使用 R 使用特定方程将曲线拟合到数据集
Using R to fit a curve to a dataset using a specific equation
我正在使用 R。
我想用一个特定的方程来拟合一条曲线到我的一个数据集（附上）
> dput(data)
structure(list(Gossypol = c(1036.331811, 4171.427741, 6039.995102,
5909.068158, 4140.242559, 4854.985845, 6982.035521, 6132.876396,
948.2418407, 3618.448997, 3130.376482, 5113.942098, 1180.171957,
1500.863038, 4576.787021, 5629.979049, 3378.151945, 3589.187889,
2508.417927, 1989.576826, 5972.926124, 2867.610671, 450.7205451,
1120.955, 3470.09352, 3575.043632, 2952.931863, 349.0864019,
1013.807628, 910.8879471, 3743.331903, 3350.203452, 592.3403778,
1517.045807, 1504.491931, 3736.144027, 2818.419785, 723.885643,
1782.864308, 1414.161257, 3723.629772, 3747.076592, 2005.919344,
4198.569251, 2228.522959, 3322.115942, 4274.324792, 720.9785449,
2874.651764, 2287.228752, 5654.858696, 1247.806111, 1247.806111,
2547.326207, 2608.716056, 1079.846532), Treatment = structure(c(2L,
3L, 4L, 5L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 5L, 1L, 2L, 3L, 4L, 5L,
1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L,
2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L,
3L, 4L, 5L, 1L, 2L, 3L, 1L), .Label = c("C", "1c_2d", "3c_2d",
"9c_2d", "1c_7d"), class = "factor"), Damage_cm = c(0.4955, 1.516,
4.409, 3.2665, 0.491, 2.3035, 3.51, 1.8115, 0, 0.4435, 1.573,
1.8595, 0, 0.142, 2.171, 4.023, 4.9835, 0, 0.6925, 1.989, 5.683,
3.547, 0, 0.756, 2.129, 9.437, 3.211, 0, 0.578, 2.966, 4.7245,
1.8185, 0, 1.0475, 1.62, 5.568, 9.7455, 0, 0.8295, 2.411, 7.272,
4.516, 0, 0.4035, 2.974, 8.043, 4.809, 0, 0.6965, 1.313, 5.681,
3.474, 0, 0.5895, 2.559, 0)), .Names = c("Gossypol", "Treatment",
"Damage_cm"), row.names = c(NA, -56L), class = "data.frame")
等式是：y~yo+a*(1-b^x)
在哪里：
y = Gossypol（来自我的数据集）
x = Damage_cm（来自我的数据集）
其他3个参数未知：
yo = Intercept、a = asymptote 和 b = slope
我想我必须使用包 nls2。到目前为止，我写了以下代码：
data<-read.csv("Regression_exp2.csv",header=T, sep = ",")
library(nls2)
attach(data)
m<-nls(Gossypol~Y+A*(1-B^Damage_cm),data=data,start = list(Y=1700,A=4000,B=1))
这给了我错误信息：
Error in nlsModel(formula, mf, start, wts): singular gradient matrix
at initial parameter estimates
最后我想用方程画曲线（有SE区间，我一般用ggplot2）
另外，我想知道R2和p值。
我也会对参数感兴趣 yo ， a 和 b
我以前从未这样做过，如果有人能帮助我或提示我如何在 R 中做到这一点，我将不胜感激？我想我必须使用非线性方法 (glm(...))
非常感谢，
麦克
您必须稍微调整一下起始值：
> data
Gossypol Treatment Damage_cm
1 1036.3318 1c_2d 0.4955
2 4171.4277 3c_2d 1.5160
3 6039.9951 9c_2d 4.4090
4 5909.0682 1c_7d 3.2665
5 4140.2426 1c_2d 0.4910
...
