Count the Number of Words that Start with each Letter in a File

Question

I need to take a file and print the number of times each letter of the alphabet starts a word, in a descending order of word count. For example if the file is:

my nice name is Mike Meller 

then the output should be:

3 M
2 N
1 I

I need to do this in a single line. I know commands like wc -m and wc -w but I'm not sure how to iterate over each character and print it in the same way and then sort it like they want.

Answer 1

One way... (edited to avoid counting the same word twice)

$ echo "my nice name is Mike Meller" | tr ' ' '\n' | sort -f | uniq -i | sed -nr 's/^([a-z]).*/\U\1/Ip' | uniq -c | sort -r
  3 M
  2 N
  1 I

• tr ' ' '\n' change spaces into newlines
• sort -f sort the lines so same entries are together, even if case is different
• uniq -i remove duplicate words, ignoring case
• sed -nr 's/^([a-z]).*/\U\1/Ip' remove everything but the first letter, change all letters to upper case, and don't print the line if it doesn't start with a letter
• uniq -c count the lines that are the same
• sort -r sort descending

(replace echo "my nice name is Mike Meller" with cat name-of-your-file)

Answer 2

With perl:

perl -Mopen=locale -lne '
  $c{uc $_}++ for /\b\p{Alpha}/g;
  END{for (sort {$c{$b} <=> $c{$a}} keys %c) {print "$c{$_} $_"}}'

Note that if some letters appear in a decomposed form. For instance, if É is entered as É (that is E followed by a U+0301 combining accent) instead of the pre-composed É (U+00E9), then it will be counted as E, not É nor É.

If that's a concern, then probably the best approach is to first decompose the text (since some graphems don't have a pre-composed form) and work on a graphem cluster basis. There are some like fi that you'd probably want to break down anyway:

Compare:

$ printf 'my fine name is \uc9ric, maybe E\u301ric, certainly not Eric\n' |
  perl -Mopen=locale -lne '
    $c{uc $_}++ for /\b\p{Alpha}/g;
    END{for (sort {$c{$b} <=> $c{$a}} keys %c) {print "$c{$_} $_"}}'
2 E
2 N
2 M
1 C
1 FI
1 É
1 I

with:

$ printf 'my fine name is \uc9ric, maybe E\u301ric, certainly not Eric\n' |
  perl -Mopen=locale -MUnicode::Normalize -lne '
    $c{uc $_}++ for NFKD($_) =~ /\b(?=\p{Alpha})\X/g;
    END{for (sort {$c{$b} <=> $c{$a}} keys %c) {print "$c{$_} $_"}}'
2 É
2 M
2 N
1 E
1 I
1 C
1 F

Answer 3

GNU awk:

gawk '
  { for (i=1; i<=NF; i++) count[toupper(substr($i,1,1))]++ } 
  END {
    PROCINFO["sorted_in"] = "@val_num_desc"
    for (key in count) print count[key], key
  }
' file 

Answer 4

Not a homework project, I hope? ;-) The tricky part is that you don't want to count the "L" in Meller twice, right? Hence the "distinct".

$cat t
my nice name is Mike Meller

And then a pipeline of commands to do the transforming:

$tr '[a-z]' '[A-Z]' < t |     # Convert all to upper case
fold -b -w 1 |                # Break into one letter per line
awk -f t.awk |                # Pipe the whole mess to awk to count
sort -r -n                    # Sort in reverse numeric order

The awk script is best broken into a separate file, though you can just put it all into a bash one-liner:

$cat t.awk    
/ / {                         # Match spaces,
  for (c in wc) {dc[c]+=1}    #  Accumulate word count (wc) into doc count (dc)
  split("",wc)                #  Reset the word count
}

!/ / {                        # Match non-spaces,
  if (wc[$1] == "") wc[$1]=1  #  If haven't already seen char in this word, mark it Donny
}

# Finally, output the count and the letter
END {
  for (c in wc) {dc[c]+=1}    # Accumulate one last time, in case there is no trailing space
  for (c in dc) {print c, dc[c]}
}

Which produces (for me) this output:

$tr '[a-z]' '[A-Z]' < t | fold -b -w 1 | awk -f t.awk  | sort -r -n
4 M
4 E
3 I
2 N
1 Y
1 S
1 R
1 L
1 K
1 C
1 A