data.frame的复杂子集
[英]Complex subset of a data.frame
问题描述
I have a data frame with close to a million objects in it. 
我有一个包含近一百万个对象的数据框。 I need an efficient to way to subset the data based on multiple criteria. 我需要一种有效的方法来基于多个条件对数据进行子集化。 I can do this is a for loop but was wondering if there is a more elegant way to do this. 我可以这样做是一个for循环，但是想知道是否有更优雅的方法可以做到这一点。
Time Instance Server Metric Value
17/08/2014 04:00:00 PM ID1 Server888 disk.commandsaveraged.average 0
17/08/2014 04:00:00 PM ID1 Server999 disk.commandsaveraged.average 0
17/08/2014 04:00:00 PM ID1 Server777 disk.commandsaveraged.average 0
17/08/2014 04:05:00 PM ID1 Server888 disk.commandsaveraged.average 0
17/08/2014 04:05:00 PM ID1 Server999 disk.commandsaveraged.average 0
17/08/2014 04:05:00 PM ID1 Server777 disk.commandsaveraged.average 0
17/08/2014 04:00:00 PM ID2 Server888 disk.commandsaveraged.average 0
17/08/2014 04:05:00 PM ID2 Server888 disk.commandsaveraged.average 0
17/08/2014 04:00:00 PM ID3 Server999 disk.commandsaveraged.average 0
17/08/2014 04:05:00 PM ID3 Server999 disk.commandsaveraged.average 0
17/08/2014 04:00:00 PM ID3 Server777 disk.commandsaveraged.average 0
17/08/2014 04:05:00 PM ID3 Server777 disk.commandsaveraged.average 0
17/08/2014 04:00:00 PM ID1 Server888 disk.numberreadaveraged.average 0
17/08/2014 04:00:00 PM ID1 Server999 disk.numberreadaveraged.average 0
17/08/2014 04:00:00 PM ID1 Server777 disk.numberreadaveraged.average 0
17/08/2014 04:05:00 PM ID1 Server888 disk.numberreadaveraged.average 0
17/08/2014 04:05:00 PM ID1 Server999 disk.numberreadaveraged.average 0
17/08/2014 04:05:00 PM ID1 Server777 disk.numberreadaveraged.average 0
17/08/2014 04:00:00 PM ID2 Server888 disk.numberreadaveraged.average 0
17/08/2014 04:05:00 PM ID2 Server888 disk.numberreadaveraged.average 0
17/08/2014 04:00:00 PM ID3 Server999 disk.numberreadaveraged.average 0
17/08/2014 04:05:00 PM ID3 Server999 disk.numberreadaveraged.average 0
17/08/2014 04:00:00 PM ID3 Server777 disk.numberreadaveraged.average 0
17/08/2014 04:05:00 PM ID3 Server777 disk.numberreadaveraged.average 0
17/08/2014 04:00:00 PM ID1 Server888 disk.numberwriteaveraged.average 0
17/08/2014 04:00:00 PM ID7 Server999 disk.numberwriteaveraged.average 0
17/08/2014 04:00:00 PM ID1 Server777 disk.numberwriteaveraged.average 0
17/08/2014 04:05:00 PM ID1 Server888 disk.numberwriteaveraged.average 0
17/08/2014 04:05:00 PM ID1 Server999 disk.numberwriteaveraged.average 0
17/08/2014 04:05:00 PM ID7 Server777 disk.numberwriteaveraged.average 0
17/08/2014 04:00:00 PM ID2 Server888 disk.numberwriteaveraged.average 0
17/08/2014 04:05:00 PM ID5 Server888 disk.numberwriteaveraged.average 0
17/08/2014 04:00:00 PM ID3 Server999 disk.numberwriteaveraged.average 0
17/08/2014 04:05:00 PM ID4 Server999 disk.numberwriteaveraged.average 0
17/08/2014 04:00:00 PM ID3 Server777 disk.numberwriteaveraged.average 0
17/08/2014 04:05:00 PM ID3 Server777 disk.numberwriteaveraged.average 0
What I want to do is create a subset where metric == disk.numberwriteaveraged.average , Server == Server999 & Server == Server888 AND WHERE both servers have the same instance ID's in common. 我想做的是创建一个子集，其中metric == disk.numberwriteaveraged.average ， Server == Server999 & Server == Server888两台服务器的实例ID相同。
NOTE, I use the term subset purely because I don't know of any other way to filter data i R, still learning. 注意，我纯粹使用术语“子集”是因为我不知道仍然可以学习的其他任何方法来过滤数据i R。 I am looking for speed and I will be generating data sets much larger than my current one. 我正在寻找速度，并且我将生成比当前数据集大得多的数据集。
1 个解决方案
解决方案1
2 2014-08-21 20:45:30
(If I understand your question correctly) In your case, data.table is your friend. （如果我正确理解了您的问题），在您的情况下， data.table是您的朋友。 Try (assuming df is your data set): 尝试（假设df是您的数据集）：
library(data.table)
df2 <- setDT(df)[, .SD[Metric == "disk.commandsaveraged.average" &
(Server == "Server999" | Server == "Server888")], by = Instance]
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