bash 用出现次数列出文本文件中的所有单词？
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List all the words in a text file with occurrence counts?
提问by HymanWM
Suppose I have file text.txtas below:
假设我有text.txt如下文件：
she likes cats, and he likes cats too.
she likes cats, and he likes cats too.
I'd like my result to look like:
我希望我的结果看起来像：
she 1
likes 2
cats 2
and 1
he 1
too 1
If putting space , .into it would make the scripts easier, that would be fine.
如果放入space , .它会使脚本更容易，那就没问题了。
Is there a simple shell pipeline that could achieve this?
是否有一个简单的 shell 管道可以实现这一目标？
回答by phs
Here's a one-liner near and dear to my heart:
这是我心中最亲近的单行诗：
cat text.txt | sed 's|[,.]||g' | tr ' ' '\n' | sort | uniq -c
The sed strips punctuation (tune regex to taste), the tr puts the results one word per line.
sed 去掉标点符号（根据口味调整正则表达式），tr 将结果每行一个字。
回答by Ed Morton
With GNU awk you can just specify the Record Separator (RS) to be any sequence of non-alphabetic characters:
使用 GNU awk，您可以将记录分隔符 (RS) 指定为任何非字母字符序列：
$ gawk -v RS='[^[:alpha:]]+' '{sum[##代码##]++} END{for (word in sum) print word,sum[word]}' file
she 1
likes 2
and 1
too 1
he 1
cats 2
but that won't solve your problem of how to identify "words" in general.
但这并不能解决您一般如何识别“单词”的问题。
