char countiing script help...

[Manuel Contreras]

I have a testFILE that with say 100 lines.

how would I be able to provide the number of char. or words for each line in a file?

like wc, but would be open to awk,sed,etc...

number of words:
$ head -1 testFILE | tail -1 | wc -w
5

number of characters:
$ head -1 testFILE | tail -1 | wc -c
90

any assistance is great appreciated,
manuel contreras

[Mel Burslan · Accepted solution]

cat file | while read line
do
chars=`echo $line | wc -c`
words=`echo $line | wc -w`
printf $chars:$words:; echo $line
done

this will count the characters and words on each line and present them like this

41:8:quick brown fox jumped over the lazy dog


hope this is what you are looking for

[curt larson_1]

awk '{
printf("%s:%s %s\n",length($0)+1,NF,$0);
}' yourfile

[Hein van den Heuvel]

perl -pe 'print split.":".length.":"' /etc/hosts

awk '{print NF ":" length ":" $0}' /etc/hosts

[H.Merijn Brand (procura]

Hein, that perl snippet is a nice thought, but not OK. You use split in list context, so it returns all characters in the line.

`Words' is an arguable definition. Where do you want to split on? Is "user:xhdffw2:203:200" 1 word, or are there 7 words, or 4

If it is 7, use

# perl -nle 'print"$.:",scalar(split/\b/),":".length' /etc/hosts

which will split on word bounds and print the line number, the number od `words', and the number of characters (including white space) for each line.

If `words' are split on white space, like in text files ("which," being ONE words, including the comma), you could use

# perl -nle 'print"$.:",scalar(split/\s+/),":".length' /etc/hosts

If you dont want to count the white space in the character count,

# perl -nle '$"="";@x=split/\s+/;print"$.:",scalar@x,":".length"@x"' /etc/hosts

As you can see, the possibilities are endless. It all depends on /your/ defenition of the truth ...

Enjoy, Have FUN! H.Merijn

[H.Merijn Brand (procura]

Sorry, Hein, I misread the "." for a "," (/me blames too small font)

No points for this one please.

Enjoy, Have FUN! H.Merijn

[Muthukumar_5]

Hai,

Do you want to have command as like wc. We can do it with awk command as like as ,

cat filename | awk '{ print "line: " NR " words: " NR " chars: "length($0) }'

It will print line number, number of words and length of the line.

Regards,
Muthukumar.

[Manuel Contreras]

this may be a dumb question, but why do some of the commands recognize/display different
number of chars in a line (there is no space after the last char)?

(this command does not count blank spaces)
$ cat man99 | while read line
> do
> chars=`echo $line | wc -c`
> words=`echo $line | wc -w`
> printf $chars:$words:; echo $line
> done | head -5
90:1:12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789
66:5:A20040601FWL1085 000000000.095940 000306275.000000FG1085 A0406 EA
66:5:A20040601FWS1085 000000000.077005 000308043.000000FG1085 A0406 EA
67:5:A20040601FWV05600 000000000.008854 000308043.000000FG1085 A0406 EA
67:5:A20040601FWV07019 000000000.679378 000004611.172617FG1085 A0406 LB

(this command seems to count 90 char. on first line)
$ awk '{
> printf("%s:%s %s\n",length($0)+1,NF,$0);
> }' man99 | head -5
90:1 12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789
90:5 A20040601FWL1085 000000000.095940 000306275.000000FG1085 A0406 EA
90:5 A20040601FWS1085 000000000.077005 000308043.000000FG1085 A0406 EA
90:5 A20040601FWV05600 000000000.008854 000308043.000000FG1085 A0406 EA
90:5 A20040601FWV07019 000000000.679378 000004611.172617FG1085 A0406 LB

(this one also recognizes 90 char. on first line)
$ perl -pe 'print split.":".length,":"' man99 | head -5
1:90:12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789
5:90:A20040601FWL1085 000000000.095940 000306275.000000FG1085 A0406 EA
5:90:A20040601FWS1085 000000000.077005 000308043.000000FG1085 A0406 EA
5:90:A20040601FWV05600 000000000.008854 000308043.000000FG1085 A0406 EA
5:90:A20040601FWV07019 000000000.679378 000004611.172617FG1085 A0406 LB

(this one works correctly)
$ awk '{print NF ":" length ":" $0}' man99 | head -5
1:89:12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789
5:89:A20040601FWL1085 000000000.095940 000306275.000000FG1085 A0406 EA
5:89:A20040601FWS1085 000000000.077005 000308043.000000FG1085 A0406 EA
5:89:A20040601FWV05600 000000000.008854 000308043.000000FG1085 A0406 EA
5:89:A20040601FWV07019 000000000.679378 000004611.172617FG1085 A0406 LB

(as well as this one)
$ cat man99 | awk '{ print "line: " NR " words: " NR " chars: "length($0) }' | head -5
line: 1 words: 1 chars: 89
line: 2 words: 2 chars: 89
line: 3 words: 3 chars: 89
line: 4 words: 4 chars: 89
line: 5 words: 5 chars: 89





thank you all...I appreciate the education,

manuel contreras

[Muthukumar_5]

Hai,

Use the format of
cat filename | awk '{ print "line: " NR " words: " NF " chars: "length($0) }'

NF for NR in the words list.

You will see a difference between the wc and awk command for computing the characters.

echo "test" | wc -c will give the 5 for characters count.

You can see the contents of string with more descriptive manner with od command as,

echo "test" | od -dc as
0000000 24936 2665
h a i \n
0000004

awk is working by taking the space,tabs and newlines characters for field separtor (fs).

If you do,

echo "test" | awk '{ length($0) }'
will give only 4 for string length.

It won't take the \n in that. Because it is used to separate the fields on the awk call.

Regards,
Muthukumar.