[英]Count value in shell script
I need help , i have a file output.txt
我需要帮助,我有一个文件output.txt
ttl 128
ttl 128
ttl 128
ttl 128
ttl 1
ttl 128
ttl 255
ttl 1
ttl 64
ttl 128
ttl 128
ttl 1
i need count how many times appear the same value in the lines of the file. 我需要计算在文件行中出现相同值的次数。 The final result must be something like this: 最终结果必须是这样的:
ttl 128 - 7 times
ttl 64 - 1 time
ttl 255 - 1 time
ttl 1 - 3 times
I hope you can help me. 我希望你能帮助我。 I'm trying to use grep command. 我正在尝试使用grep命令。
Thanks a lot 非常感谢
You can do it with uniq
and sort
: 您可以使用uniq
和sort
做到这一点:
<output.txt sort -V | uniq -c
Output: 输出:
3 ttl 1
1 ttl 64
7 ttl 128
1 ttl 255
sort, uniq are sufficient for this job. 排序,uniq足以胜任这项工作。 however to get the same output format as described in question, try this awk line 但是,要获得与上述问题相同的输出格式,请尝试以下awk行
awk '{a[$0]++}END{for(x in a){t=a[x]>1?"times":"time";print x " - "a[x],t}} file
for example 例如
times
, otherwise time
without s. 如果count> 1显示times
,否则time
不带s。 output is: 输出为:
ttl 1 - 3 times
ttl 64 - 1 time
ttl 128 - 7 times
ttl 255 - 1 time
:) :)
You can use the sort
command to sort your file and uniq
to make it unique and count the occurrences: 您可以使用sort
命令对文件和uniq
进行排序,以使其唯一并计算出现次数:
sort output.txt | uniq -c
Note: Only 'problem' is that the counts are in front of each occurrence and in your example after it. 注意:唯一的“问题”是计数在每个事件的前面,而在您的示例中是在此之后。
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