我想对一个分开的空间进行分类 table, ,带有第二个字段上发现的数值。我可以假设第二个字段始终是foon,但n的长度未知:

antiq. foo11 girls
colleaguing foo2 Leinsdorf
Cousy foo0 Montgomeryville
bowlegged foo1 pollack
Chevrier foo10 ill-conceived
candlebomb foo3 seventieths
autochthony foo101 re-enable
beneficiate foo100 osteometric

我读 man sort(1) 并使用各种选项。在我的系统上,我找到了行:

sort -n -k2.5 table

去工作。

我的问题是 为什么?

根据人页面:

-k, --key=POS1[,POS2]
   start a key at POS1, end it at POS 2 (origin 1) 
...
POS is F[.C][OPTS], where F is the field number and C the characterposition in the
field. OPTS is one or more single-letter ordering options, which override global
ordering options for that key. If no key is given, use the entire line as the key. 

所以为什么 sort -n -k2.4 表不起作用, sort -n -k2.5 做?

有帮助吗?

解决方案 2

答案是: 领先的空间被计为现场的一部分, , 除非:

sort -b -n -k2.4 table

或好奇:

LC_ALL=C sort -t" " -n -k2.4 table

这也会产生正确的结果。


... 还有一件事情 ...

似乎最好使用:

sort -b -n -k2.4,2 table

因此将排序限制在第二场的末尾。

其他提示

我不知道它是否有帮助,但是 信息排序 指出以下内容:

排序-t:-k 2,2n -k 5.3,5.4

 Note that if you had written `-k 2' instead of `-k 2,2' `sort'
 would have used all characters beginning in the second field and
 extending to the end of the line as the primary _numeric_ key.
 For the large majority of applications, treating keys spanning
 more than one field as numeric will not do what you expect.

也许尝试在之间添加一个空间 -k2, ,或尝试设置 POS2?

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