确定陈旧数据

Question

说我有这种格式

12:04:21 .3 
12:10:21 1.3 
12:13:21 1.4 
12:14:21 1.3 
..and so on 

我想，比如说，10个因之时间戳第二栏找到重复号的文件，从而找到的陈旧。

12:04:21 .3 
12:10:21 1.3 
12:14:21 1.3 
12:10:21 1.3 
12:14:21 1.3 
12:12:21 1.3 
12:24:21 1.3 
12:30:21 1.3 
12:44:21 1.3 
12:50:21 1.3 
13:04:21 1.3 
13:24:21 1.7 

应打印12点10分21秒至13时04分21秒1.3

，我想输出的开始和结束，并在陈旧的时间戳范围的

有人可以帮助我想出来吗？

您可以用awk，bash的

感谢

Answer 1

awk 'BEGIN { count = 1} { if ($2 == prev) { ++count; if (! start) {start = prevtime} end = $1 } 
     else if (count >= 10) { print start, end, prev; count = 1; start = "" } 
     else { start = "" }; 
     prev = $2; prevtime = $1 }' file.dat 

编辑2：

找到并修复另一个bug。

Answer 2

这里是我的版本，这是更详细：

# This function prints out the summary only when count >= 10 
function print_summary(count, first, last, value) { 
    if (count >= 10) { 
     printf "%s through %s %s (%d)\n", first, last, last_value, count 
    } 
} 

$2 == last_value { 
    last_occurance = $1 
    count++ 
} 

$2 != last_value { 
    print_summary(count, first_occurance, last_occurance, last_value) 
    first_occurance = $1 
    last_value = $2 
    count = 1 
} 

END { 
    print_summary(count, first_occurance, last_occurance, last_value) 
}