我试图用单指数“衰减”来拟合下面数据中的一些曲线。图形显示并不像提取时间常数那么重要。 y 轴为 pA，x 轴为时间(以秒为单位)。
dput(stackover_data)
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191000L, 192000L, 193000L, 194000L, 195000L, 196000L, 197000L,
198000L, 199000L, 200000L), class = "data.frame")
我尝试过 lm(y~x) 但似乎没有得到正确的答案(在 Igor 中验证了正确的答案)，显然这是因为它是线性模型而不是指数模型。欢迎任何建议。我正在为此苦苦挣扎!
谢谢大家!
最佳答案
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))
关于r - 在 R 中拟合指数衰减并提取时间常数值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34595190/