Cut and replace Nth character on every row

Question

I have predictable piped input, I want to iterate over each row and change two characters. the character positions are 19 and 20 (on every row.) The 19th character is a comma, I want to cut that. The 20th charater is a space, I want to replace it with an 'x'

775448167763476486, 783834143007506433, 35972, 35972,
775448167763476486, 844395243412914178, 408008, 408008,
775448167763476486, 891964514511355905, 8003, 8003,
783834143007506433, 891655551753846784, 66633, 66633,

Should become

715448167763476486x783834143007506433, 35972, 35972,
775448167763476486x844395243412911178, 408008, 408008,
705448167763476486x891964513511335905, 8003, 8003,
723834143007506433x891655551753846784, 66633, 66633

If there is an even simpler way to achieve the same effect, that'd be even better.

Edit.

So far I have tried a multitude of different approaches with

sed primarily..

sed 's/^ *([^,]*) //;s/, \([^,]*\),/\1x,/g'

As I understand is useful as looking for the first instance of a comma *([^,]*) I'm not sure of how the rest of the expression follows on after the ;

sed 's/, \([^,]*\),/\1x,/g'

I tried modifying the command this way but found an appended 'x' at the end of each row.

sed 's/\(.*\)\(.\{18\}\)\(.*\)/\1x\3/'

Same problem here. Two other expressions I tried to look at..

sed -E 's/^ *([^,]*) // 

sed -e 's/\(.\)$/,\1/'

with the 'regular expression flag '-e' but to be honest I don't really understand what that means yet.

cut

cut -c-18 -c21-  

When I tried this, the error I got was:

cut: only one type of list may be specified Try 'cut --help' for more information.

As I understand this is condensing or compressing with the '-c' flag, I'm assuming that the error might be related to the piped input coming from sed but I'm not sure, have to investigate with the man pages.

awk

awk '{gsub(/ /, "", $1); gsub(/ /, "x", $2); print}'

I felt this is a pretty clear way of seeing the idea of replacing whitespace with the desired charater, but had no effect on the input (unchanged.)

Which is why I posted here, not just looking for answers (sorry if the original post sounded like that, I just try to be brief and to the point.) Really just looking for ideas on how to approach such a problem.

Answer 1

The simplest way given your example input is to just replace the first comma-and-space with an x (here, file has your example):

$ sed 's/, /x/' file 
775448167763476486x783834143007506433, 35972, 35972,
775448167763476486x844395243412914178, 408008, 408008,
775448167763476486x891964514511355905, 8003, 8003,
783834143007506433x891655551753846784, 66633, 66633,

That has the benefit of working irrespective of the length of the first field. If you must change the 19th and 20th characters, you could do:

$ sed 's/../x/10' file 
775448167763476486x783834143007506433, 35972, 35972,
775448167763476486x844395243412914178, 408008, 408008,
775448167763476486x891964514511355905, 8003, 8003,
783834143007506433x891655551753846784, 66633, 66633,

The trick here is that the /10 means "repace the tenth occurrence of the pattern" and since the pattern is .., so two characters, it will change the 19th and 20th.

Answer 2

If you need to replace the 19^th and 20^th character of each line with x if and only if they are , and space respectively, you can do:

sed 's/^\(.\{18\}\), /\1x/'

With most sed implementations (all those compliant to POSIX 2024), that can be made more legible by switching from basic to extented regexp with -E:

sed -E 's/^(.{18}), /\1x/'

Answer 3

As sed is neither in the title or tagged, and the input is predictable, I would offer:

awk '{ print substr ($0, 1, 18) "x" substr ($0, 21); }'

Has the advantage of being explicit, and the character positions can be passed as args to awk without substituting into the syntax of a sed pattern.

Answer 4

Using Raku (formerly known as Perl_6)

With sed-like code (using Raku's -pe autoprinting flags):

~$ raku -pe 's/ \,\h /x/;'  file

#OR

~$ raku -pe 's:10th/ .. /x/;'  file

#OR

~$ raku -pe ' $_.=subst( / \,\h /, "x");'  file

OR:

With awk-like code (using Raku's -ne non-autoprinting flags):

~$ raku -ne 'put S/ \,\h /x/;'  file

#OR

~$ raku -ne 'put S:10th/ .. /x/;'  file

#OR

~$ raku -ne ' .subst( / \,\h /, "x").put;'  file

Above are answers written in Raku, a member of the Perl-family of programming languages. Raku features a powerful regex engine and high-level support for Unicode, built-in.

Above 'escaping' rules are simplified in Raku, such that all non-alnum characters must be escaped/backslashed to be understood literally. So \, represents a literal comma, while \h represents a single horizontal whitespace character. As with PCRE engines, the unescaped . dot represents "any-character" (a metacharacter). There are a plethora of regex-modifiers, such as :g or :global. To replace the first "x" times a pattern is seen use the :x() modifier.

To only replace an "nth" occurrence use the :nth() modifier, which accepts very readable forms such as :1st(or variant :st(1) ), :2nd (or variant :nt(2) ), :3rd (or variant :rd(3) ) or even :10th (acceptable variants are :nth(10) or just :th(10) ).

Sample Input:

775448167763476486, 783834143007506433, 35972, 35972,
775448167763476486, 844395243412914178, 408008, 408008,
775448167763476486, 891964514511355905, 8003, 8003,
783834143007506433, 891655551753846784, 66633, 66633,
783834143007506433, 891655551753846784, 66633, 66633,

Sample Output:

775448167763476486x783834143007506433, 35972, 35972,
775448167763476486x844395243412914178, 408008, 408008,
775448167763476486x891964514511355905, 8003, 8003,
783834143007506433x891655551753846784, 66633, 66633,
783834143007506433x891655551753846784, 66633, 66633,

Once you munge your file with Raku, you canchop off trailing characters, split/join on different characters, or even use an authentic CSV parser such as Raku's Text::CSV module. More info below.

https://docs.raku.org/language/regexes
https://raku.land/zef:Tux/Text::CSV
https://raku.org

Answer 5

Using GNU awk:

The following command is used taking cue from the answer:

$ awk '{print gensub(/(^.{18})..(.*)/, "\\1x\\2", "g")}' file

If input hasn't TAB as a character in input, then the cut may be used:

$ cut --output-delimiter="x" -c -18,21- file

---

Using perl:

$ perl -pe 's/^.{18}\K, /x/' file
$ perl -pe 'substr $_,18,2,"x"' file

In the first command sub() function will Keep what is matched left to it i.e., first 18 characters and then replaces comma and a space with x.

In the latter, substr() replaces 19^th and 20^th by substr(EXPR,OFFSET,LENGTH,REPLACEMENT). $_,18,2 is because counting of characters start from 0.