是否有可能找到與grep相同的單詞多個出現的行？

Question

是否可以編寫（單行）grep表達式來查找包含三個相同單詞的行？請注意，我們並不知道這個詞先驗。下面的代碼片段捕獲大多數情況下：

$ grep -E '(\w+)[[:space:]]+\1[[:space:]]+\1' test_data.txt 

然而，這不趕下正面例子：

午餐晚餐晚飯晚飯午餐

另外請注意，我們只尋找完整的單詞，而不是簡單的字符重複。因此，一個反面例子的一個例子是：

他彈撥牛逼他花牛逼他重新

EDIT（感謝@列弗-levitsky）：

上面的正面例子實際上被抓到了，但以下不是：

午餐午餐晚餐晚餐午餐

Answer 1

這應該爲你工作：

grep -E "[[:<:]](\w+)[[:>:]].*[[:<:]]\1[[:>:]].*[[:<:]]\1[[:>:]]" testfile 

例如：

paul@horus:~/src/sandbox$ cat testfile 
how is summer summer summer ha ha 
this summer is a hot summer of summers yes it is 
summer summer summer 
there is only one summer in this sentence 
summer appears as the first and last summer words in this summer 
the summertime is always in summer, one of several summers 
the summer of which we speak is summery but is a real summer summer, yes 
this also works with cats, since there are three cats in these cats, ha! 
paul@horus:~/src/sandbox$ grep -E "[[:<:]](\w+)[[:>:]].*[[:<:]]\1[[:>:]].*[[:<:]]\1[[:>:]]" testfile 
how is summer summer summer ha ha 
summer summer summer 
summer appears as the first and last summer words in this summer 
the summer of which we speak is summery but is a real summer summer, yes 
this also works with cats, since there are three cats in these cats, ha! 
paul@horus:~/src/sandbox$ 

[[:<:]]和[[:>:]]比賽在一個字，分別的開始和結束的空字符串，這樣你就可以使用它們確定單詞的邊界，而不必假設單詞之間用空格分隔，而不是用標點符號等。

Answer 2

這不是grep也不regex，但它可能工作：

awk -F"[,. \t]*" '{for (i=1;i<=NF;i++) {if (++a[$i]==3) {printf "%s ",$i;f=1}} if (f) print "";f=0;delete a}' file 

它計算在每一行字，並打印就行了字，如果有三個或更多的發現了它的。