Replace range of numbers with certain number - bash

Question

I need to replace a range of numbers with a certain number. I really tried it hard to code it myself with sed (like sed "s/[33-64]/64/") or awk, but always get wrong results. It tends to replace single digits instead of numbers... What I need would be: Replacing 0-32 -> 32, 33-64 -> 64, 65-128 -> 128, 129-255 -> 255. In between these numbers are IPs, which should stay untouched. I think this command is selecting all, but IPs:
sed '/[0-9]\{1,3\}\.[0-9]\{1,3\}\.[0-9]\{1,3\}\.[0-9]\{1,3\}/! ... '
So I have a file like this:
65.74.16.161
232
10.128.8.72
63
10.128.14.13
100
10.128.8.58
32
10.128.4.129
60
10.128.240.18
59
and it should look like this:
65.74.16.161
255
10.128.8.72
64
10.128.14.13
128
10.128.8.58
32
10.128.4.129
64
10.128.240.18
64

Answer 1

The [33-64] defines a character class and is a funny way of writing [3-6] and does indeed only match a single character — any single digit from 3, 4, 5 or 6. If you really want to do it with sed, and you're concerned with values from 33 to 64, then you have to write it out differently — and much more verbosely.
In part it depends on which version of sed you have. A solution that will work with classic sed is:
sed -e 's/^[0-9]$/32/' \
-e 's/^[12][0-9]$/32/' \
-e 's/^3[012]$/32/' \
-e 's/^3[3-9]$/64/' \
-e 's/^[45][0-9]$/64/' \
-e 's/^6[0-4]$/64/' \
-e 's/^6[5-9]$/128/' \
-e 's/^[7-9][0-9]$/128/' \
-e 's/^1[01][0-9]$/128/' \
-e 's/^12[0-8]$/128/' \
-e 's/^129$/255/' \
-e 's/^1[3-9][0-9]$/255/' \
-e 's/^2[0-4][0-9]$/255/' \
-e 's/^25[0-5]$/255/'
But, as you can see, it is quite painful. If you have GNU sed, you can use the -r option to enable extended regular expressions; if you have Mac OS X or BSD sed, you can use the -E option to enable extended regular expressions. Then you can reduce the code above to:
sed -E \
-e 's/^([0-9]|[12][0-9]|3[012])$/32/' \
-e 's/^(3[3-9]|[45][0-9]|6[0-4])$/64/' \
-e 's/^(6[5-9]|[7-9][0-9]|1[01][0-9]|12[0-8])$/128/' \
-e 's/^(129|1[3-9][0-9]|2[0-4][0-9]|25[0-5])$/255/'
However, you might do better using awk:
awk '/^[0-9][0-9]*$/ { if ($1 <= 32) print 32
else if ($1 <= 64) print 64
else if ($1 <= 128) print 128
else if ($1 <= 255) print 255
else print $1
next
}
{ print }'
The final else clause accurately prints any unexpected values, such as 256 or 999 or, indeed, 123456789. There are those who would write 1 in place of { print } — the part of the awk script that matches and prints the IP addresses.

Answer 2

You can use this awk with some arithmetic:
awk '$1 == ($1+0) && $1<=255{$1 = ($1>128)?255:($1>64?128:32 * int(($1+31)/32))} 1' file
65.74.16.161
255
10.128.8.72
64
10.128.14.13
128
10.128.8.58
32
10.128.4.129
64
10.128.240.18
64
$1 == ($1+0) is a check to determine $1 is an integer.

Answer 3

Using awk:
awk -vFS=. ' NF == 1 { v=2^int((log($1)/log(2))+0.5); $1 = v>255?255:v; }1' input
Gives:
65.74.16.161
255
10.128.8.72
64
10.128.14.13
128
10.128.8.58
32
10.128.4.129
64
10.128.240.18
64