我有日期,时间,值格式的数据。这是一个示例:
04/01/2010,07:10,17159
04/01/2010,07:20,4877
04/01/2010,07:30,6078
04/01/2010,07:40,3105
04/01/2010,07:50,4073
04/01/2010,08:00,6986
04/01/2010,08:10,7906
04/01/2010,08:20,7681
04/01/2010,08:30,5665
04/01/2010,08:40,6631
04/01/2010,08:50,4633
04/01/2010,09:00,6346
04/01/2010,09:10,6444
04/01/2010,09:20,6324
04/01/2010,09:30,11696
04/01/2010,09:40,7667
04/01/2010,09:50,6375
04/01/2010,10:00,5934
04/01/2010,10:10,12626
04/01/2010,10:20,11674
04/01/2010,10:30,4660
04/01/2010,10:40,3831
04/01/2010,10:50,7089
04/01/2010,11:00,4548
04/01/2010,11:10,2590
04/01/2010,11:20,3334
04/01/2010,11:30,5171
我想将其转换为保持相同格式的时间值序列。即我也需要能够存储日期和时间组件。这是因为我想对数据进行“反季节化”。
我试过了
z <- read.csv("fileName", header=TRUE,sep=",")
但不确定从这里做什么。谁能告诉我如何正确加载到时间序列对象中?还是有其他方法可以做到这一点?
提前致谢
您可以使用该zoo
程序包。下面的代码被编写为可复制的,但实际上text="Lines"
将替换为file="fileName"
。同样,如问题中所示,“日期”字段不明确,如果不是天/月/年,则可能需要调整百分比代码。
library(zoo)
Lines <- "Date,Time,Value
04/01/2010,07:10,17159
04/01/2010,07:20,4877
04/01/2010,07:30,6078
04/01/2010,07:40,3105
"
z <- read.zoo(text = Lines, sep = ",", header = TRUE,
index = 1:2, tz = "", format = "%d/%m/%Y %H:%M")
这使:
> z
2010-01-04 07:10:00 2010-01-04 07:20:00 2010-01-04 07:30:00 2010-01-04 07:40:00
17159 4877 6078 3105
本文收集自互联网,转载请注明来源。
如有侵权,请联系[email protected] 删除。
我来说两句