How to replace the third comma in a csv file with a space in linux using bash?

preg_replace('/[\s,]+/', 'YES', $str);

sed -i 's/AA BB/CC DD/g' file

$ echo 'xxxx AA BB xxx' | sed 's/AA BB/CC DD/g'
xxxx CC DD xxx

awk '{sub(/AA BB/,"CC DD")}1' infile > outfile

$ echo 'xxxx AA BB xxx' | awk '{sub(/AA BB/,"CC DD")}1'
xxxx CC DD xxx

sed 's/,/ : /g' testfile

sed -i.bak 's/,/ : /g' testfile

mv testfile testfile.bak
sed 's/,/ : /g' testfile.bak >testfile

tr ',' ':' <testfile

mv testfile testfile.bak
tr ',' ':' <testfile.bak >testfile

Sub foo()
Dim objFSO
Const ForReading = 1
Const ForWriting = 2
Dim objTS 'define a TextStream object
Dim strContents As String
Dim fileSpec As String

fileSpec = "C:\Test.txt" 'change the path to whatever yours ought to be
Set objFSO = CreateObject("Scripting.FileSystemObject")
Set objTS = objFSO.OpenTextFile(fileSpec, ForReading)

Do While Not objTS.AtEndOfStream
    strContents = strContents & " " & objTS.ReadLine 'Read line by line and store all lines in strContents
Loop

strContents = Replace(strContents, ",", " ")
objTS.Close

Set objTS = objFSO.OpenTextFile(fileSpec, ForWriting)
objTS.Write strContents
objTS.Close
End Sub

sed -E 's/(^[a-zA-Z]+).*/\1/' Input_file

awk 'match($0,/^[a-zA-Z]+/){print substr($0,RSTART,RLENGTH)}' Input_file

awk 'match($0,/^[a-zA-Z]+\(\)-/){print substr($0,RSTART,RLENGTH)}' Input_file

awk 'match($0,/^[[:alpha:]()-]+/){print substr($0,RSTART,RLENGTH)}' Input_file