namelen.sh

In a recent discussion on LKML (Link) the question came up how long an average filename is. To answer that, I wrote a small bash script. 94.15 Percent of the filenames in my home directory are shorter than 25 characters. Here is a sample output and the source code. Sample Output:

Searching...
Getting name length...
350932 Files considered
Length Files Percent Cumulative
 1   1761   0.50   0.50
 2   3315   0.94   1.45
 3  14846   4.23   5.68
 4  16595   4.73  10.41
 5   8413   2.40  12.80
 6  12311   3.51  16.31
 7  30322   8.64  24.95
 8  31808   9.06  34.02
 9  29943   8.53  42.55
10  28826   8.21  50.76
11  24336   6.93  57.70
12  21693   6.18  63.88
13  10922   3.11  66.99
14   9091   2.59  69.58
15  17178   4.89  74.48
16   8565   2.44  76.92
17   4553   1.30  78.21
18   3840   1.09  79.31
19   4596   1.31  80.62
20   3536   1.01  81.63
21   5015   1.43  83.05
22   4216   1.20  84.26
23   4839   1.38  85.63
24   6868   1.96  87.59
25   5306   1.51  89.10
26  17719   5.05  94.15
27    468   0.13  94.29
28    523   0.15  94.44
29    386   0.11  94.55
30    325   0.09  94.64
31    283   0.08  94.72
32   4577   1.30  96.02
33    187   0.05  96.08
34    139   0.04  96.12
35    162   0.05  96.16
36  12359   3.52  99.68

Source:

#!/bin/sh
FN="/tmp/filenames"
FNL="/tmp/filenames1"
echo Searching...
find . ! -name '.' ! -name '..' > $FN
echo "Getting name length..."
echo -n > $FNL
for i in `cat $FN`; do
        x=${i##*/}
        echo ${#x} >> $FNL
done
TOTAL=`cat $FNL | wc -l`
echo $TOTAL Files considered
echo "Length Files Percent Cumulative"
cum=0
totcnt=0
for i in `seq 1 100`; do
        cnt=`grep "^$i$" $FNL | wc -l`
        totcnt=$((cnt + totcnt))
        perc=$(echo "scale=8; $cnt * 100 / $TOTAL" | bc)
        cum=$(echo "scale=8; $perc + $cum" | bc)
        echo $i $cnt $perc $cum | awk -F' ' '{printf "%2d %6d %6.2f %6.2f\n", $1, $2, $3, $4}'
done
rm -f $FN
rm -f $FNL
echo "Done."