bash - 如何用sed用空格替换不是字母的任何东西？

Question

我需要从文件中删除不是字母、小写或大写的所有内容，并将其替换为空格，例如：

The bear ate 3 snakes, then ate 50% of the fish from the river.

这变成：

The bear ate   snakes  then ate     of the fish from the river 

• 有时文件包含不寻常的字符。它保存为 UTF-8。

如何用空格替换任何非字母？

Answer 1

如果你想支持 unicode 字母(as mentioned in your question)，那么这个 perl 命令替换所有unicode non-letters：

echo $line | perl -pe 's/[^\p{L}\s]+/ /g;'

参考：http ://www.regular-expressions.info/unicode.html

Answer 2

$ echo "The bear ate 3 snakes, then ate 50% of the fish from the river." | sed "s/[^a-zA-Z]/ /g"
The bear ate   snakes  then ate     of the fish from the river 

Answer 3

这可能对您有用：

echo 'The bear ate 3 snakes, then ate 50% of the fish from the river.' | 
tr -c '[:alpha:]' ' '
The bear ate   snakes  then ate     of the fish from the river

或者：

echo 'The bear ate 3 snakes, then ate 50% of the fish from the river.' |
sed 's/[^[:alpha:]]/ /g'
The bear ate   snakes  then ate     of the fish from the river

Answer 4

尝试：

sed 's/[^A-Za-z]/ /g;' myfile.txt