linux - 如何使用 bash 脚本计算单词中出现次数最多的 3 个字母序列

Question

这个问题已经有答案了:

How can I count most occuring sequence of 3 letters within a word with a bash script (3 个回答)

已关闭 3 年前。

Indo Cheap 有一个示例文件，例如

XYZAcc
ABCAccounting
Accounting firm
Accounting Aco
Accounting Acompany
Acoustical consultant

他需要获取单词中出现次数最多的 3 个字母序列。

输出应该是

acc = 5 aco = 3

他询问这在 bash 中是否可行。

他说:“我完全不知道如何使用 awk、sed、grep 来完成它。

任何线索怎么可能......”

Answer 1

这对于 bash、sed 和 awk 来说绝对是可能的，具体方法如下:

#!/bin/bash

for line in $(cat sample | tr 'A-Z' 'a-z' | tr -s ' ' '\n'); do
  ll=${#line}
  for ((i = 0; i < ll - 2; i++)) ; do   # for each word
    echo ${line:i:3}                    # print all sequences of 3 letters
  done
done | 
  sort |                                # sort the sequences of three letters
  uniq -c |                             # count the sequences
  sed '/^ *1 /d' |                      # filter out the not repeated sequences
  sort -n -r |                          # most frequent sequences first
  awk -F ' ' '{print $2" = "$1}' |      # format output as asked
  tr '\n' ' '                           # put all results on one line 
echo                                    # add a new line at the end

上面示例的输出是:

cou = 5 acc = 5 unt = 4 tin = 4 oun = 4 nti = 4 ing = 4 cco = 4 aco = 3

如果需要其他格式的输出，我们可以根据需要轻松调整脚本代码。