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Given there is 40 columns in a record. I want to replace 35th column so that the 35th column will be replace with the content of 35th column and a "$" symbol. What came up my mind is like:

awk '{print $1" "$2" "...$35"$ "$36...$40}'

I works but because it is infeasible when the number of column is as huge as 10k. I need a better way to do this.

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3 Answers 3

up vote 8 down vote accepted

You can do like this:

awk '$35=$35"$"'
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There are probably more efficient ways to do this. With that caveat:

awk '{$35 = $35"$"; print}' infile > outfile
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If the field delimiter is <space>:

sed 's/  */$&/35'
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If the field delimiter is unknown: sed 's/./$&/35' –  Underverse Jul 6 at 1:38
@Underverse - I don't think that's the same thing. That should prefix the 35th char on an input line w/ the char $ - delimited or not. The thing above should affix the 35th occurrence of any number of delimiter chars - in other words, the 35th field - with the char $ - no matter how chars are in each field. –  mikeserv Jul 6 at 1:43
Ah, "40 columns in a record". I read 'the 35th column' as literally 'the 35th character column of the text file'. –  Underverse Jul 6 at 1:57

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