variable prev and set its value to current line value.
next ##next will skip all further statements from here.
}
prev{ ##Checking condition if prev variable is NOT NULL then do following.
$0=prev OFS $0 ##Setting current line value to prev OFS and current line.
prev="" ##Nullifing variable prev here.
}
1 ##1 will print the edited/non-edited line here.
END{ ##Starting END block for this awk program here.
if(prev){ ##Checking condition if variable prev is NOT NULL then do following.
print prev ##Printing variable prev here.
} ##Closing BLOCK for if condition here.
} ##Closing END block of this awk program here.
' Input_file ##Mentioning Input_file here.
Output will be as follows.
1 7.75
2
1.27
X
5.10

Answer 4

Using any awk in any shell on every UNIX box:
$ awk '{printf "%s%s", $0, ($1==1"" ? OFS : ORS)}' file
name cat
1 7.75
2
1.27
X
5.10
FYI some (all?) of the sed solutions posted so far are relying on non-POSIX functionality and so YMMV depending on what they do depending on which sed you use.