我有以下数值:
Day 1: X11 X12 X13 X14 X15 ...
Day 2: X21 X22 X23 X24 X25 ...
Day 3: X31 X32 X33 X34 X35 ...
...
...
我可以使用每天的值来绘制不同的CDF,但有没有一种方法可以将所有日期的CDF可视化,以了解分布在不同时间的变化情况如何?有没有像我可以绘制的3D变体?
实际值:
structure(list(Time = structure(c(1354348800, 1354435200, 1354521600,
1354608000, 1354694400, 1354780800, 1354867200, 1354953600, 1355040000,
1355126400, 1355212800, 1355299200, 1355385600, 1355472000, 1355558400,
1355644800, 1355731200, 1355817600, 1355904000, 1355990400, 1356076800,
1356163200, 1356249600, 1356336000, 1356422400, 1356508800, 1356595200,
1356681600, 1356768000, 1356854400, 1356940800, 1354348800, 1354435200,
1354521600, 1354608000, 1354694400, 1354780800, 1354867200, 1354953600,
1355040000, 1355126400, 1355212800, 1355299200, 1355385600, 1355472000,
1355558400, 1355644800, 1355731200, 1355817600, 1355904000, 1355990400,
1356076800, 1356163200, 1356249600, 1356336000, 1356422400, 1356508800,
1356595200, 1356681600, 1356768000, 1356854400, 1356940800, 1354348800,
1354435200, 1354521600, 1354608000, 1354694400, 1354780800, 1354867200,
1354953600, 1355040000, 1355126400, 1355212800, 1355299200, 1355385600,
1355472000, 1355558400, 1355644800, 1355731200, 1355817600, 1355904000,
1355990400, 1356076800, 1356163200, 1356249600, 1356336000, 1356422400,
1356508800, 1356595200, 1356681600, 1356768000, 1356854400, 1356940800,
1354348800, 1354435200, 1354521600, 1354608000, 1354694400, 1354780800,
1354867200, 1354953600, 1355040000, 1355126400, 1355212800, 1355299200,
1355385600, 1355472000, 1355558400, 1355644800, 1355731200, 1355817600,
1355904000, 1355990400, 1356076800, 1356163200, 1356249600, 1356336000,
1356422400, 1356508800, 1356595200, 1356681600, 1356768000, 1356854400,
1356940800, 1354348800, 1354435200, 1354521600, 1354608000, 1354694400,
1354780800, 1354867200, 1354953600, 1355040000, 1355126400, 1355212800,
1355299200, 1355385600, 1355472000, 1355558400, 1355644800, 1355731200,
1355817600, 1355904000, 1355990400, 1356076800, 1356163200, 1356249600,
1356336000, 1356422400, 1356508800, 1356595200, 1356681600, 1356768000,
1356854400, 1356940800, 1354348800, 1354435200, 1354521600, 1354608000,
1354694400, 1354780800, 1354867200, 1354953600, 1355040000, 1355126400,
1355212800, 1355299200, 1355385600, 1355472000, 1355558400, 1355644800,
1355731200, 1355817600, 1355904000, 1355990400, 1356076800, 1356163200,
1356249600, 1356336000, 1356422400, 1356508800, 1356595200, 1356681600,
1356768000, 1356854400, 1356940800), class = c("POSIXct", "POSIXt"
), tzone = ""), Value = c(430664.239261698, 490234.194921927,
526998.520971122, 536602.462982633, 258691.669906957, 11829.8290116318,
12650.9461086689, 2720.1227453922, 6148.95630258592, 2680.86993550808,
2678.03258008561, 2665.03710105507, 2704.83781604574, 2704.99305811391,
2742.7040489269, 2802.16126409598, 2835.37392203882, 2822.02107441536,
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2745.58482436398, 2729.8646537471, 2706.30089379909, 2657.97901504968,
2755.8939918735, 2756.32635948648, 2703.73644754399, 2595.09862261747,
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2728.37598724549, 2732.36209370411, 2695.9513678436, 2686.83499265917,
2681.5743717285, 2720.45290857566, 2732.73841594837, 2760.53934947503,
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2807.55312109634, 2804.11955879203, 2757.32663905538, 2745.86930679521,
2739.23425025641, 2715.98260303707, 2692.68857278371, 3527.71521116871,
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58180354.3542921, 50719237.2638962, 33168848.2017861, 33850765.333152,
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54204572.2980234, 58898821.9609343, 62180038.7679437, 53059333.3332883,
43914570.1205393, 79475697.2825237, 64196704.372024, 66182175.1487631,
80851142.6092836, 65475805.206847, 68931230.2593625, 59859872.3417313,
65228418.926433, 61498299.9737327, 60767088.8695188, 62752189.9148476,
56770234.222601, 40910036.2385198, 39235601.7774785, 37635952.1705463
)), .Names = c("Time", "Value"), row.names = c(NA, -186L), class = "data.frame")
编辑:产生的图片
答案 0 :(得分:5)
不确定。检查包rgl.surface()
中的功能rgl
:
library(rgl)
data <- lapply(seq(0,2,by=0.1),function(i){rnorm(100)+i}) # a list with example data vectors
m <- sapply(data,function(i){ecdf(i)(seq(-3,5,by=0.1))}) # a matrix with the empirical CDF for each vector in the list, over a given range -3 to 5
rgl.surface(1:nrow(m), 5*(1:ncol(m)), m*20) # play with the coefficients to get a better result
你甚至可以旋转这个表面!有关更多信息,请查看this question。
更新:以下是您可以对数据执行的操作:
values.per.day <- lapply(unique(data$Time), function(x)data$Value[data$Time==x])
m <- sapply(values.per.day, function(x) ecdf(x)(seq(min(data$Value), max(data$Value), length.out=1000)) )
rgl.surface(1:nrow(m), 5*(1:ncol(m)), m*20)