54 2547.3262 1c_2d 0.5895
55 2608.7161 3c_2d 2.5590
56 1079.8465 C 0.0000
那么您可以拨打：
m<-nls(data$Gossypol~Y+A*(1-B^data$Damage_cm),data=data,start = list(Y=1000,A=3000,B=0.5))
打印 m 给你：
> m
Nonlinear regression model
model: data$Gossypol ~ Y + A * (1 - B^data$Damage_cm)
data: data
Y A B
1303.4450 2796.0385 0.4939
residual sum-of-squares: 1.03e+08
现在可以根据拟合得到数据了：
fitData <- 1303.4450 + 2796.0385*(1-0.4939^data$Damage_cm)
绘制数据以比较拟合和原始数据：
plot(data$Damage_cm, data$Gossypol, col='black')
par(new=T)
plot(data$Damage_cm,fitData, col='red', ylim=c(0,8000), axes=F, ylab='')
这给你：
如果你想使用 nls2 确保它已安装，如果没有你可以使用
install.packages('nls2')
这样做。
library(nls2)
m2<-nls2(data$Gossypol~Y+A*(1-B^data$Damage_cm),data=data,start = list(Y=1000,A=3000,B=0.5))
与 nls:
的值相同
> m2
Nonlinear regression model
model: data$Gossypol ~ Y + A * (1 - B^data$Damage_cm)
data: structure(list(Gossypol = c(1036.331811, 4171.427741, 6039.995102, 5909.068158, 4140.242559, 4854.985845, 6982.035521, 6132.876396, 948.2418407, 3618.448997, 3130.376482, 5113.942098, 1180.171957, 1500.863038, 4576.787021, 5629.979049, 3378.151945, 3589.187889, 2508.417927, 1989.576826, 5972.926124, 2867.610671, 450.7205451, 1120.955, 3470.09352, 3575.043632, 2952.931863, 349.0864019, 1013.807628, 910.8879471, 3743.331903, 3350.203452, 592.3403778, 1517.045807, 1504.491931, 3736.144027, 2818.419785, 723.885643, 1782.864308, 1414.161257, 3723.629772, 3747.076592, 2005.919344, 4198.569251, 2228.522959, 3322.115942, 4274.324792, 720.9785449, 2874.651764, 2287.228752, 5654.858696, 1247.806111, 1247.806111, 2547.326207, 2608.716056, 1079.846532), Treatment = structure(c(2L, 3L, 4L, 5L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 1L), .Label = c("C", "1c_2d", "3c_2d", "9c_2d", "1c_7d"), class = "factor"), Damage_cm = c(0.4955, 1.516, 4.409, 3.2665, 0.491, 2.3035, 3.51, 1.8115, 0, 0.4435, 1.573, 1.8595, 0, 0.142, 2.171, 4.023, 4.9835, 0, 0.6925, 1.989, 5.683, 3.547, 0, 0.756, 2.129, 9.437, 3.211, 0, 0.578, 2.966, 4.7245, 1.8185, 0, 1.0475, 1.62, 5.568, 9.7455, 0, 0.8295, 2.411, 7.272, 4.516, 0, 0.4035, 2.974, 8.043, 4.809, 0, 0.6965, 1.313, 5.681, 3.474, 0, 0.5895, 2.559, 0)), .Names = c("Gossypol", "Treatment", "Damage_cm"), row.names = c(NA, -56L), class = "data.frame")
Y A B
1303.4450 2796.0385 0.4939
residual sum-of-squares: 1.03e+08
Number of iterations to convergence: 2
Achieved convergence tolerance: 4.936e-06
如果你喜欢ggplot2:
ggplot(data, aes(x = Damage_cm, y = Gossypol)) +
geom_point() +
geom_smooth(method = "nls",
formula = y ~ Y + A * (1 - B^x),
start = c(Y=1000, A=3000, B=0.5), se = F)
尽管恐怕必须在 ggplot 之外模拟标准错误。
我正在使用 R。 我想用一个特定的方程来拟合一条曲线到我的一个数据集（附上）
> dput(data)
structure(list(Gossypol = c(1036.331811, 4171.427741, 6039.995102,
5909.068158, 4140.242559, 4854.985845, 6982.035521, 6132.876396,
948.2418407, 3618.448997, 3130.376482, 5113.942098, 1180.171957,
1500.863038, 4576.787021, 5629.979049, 3378.151945, 3589.187889,
2508.417927, 1989.576826, 5972.926124, 2867.610671, 450.7205451,
1120.955, 3470.09352, 3575.043632, 2952.931863, 349.0864019,
1013.807628, 910.8879471, 3743.331903, 3350.203452, 592.3403778,
1517.045807, 1504.491931, 3736.144027, 2818.419785, 723.885643,
1782.864308, 1414.161257, 3723.629772, 3747.076592, 2005.919344,
4198.569251, 2228.522959, 3322.115942, 4274.324792, 720.9785449,
2874.651764, 2287.228752, 5654.858696, 1247.806111, 1247.806111,
2547.326207, 2608.716056, 1079.846532), Treatment = structure(c(2L,
3L, 4L, 5L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 5L, 1L, 2L, 3L, 4L, 5L,
1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L,
2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L,
3L, 4L, 5L, 1L, 2L, 3L, 1L), .Label = c("C", "1c_2d", "3c_2d",
"9c_2d", "1c_7d"), class = "factor"), Damage_cm = c(0.4955, 1.516,
4.409, 3.2665, 0.491, 2.3035, 3.51, 1.8115, 0, 0.4435, 1.573,
1.8595, 0, 0.142, 2.171, 4.023, 4.9835, 0, 0.6925, 1.989, 5.683,
3.547, 0, 0.756, 2.129, 9.437, 3.211, 0, 0.578, 2.966, 4.7245,
1.8185, 0, 1.0475, 1.62, 5.568, 9.7455, 0, 0.8295, 2.411, 7.272,
4.516, 0, 0.4035, 2.974, 8.043, 4.809, 0, 0.6965, 1.313, 5.681,
3.474, 0, 0.5895, 2.559, 0)), .Names = c("Gossypol", "Treatment",
"Damage_cm"), row.names = c(NA, -56L), class = "data.frame")
等式是：y~yo+a*(1-b^x)
在哪里：
y = Gossypol（来自我的数据集）
x = Damage_cm（来自我的数据集）
其他3个参数未知：
yo = Intercept、a = asymptote 和 b = slope
我想我必须使用包 nls2。到目前为止，我写了以下代码：
data<-read.csv("Regression_exp2.csv",header=T, sep = ",")
library(nls2)
attach(data)
m<-nls(Gossypol~Y+A*(1-B^Damage_cm),data=data,start = list(Y=1700,A=4000,B=1))
这给了我错误信息：
Error in nlsModel(formula, mf, start, wts): singular gradient matrix at initial parameter estimates
最后我想用方程画曲线（有SE区间，我一般用ggplot2）
另外，我想知道R2和p值。
我也会对参数感兴趣 yo ， a 和 b
我以前从未这样做过，如果有人能帮助我或提示我如何在 R 中做到这一点，我将不胜感激？我想我必须使用非线性方法 (glm(...))
