Peta Hitchens - 1 year ago 48

R Question

I have a data frame

`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

I am trying to calculate the numerical integration from

`time = 0`

`\integral_{u=0}^t Qfr du`

My data set looks like

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

So far I have only been able to calculate the integral in total, using the package

`pracma`

`require(pracma)`

trapz(x$time, x$Qfr)

# [1] 0.1536843

How do I code for integral from the origin to the time given in that row?

Any help greatly appreciated!

`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 Source

Since the other answer shows you how to use `pracma::trapz`

to achieve your purpose, I can't do that way. I had planned to wrote an answer that way, but since I spent a great deal of time editing your question, @shayaa took the first place. Luckily, I have a much better idea.

Trapezoidal numerical integral is nothing complicated. You already have `time`

on a regular grid 1, 2, 3, ... 100 with bin size 1, as well as known function values `Qfr`

on the grid. The numerical integral on each bin is just the area of the trapezoid. So, you could compute:

```
## integration on each bin cell
cell <- with(x, (Qfr[1:99] + Qfr[2:100]) / 2)
```

Then, the cumulative integral value you want is just:

```
cumsum(c(0, cell))
```

**This method is supper fast!** Suppose you have `N`

data points, this method has computational costs of `O(N)`

, yet it is fully vectorized. The other answer using `sapply`

is not vectorized, and will cost you `O(N^2)`

computation.