在R中找到零交叉

Question

如果我有以下数据：

df <- structure(list(x = c(1.63145539094563, 1.67548187017034, 1.71950834939504, 
1.76353482861975, 1.80756130784445, 1.85158778706915, 1.89561426629386, 
1.93964074551856, 1.98366722474327, 2.02769370396797, 2.07172018319267, 
2.11574666241738, 2.15977314164208, 2.20379962086679, 2.24782610009149, 
2.2918525793162, 2.3358790585409, 2.3799055377656, 2.42393201699031, 
2.46795849621501, 2.51198497543972, 2.55601145466442, 2.60003793388912, 
2.64406441311383, 2.68809089233853, 2.73211737156324, 2.77614385078794, 
2.82017033001265, 2.86419680923735, 2.90822328846205, 2.95224976768676, 
2.99627624691146, 3.04030272613617, 3.08432920536087, 3.12835568458557, 
3.17238216381028, 3.21640864303498, 3.26043512225969, 3.30446160148439, 
3.3484880807091, 3.3925145599338, 3.4365410391585, 3.48056751838321, 
3.52459399760791, 3.56862047683262, 3.61264695605732, 3.65667343528202, 
3.70069991450673, 3.74472639373143, 3.78875287295614), y = c(24.144973858154, 
18.6408277478876, 21.9174270206615, 22.8017876727379, 20.9766270378248, 
18.604384256745, 18.4805250429826, 15.8436744335752, 13.6357170277296, 
11.6228806771368, 9.4065868126964, 6.81644596802601, 4.41187500831424, 
4.31911614349431, 0.678259284890563, -1.18632719250877, -2.32986407762089, 
-3.84480566043122, -5.24738510499144, -5.20160089844013, -5.42094587600499, 
-5.39886757202858, -5.26753920575326, -4.68727963638973, -2.73267203102102, 
0.296905237887623, 2.45725152489283, 5.12102449689086, 7.13986218237411, 
10.2044876281093, 14.4358946463429, 19.0643081865458, 22.8920445618834, 
26.7229418763085, 31.3776791707576, 36.19058349817, 41.2843224331918, 
46.3396522631345, 51.4321502764393, 56.4080998038294, 61.5215778808583, 
66.6845421308734, 71.3912749310486, 76.0856977880158, 80.7039319129457, 
84.4095953723555, 88.0163019647757, 89.918078622734, 91.6341473685881, 
94.0404562451352)), class = c("tbl_df", "tbl", "data.frame"), .Names = c("x", 
"y"), row.names = c(NA, -50L))

情节：

在x时如何找到确切的y == 0值？我尝试进行内插，但不一定给我一个y等于零的值。有人知道过零的功能吗？

Answer 1

首先，可以使用

定义相应的（线性）插值函数

approxfun(df$x, df$y)

结果看起来像

curve(approxfun(df$x, df$y)(x), min(df$x), max(df$x))

那么那些零交叉可以看作是这个函数的根源。在基数R中有一个函数uniroot，但它查找单个根，而在您的情况下，我们有两个。因此，一种选择就是像

一样的rootSolve软件包

library(rootSolve)
uniroot.all(approxfun(df$x, df$y), interval = range(df$x))
# [1] 2.263841 2.727803