Linux Bash 输出具有最高值的行
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Bash output the line with highest value
提问by teutara
my question is pretty much like this onebut with one difference; i want the output the line that has highest score on the 3rd tab. my data is like:
我的问题很像这个，但有一点不同；我希望输出在第三个选项卡上得分最高的行。我的数据是这样的：
1.gui Qxx 16
2.gui Qxy 23
3.guT QWS 11
and i want to get this:
我想得到这个：
1.gui Qxy 23
3.guT QWS 11
I used:
我用了：
cat file.f | uniq | cut -d" " -f3 | sort | uniq -d >>out.f
but did not get what i want!?
但没有得到我想要的！？
采纳答案by Chris Seymour
With sort:
与sort：
$ sort -rk3 file # Sort on column 3, display all results
2.gui Qxy 23
1.gui Qxx 16
3.guT QWS 11
$ sort -rk3 file | head -2 # Sort on column 3, filter number of results
2.gui Qxy 23
1.gui Qxx 16
$ sort -rk3 file | uniq # Sort on column 3, on display unique results
2.gui Qxy 23
1.gui Qxx 16
3.guT QWS 11
-rreverse sort, highest first.
-r反向排序，最高的在前。
-k3sort on the 3rd column.
-k3对第 3 列进行排序。
If you only want to display line which the 3rd column is greater than some value (i.e. 15)then try this using awk:
如果您只想显示第 3 列大于某个值（即 15）的行，请尝试使用awk：
awk '>15' file | sort -rk3 # Display line where column 3 > 15 and sort
2.gui ?Qxy ?23
1.gui ?Qxx ?16
回答by teutara
for future users with same question:
对于有相同问题的未来用户：
do not forget to introduce -n switch to the -sortcommand, or your values are ordered starting from 9999's and followed by 999's etc.. so use
不要忘记在-sort命令中引入 -n 开关，或者您的值从 9999 开始排序，然后是 999 等。所以使用
sort -rnk3 file
and if you want to get only one line with highest value (remove duplicates) use this:
如果您只想获得具有最高值的一行（删除重复项），请使用以下命令：
sort -rnk3 file | awk '!x[]++'
and if you have an usual delimiter you can tell -awkto notice:
如果你有一个常用的分隔符，你可以告诉-awk注意：
sort -rnk3 file | awk -F"[. ]" '!x[]++'
回答by Eric González
This must give you the highest value for those lines where the name is repeated and conserve those lines that have not repeated names.
这必须为那些名称重复的行提供最高值，并保留那些没有重复名称的行。
sort -rk3 file | awk '!seen[]++' > file_filtered.txt
