拟合指数衰减并拉出R中的时间常数值

Question

我试图通过单指数＆＃34;衰减＆＃34;来拟合下面数据中的一些曲线。图形显示并不像拉出时间常数那么重要。 y轴是pA，x是以秒为单位的时间。

dput(stackover_data)
structure(list(Time = c(0.09990001, 0.19990001, 0.29990001, 0.39990001, 
0.49990001, 0.59990001, 0.69990001, 0.79990001, 0.89990001, 0.99990001, 
1.09990001, 1.19990001, 1.29990001, 1.39990001, 1.49990001, 1.59990001, 
1.69990001, 1.79990001, 1.89990001, 1.99990001, 2.09990001, 2.19990001, 
2.29990001, 2.39990001, 2.49990001, 2.59990001, 2.69990001, 2.79990001, 
2.89990001, 2.99990001, 3.09990001, 3.19990001, 3.29990001, 3.39990001, 
3.49990001, 3.59990001, 3.69990001, 3.79990001, 3.89990001, 3.99990001, 
4.09990001, 4.19990001, 4.29990001, 4.39990001, 4.49990001, 4.59990001, 
4.69990001, 4.79990001, 4.89990001, 4.99990001, 5.09990001, 5.19990001, 
5.29990001, 5.39990001, 5.49990001, 5.59990001, 5.69990001, 5.79990001, 
5.89990001, 5.99990001, 6.09990001, 6.19990001, 6.29990001, 6.39990001, 
6.49990001, 6.59990001, 6.69990001, 6.79990001, 6.89990001, 6.99990001, 
7.09990001, 7.19990001, 7.29990001, 7.39990001, 7.49990001, 7.59990001, 
7.69990001, 7.79990001, 7.89990001, 7.99990001, 8.09990001, 8.19990001, 
8.29990001, 8.39990001, 8.49990001, 8.59990001, 8.69990001, 8.79990001, 
8.89990001, 8.99990001, 9.09990001, 9.19990001, 9.29990001, 9.39990001, 
9.49990001, 9.59990001, 9.69990001, 9.79990001, 9.89990001, 9.99990001, 
10.09990001, 10.19990001, 10.29990001, 10.39990001, 10.49990001, 
10.59990001, 10.69990001, 10.79990001, 10.89990001, 10.99990001, 
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11.59990001, 11.69990001, 11.79990001, 11.89990001, 11.99990001, 
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12.59990001, 12.69990001, 12.79990001, 12.89990001, 12.99990001, 
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17.59990001, 17.69990001, 17.79990001, 17.89990001, 17.99990001, 
18.09990001, 18.19990001, 18.29990001, 18.39990001, 18.49990001, 
18.59990001, 18.69990001, 18.79990001, 18.89990001, 18.99990001, 
19.09990001, 19.19990001, 19.29990001, 19.39990001, 19.49990001, 
19.59990001, 19.69990001, 19.79990001, 19.89990001, 19.99990001
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184000L, 185000L, 186000L, 187000L, 188000L, 189000L, 190000L, 
191000L, 192000L, 193000L, 194000L, 195000L, 196000L, 197000L, 
198000L, 199000L, 200000L), class = "data.frame")

我尝试过做lm（y~x），但它似乎没有得到正确的答案（在Igor中验证了正确的答案），显然这是因为它是一个线性模型，而不是一个指数。欢迎任何所有建议。我在这方面苦苦挣扎！

谢谢大家！

Answer 1

names(DF) <- make.names(names(DF))
plot(Trace.1 ~ Time, data = DF)
#remove the initial values that clearly don't follow the model
DF1 <- DF[-seq_len(which((diff(DF$Trace.1) < -1e3))),]

plot(Trace.1 ~ Time, data = DF1)
fit <- nls(Trace.1 ~ SSasymp(Time, Asym, R0, lrc), data = DF1)
summary(fit)
coef(fit)
help("SSasymp") #for an explanation of the parameters
lines(DF1$Time, predict(fit))