bash Sed - 删除字符串前的 x 个字符
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Sed - Delete x numbers of characters before a string
提问by user3145872
I have a txt file and I want to delete x number of characters before a specific string
我有一个 txt 文件，我想删除特定字符串之前的 x 个字符
Example :
例子 ：
52.451878252 foo 845.851878212 foo
98.444854655 foo 458.344854656 foo
I want to delete 8 characters before the string ' foo' in order to get this :
我想删除字符串 'foo' 之前的 8 个字符以获得这个：
52.45 foo 845.66 foo
98.44 foo 458.66 foo
How should I do with the sed command (or something else) ?
我应该如何处理 sed 命令（或其他命令）？
Thank you :)
谢谢 ：）
回答by janos
Here you go:
干得好：
sed -e 's/........foo/ foo/g' file.txt
Or equivalently:
或等效地：
sed -e 's/.\{8\}foo/ foo/g' file.txt
Or even:
甚至：
sed -e 's/.\{8\}\(foo\)/ /g' file.txt
Or a different approach, with the same output for your sample:
或者采用不同的方法，对您的样本使用相同的输出：
sed -e 's/[0-9]\{7\} / /g' file.txt
回答by John B
If your goal is to round each float down to the second decimal point, you could use printf.
如果您的目标是将每个浮点数四舍五入到第二个小数点，您可以使用printf.
while read line; do
printf "%0.2f %s %0.2f %s\n" $line
done < file.txt
