如何使用grep计数txt文件中五个字母单词的数量？

Question

我不太擅长linux，并且正在尝试使用grep计数五个字母单词。

Answer 1

Use the c flag to count, look for patterns containing five characters:

 $ cat file
 some text file containing many words and sentences.
 $ tr ' ' '\n' < file | grep -c '^[ \t]*[a-zA-Z]\{5\}[ \t]*$'
 1

Answer 2

You can use:

grep -o -w "\w\{5\}" your_file | wc -w

With -o only matched words will be printed, -w denotes that regex is searched as a word, \\w\\{5\\} - regex string itself (matches 5 continuous word characters). So, with your_file containing

word1 word2 word3
long_word 123 word4

Output of grep -o -w "\\w\\{5\\}" your_file will be

word1
word2
word3
word4

Piped wc -w just counts this.

Note : If you don't want to match all alphanumeric characters - replace \\w meta-character by something more specific. For example [az] - lowercase English letters.

Answer 3

This gnu awk (due to mulitple characters in Record Selector) does count how many word have 5 letters. It does ignore ., etc.

awk -v RS="[ .,?!]|\n" 'length($0)==5 {a++} END {print a}' file