计算从原点到R中每个时间点的梯形积分

Question

我有一个数据框x：

head(x)
#  time         Qfr
#1    1 0.004751271
#2    2 0.005405618
#3    3 0.005785781
#4    4 0.006028213
#5    5 0.006179973
#6    6 0.006263814

我正在尝试计算从time = 0到每个时间点的数值积分，即积分：

\integral_{u=0}^t Qfr du

我的数据集看起来像

plot(x, type = "p", cex = 0.2)

到目前为止，我只能使用包pracma来计算总积分：

require(pracma)
trapz(x$time, x$Qfr)
# [1] 0.1536843

如何编码从原点到该行中给定时间的积分？

任何帮助都非常感谢！

x <- 
structure(list(time = 1:100, Qfr = c(0.00475127142639315, 0.00540561802578535, 
0.00578578141896237, 0.00602821304872631, 0.00617997318815436, 
0.00626381438010966, 0.0062930341038365, 0.00627650284793016, 
0.00622076547748955, 0.00613104312485634, 0.00601175416200995, 
0.00586680072681021, 0.00569973138194467, 0.00551383427584607, 
0.00531218958660475, 0.00509769744944577, 0.00487309097312275, 
0.00464094029551979, 0.0044036514994002, 0.00416346290979426, 
0.00392244046575488, 0.00368247330791138, 0.00344527034180023, 
0.00321235826358148, 0.00298508133306843, 0.00276460302703881, 
0.00255190958997126, 0.0023478154110241, 0.00215297008955578, 
0.00196786700285879, 0.00179285315617775, 0.00162814007427384, 
0.00147381548391774, 0.00132985553610085, 0.00119613732394456, 
0.00107245146585054, 0.000958514542040229, 0.000853981195025623, 
0.000758455729566888, 0.000671503074231956, 0.000592658993812166, 
0.000521439468716574, 0.000457349183339767, 0.000399889089676022, 
0.000348563034666554, 0.00030288345957139, 0.000262376196826176, 
0.000226584404259194, 0.000195071688184451, 0.000167424475817082, 
0.000143253703811788, 0.000122195893694217, 0.000103913686771705, 
8.80959110345906e-05, 7.44572508696844e-05, 6.27375873853488e-05, 
5.27010730706422e-05, 4.4134999643845e-05, 3.68485125344791e-05, 
3.06712197100413e-05, 2.54517366998868e-05, 2.1056203851463e-05, 
1.73668062169167e-05, 1.42803211212964e-05, 1.17067134905708e-05, 
9.56779447711296e-06, 7.79595484853561e-06, 6.33298101979921e-06, 
5.1289585088234e-06, 4.14126496950611e-06, 3.33365277945052e-06, 
2.67541940141655e-06, 2.14066236056745e-06, 1.70761464370064e-06, 
1.35805559020556e-06, 1.07679186575234e-06, 8.51202848467075e-07, 
6.7084467545985e-07, 5.27107259938671e-07, 4.12918764032332e-07, 
3.22492271674051e-07, 2.51109725048683e-07, 1.94938546369442e-07, 
1.50876746740867e-07, 1.16422711484835e-07, 8.95662353428025e-08, 
6.86977528360132e-08, 5.25330624541746e-08, 4.00511738714265e-08, 
3.0443212344273e-08, 2.30705923615172e-08, 1.74309232147824e-08, 
1.31303328068787e-08, 9.86109389078818e-09, 7.38361045310242e-09, 
5.51197296231586e-09, 4.102421614833e-09, 3.044168569724e-09, 
2.25212541292965e-09, 1.66116273443706e-09)), .Names = c("time", 
"Qfr"), class = "data.frame", row.names = c(NA, -100L))

Answer 1

由于另一个答案显示了如何使用pracma::trapz来实现您的目的，我无法做到这一点。我曾计划以这种方式写一个答案，但由于我花了很多时间编辑你的问题，@ shayaa取得了第一名。幸运的是，我有一个更好的主意。

梯形数值积分并不复杂。您已经在常规网格1,2,3，... 100上拥有time，其中包含大小为1，以及网格上的已知函数值Qfr。每个箱子上的数值积分就是梯形的面积。所以，你可以计算：

## integration on each bin cell
cell <- with(x, (Qfr[1:99] + Qfr[2:100]) / 2)
## Note that precisely I should write
## cell <- with(x, (Qfr[1:99] + Qfr[2:100]) / 2 * diff(time))
## But as I said, you have equally spaced bin points with bin size 1
## `diff(time)` is always 1, hence left out
## You need to bear this in mind, once you work on more general cases.

然后，您想要的累积积分值只是：

cumsum(c(0, cell))

这种方法速度快！假设您有N个数据点，它的计算成本为O(N)，但它是完全矢量化的。使用sapply的另一个答案没有矢量化，并且会花费O(N^2)次计算费用。

Answer 2

您需要遍历索引，在本例中为i，并计算梯形直到第i个数据点。

sapply(1:100, function(i) trapz(x$time[1:i],x$Qfr[1:i]))