非常感谢， 麦克
您必须稍微调整一下起始值：
> data
Gossypol Treatment Damage_cm
1 1036.3318 1c_2d 0.4955
2 4171.4277 3c_2d 1.5160
3 6039.9951 9c_2d 4.4090
4 5909.0682 1c_7d 3.2665
5 4140.2426 1c_2d 0.4910
...
54 2547.3262 1c_2d 0.5895
55 2608.7161 3c_2d 2.5590
56 1079.8465 C 0.0000
那么您可以拨打：
m<-nls(data$Gossypol~Y+A*(1-B^data$Damage_cm),data=data,start = list(Y=1000,A=3000,B=0.5))
打印 m 给你：
> m
Nonlinear regression model
model: data$Gossypol ~ Y + A * (1 - B^data$Damage_cm)
data: data
Y A B
1303.4450 2796.0385 0.4939
residual sum-of-squares: 1.03e+08
现在可以根据拟合得到数据了：
fitData <- 1303.4450 + 2796.0385*(1-0.4939^data$Damage_cm)
绘制数据以比较拟合和原始数据：
plot(data$Damage_cm, data$Gossypol, col='black')
par(new=T)
plot(data$Damage_cm,fitData, col='red', ylim=c(0,8000), axes=F, ylab='')
这给你：
如果你想使用 nls2 确保它已安装，如果没有你可以使用
install.packages('nls2')
这样做。
library(nls2)
m2<-nls2(data$Gossypol~Y+A*(1-B^data$Damage_cm),data=data,start = list(Y=1000,A=3000,B=0.5))
与 nls:
> m2
Nonlinear regression model
model: data$Gossypol ~ Y + A * (1 - B^data$Damage_cm)
data: structure(list(Gossypol = c(1036.331811, 4171.427741, 6039.995102, 5909.068158, 4140.242559, 4854.985845, 6982.035521, 6132.876396, 948.2418407, 3618.448997, 3130.376482, 5113.942098, 1180.171957, 1500.863038, 4576.787021, 5629.979049, 3378.151945, 3589.187889, 2508.417927, 1989.576826, 5972.926124, 2867.610671, 450.7205451, 1120.955, 3470.09352, 3575.043632, 2952.931863, 349.0864019, 1013.807628, 910.8879471, 3743.331903, 3350.203452, 592.3403778, 1517.045807, 1504.491931, 3736.144027, 2818.419785, 723.885643, 1782.864308, 1414.161257, 3723.629772, 3747.076592, 2005.919344, 4198.569251, 2228.522959, 3322.115942, 4274.324792, 720.9785449, 2874.651764, 2287.228752, 5654.858696, 1247.806111, 1247.806111, 2547.326207, 2608.716056, 1079.846532), Treatment = structure(c(2L, 3L, 4L, 5L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 1L), .Label = c("C", "1c_2d", "3c_2d", "9c_2d", "1c_7d"), class = "factor"), Damage_cm = c(0.4955, 1.516, 4.409, 3.2665, 0.491, 2.3035, 3.51, 1.8115, 0, 0.4435, 1.573, 1.8595, 0, 0.142, 2.171, 4.023, 4.9835, 0, 0.6925, 1.989, 5.683, 3.547, 0, 0.756, 2.129, 9.437, 3.211, 0, 0.578, 2.966, 4.7245, 1.8185, 0, 1.0475, 1.62, 5.568, 9.7455, 0, 0.8295, 2.411, 7.272, 4.516, 0, 0.4035, 2.974, 8.043, 4.809, 0, 0.6965, 1.313, 5.681, 3.474, 0, 0.5895, 2.559, 0)), .Names = c("Gossypol", "Treatment", "Damage_cm"), row.names = c(NA, -56L), class = "data.frame")
Y A B
1303.4450 2796.0385 0.4939
residual sum-of-squares: 1.03e+08
Number of iterations to convergence: 2
Achieved convergence tolerance: 4.936e-06
如果你喜欢ggplot2:
ggplot(data, aes(x = Damage_cm, y = Gossypol)) +
geom_point() +
geom_smooth(method = "nls",
formula = y ~ Y + A * (1 - B^x),
start = c(Y=1000, A=3000, B=0.5), se = F)
尽管恐怕必须在 ggplot 之外模拟标准错误。