如何按特定列对csv进行排序 [英] How to sort csv by specific column

问题描述

我尝试按第4列对包含csv的温度进行排序。

I try to sort a csv contain temperatures by the 4th column.

Sort -n -k4 temperature.csv

结果是：

2017-06-24 11:20,23.57,19.0,16.7,0.087,3.615
2017-06-24 11:25,23.51,19.0,16.7,0.087,3.689
2017-06-24 12:45,22.03,19.0,17.1,0.096,4.152
2017-06-24 13:00,21.92,19.0,17.1,0.096,4.229
2017-06-24 14:00,22.22,19.0,17.4,0.197,4.639
2017-06-24 14:25,22.21,19.0,17.5,0.197,4.774
2017-06-24 15:10,22.30,19.0,17.1,0.134,5.472
2017-06-24 16:00,22.42,19.0,17.3,0.134,5.93
2017-06-24 17:45,22.07,21.0,17.0,0.144,6.472
2017-06-24 18:25,21.90,21.0,16.9,0.15,6.814
2017-06-24 19:40,23.01,21.0,16.9,0.318,8.503

您可以看到第4列的排序不正确。我希望第一行是17.5，最后一行是16.7。

As you can see the 4th column isn't sorted correctly. I would expect 17.5 in tbe first line and 16.7 in the last line.

我也尝试过这样做：

sort -n -t. -k4,1n temperature.csv

结果与前面的示例完全相同。
有人可以给我提示吗？

The result is exactly the same as the previous example. Can anyone give me a hint?

推荐答案

使用以下 sort 命令：

sort -t, -k4,4 -nr temperature.csv

输出：

2017-06-24 14:25,22.21,19.0,17.5,0.197,4.774
2017-06-24 14:00,22.22,19.0,17.4,0.197,4.639
2017-06-24 16:00,22.42,19.0,17.3,0.134,5.93
2017-06-24 15:10,22.30,19.0,17.1,0.134,5.472
2017-06-24 13:00,21.92,19.0,17.1,0.096,4.229
2017-06-24 12:45,22.03,19.0,17.1,0.096,4.152
2017-06-24 17:45,22.07,21.0,17.0,0.144,6.472
2017-06-24 19:40,23.01,21.0,16.9,0.318,8.503
2017-06-24 18:25,21.90,21.0,16.9,0.15,6.814
2017-06-24 11:25,23.51,19.0,16.7,0.087,3.689
2017-06-24 11:20,23.57,19.0,16.7,0.087,3.615

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• -t，-字段分隔符

  -k4,4 -仅按第4个字段排序

  -k4,4 - sort by 4th field only

  -nr -以相反的顺序对数字进行排序

  -nr - sort numerically in reverse order

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