如何计算文件中的每个字母？

Question

I have a cord.txt file as shown below,

188H,190D,245H
187D,481E,482T
187H,194E,196D
386D,388E,389N,579H
44E,60D

I need to count each letters and have to make a summary as shown below (expected output),

H,4
D,5
E,4
T,1

I know how to count each letters by using grep "<letter>" cord.txt | wc grep "<letter>" cord.txt | wc . But I have a huge file which contains more number of letters, therefore please help me to do the same.

Thanks in advance.

Answer 1

You're missing the N :-)

grep -o '[[:alpha:]]' cord.txt | sort | uniq -c

• grep -o only outputs the matching part. With the POSIX class [[:alpha:]] , it outputs all the letters contained in the input.
• sort groups the same letters together
• uniq -c reports unique lines with their counts. It needs sorted input, as it only compares the current line to the previous one.

Answer 2

The following command

1. Removes any character that is not an ASCII letter;
2. Places every character on its own line;
3. Sorts the characters;
4. Counts the number of same consecutive lines.

sed 's/[^a-zA-Z]//g' < input.txt | fold -w 1 -s | sort | uniq -c > output.txt
# ^                                ^              ^      ^
# 1.                               2.             3.     4.

Input:

188H,190D,245H
187D,481E,482T
187H,194E,196D
386D,388E,389N,579H
44E,60D

output:

 5 D
 4 E
 4 H
 1 N
 1 T

Answer 3

You might use python's collections.Counter as follows, let cord.txt content be

188H,190D,245H
187D,481E,482T
187H,194E,196D
386D,388E,389N,579H
44E,60D

and counting.py be

import collections
counter = collections.Counter()
with open("cord.txt", "r") as f:
    for line in f:
        counter.update(i for i in line if i.isalpha())
for char, cnt in counter.items():
    print("{},{}".format(char,cnt))

then python counting.py output

H,4
D,5
E,4
T,1
N,1

Note that I used for line in f where f is file-handle to avoid loading whole file into memory. Disclaimer: I used python version 3.7 , older should work but might give other order in output, as collections.Counter is subclass of dict and these do not keep order in older python versions.

Answer 4

Shortly:

tr '[0-9],' \\n <input | sort | uniq -c
     43 
      5 D
      4 E
      4 H
      1 N
      1 T

Ok, there are 43 other characters... You could drop and match your request by adding sed :

tr '[0-9],' \\n </tmp/so/input | sort | uniq -c |
     sed -ne 's/^ *\([0-9]\+\) \(.\)/\2,\1/p'
D,5
E,4
H,4
N,1
T,1