需要使用时，与从输出删除计数QUOT; uniq的-c＆QUOT;命令 [英] Need to remove the count from the output when using "uniq -c" command

问题描述

我想读一个文件，并通过特定字段的出现次数排序。假设我想找出从日志文件，然后我用的uniq -c选项，它降序排列最重复的日期。像这样

I am trying to read a file and sort it by number of occurrences of a particular field. Suppose i want to find out the most repeated date from a log file then i use uniq -c option and sort it in descending order. something like this

uniq -c | sort -nr 

这会产生一定的输出这样的 -

This will produce some output like this -

809 23/Dec/2008:19:20

这实际上是计数的第一场是我的问题....我想ONY从上面的输出日期，但中号没能得到这一点。我试图用剪切命令，这样做，

the first field which is actually the count is the problem for me .... i want to get ony the date from the above output but m not able to get this. I tried to use cut command and did this

uniq -c | sort -nr | cut -d' ' -f2 

但这只是打印空白...请有人可以帮我只得到的日期和砍掉计数。我只希望

but this just prints blank space ... please can someone help me on getting the date only and chop off the count. I want only

23/Dec/2008:19:20

感谢

推荐答案

从 uniq的计数是由空格pceded除非有超过7位在$ p $算，所以你需要做的是这样的：

The count from uniq is preceded by spaces unless there are more than 7 digits in the count, so you need to do something like:

uniq -c | sort -nr | cut -c 9-

要获得列（字符位置）9向上。或者你可以使用 SED ：

to get columns (character positions) 9 upwards. Or you can use sed:

uniq -c | sort -nr | sed 's/^.\{8\}//'

或

uniq -c | sort -nr | sed 's/^ *[0-9]* //'

这第二个选项是在1000万以上的重复次数面对强大的;如果你认为这可能是一个问题，它可能是比剪切替代更好。而且有可用的无疑是其他的选择了。

This second option is robust in the face of a repeat count of 10,000,000 or more; if you think that might be a problem, it is probably better than the cut alternative. And there are undoubtedly other options available too.

注意：这个计数通过实验在Mac OS X 10.7.3确定，但使用GNU uniq的从的coreutils 8.3。在BSD 的uniq -c 生产前一个数字计数3前导空格。该POSIX规范说从 的uniq -c <输出/ A>的格式必须象用：

Caveat: the counts were determined by experimentation on Mac OS X 10.7.3 but using GNU uniq from coreutils 8.3. The BSD uniq -c produced 3 leading spaces before a single digit count. The POSIX spec says the output from uniq -c shall be formatted as if with:

printf("%d %s", repeat_count, line);

这不会有任何前导空白。鉴于输出格式这种可能的变化，用 SED 脚本 [0-9] 正则表达式是最可靠的方法在观察和理论产量的变化处理从的uniq -c ：

which would not have any leading blanks. Given this possible variance in output formats, the sed script with the [0-9] regex is the most reliable way of dealing with the variability in observed and theoretical output from uniq -c:

uniq -c | sort -nr | sed 's/^ *[0-9]* //'

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