Linux 如何对所有少于 4 个字符的单词进行 grep？
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How do I grep for all words that are less than 4 characters?
提问by TIMEX
I have a dictionary with words separated by line breaks.
我有一本字典，里面的单词用换行符分隔。
采纳答案by Michael Goldshteyn
You can just do:
你可以这样做：
egrep -x '.{1,3}' myfile
This will also skip blank lines, which are technically not words. Unfortunately, the above reg-ex will count apostrophes in contractions as letters as well as hyphens in hyphenated compound words. Hyphenated compound words are not a problem at such a low letter count, but I am not sure whether or not you want to count apostrophes in contractions, which are possible (e.g., I'm). You can try to use a reg-ex such as:
这也将跳过空行，这在技术上不是单词。不幸的是，上面的 reg-ex 会将收缩中的撇号计算为字母以及带连字符的复合词中的连字符。在如此低的字母数下，带连字符的复合词不是问题，但我不确定您是否要计算收缩中的撇号，这是可能的（例如，I'm）。您可以尝试使用正则表达式，例如：
egrep -x '\w{1,3}' myfile
..., but this will only match upper/lower case letters and not match contractions or hyphenated compound words at all.
...，但这只会匹配大写/小写字母，而根本不匹配收缩或带连字符的复合词。
回答by Paul Tomblin
Like this:
grep -v "^...." my_file
像这样：
grep -v "^...." my_file
回答by Mark Byers
Try this regular expression:
试试这个正则表达式：
grep -E '^.{1,3}$' your_dictionary
