On 08/01/2009, Greg Snow <greg.s...@imail.org> wrote:
> If you want to just linearly interpolate, then use the functions approx or
> approxfun from the stats package (one of those that is loaded by default).
I have read the guide for approx and approxfun functions. Below is my data.
x=c(0,5,10,15,20)
y=c(16,45,77,101,125)
approx(x,y,method="linear")
$x
[1] 0.0000000 0.4081633 0.8163265 1.2244898 1.6326531 2.0408163
[7] 2.4489796 2.8571429 3.2653061 3.6734694 4.0816327 4.4897959
[13] 4.8979592 5.3061224 5.7142857 6.1224490 6.5306122 6.9387755
[19] 7.3469388 7.7551020 8.1632653 8.5714286 8.9795918 9.3877551
[25] 9.7959184 10.2040816 10.6122449 11.0204082 11.4285714 11.8367347
[31] 12.2448980 12.6530612 13.0612245 13.4693878 13.8775510 14.2857143
[37] 14.6938776 15.1020408 15.5102041 15.9183673 16.3265306 16.7346939
[43] 17.1428571 17.5510204 17.9591837 18.3673469 18.7755102 19.1836735
[49] 19.5918367 20.0000000
$y
[1] 16.00000 18.36735 20.73469 23.10204 25.46939 27.83673 30.20408
[8] 32.57143 34.93878 37.30612 39.67347 42.04082 44.40816 46.95918
[15] 49.57143 52.18367 54.79592 57.40816 60.02041 62.63265 65.24490
[22] 67.85714 70.46939 73.08163 75.69388 77.97959 79.93878 81.89796
[29] 83.85714 85.81633 87.77551 89.73469 91.69388 93.65306 95.61224
[36] 97.57143 99.53061 101.48980 103.44898 105.40816 107.36735 109.32653
[43] 111.28571 113.24490 115.20408 117.16327 119.12245 121.08163 123.04082
[50] 125.00000
So my task is to find the value of x when y=0. I couldn't find
guidance to use the xout function; any help with this please?
approxfun(x,y,rule=2,method="linear")
function (v)
.C("R_approx", as.double(x), as.double(y), as.integer(n), xout = as.double(v),
as.integer(length(v)), as.integer(method), as.double(yleft),
as.double(yright), as.double(f), NAOK = TRUE, PACKAGE = "base")$xout
<environment: 0x87e8830>
Where do I go to find out the meaning of these errors and how to resolve?
Yours,
rh...@conference.jabber.org
r 251 (27-06-07)
mandriva 2008
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