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我怎样才能有效地纠正记录器在R中的季节性漂移？

来源：花匠小妙招时间：2025-09-10 02:58

问我怎样才能有效地纠正记录器在R中的季节性漂移？
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提问于 2022-11-03 01:06:24

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我在水中安装了一组CO2记录器，在开放的水季每小时记录一次CO2。在安装之前和之后，我已经对3种不同浓度的CO2的伐木器进行了描述。

I假设季节性误差漂移为线性，假设我的特征点之间的误差为线性

。

我的脚本基于一个for循环，它遍历每个时间戳并纠正值，这是可行的，但不幸的是速度不够快。我知道这可以在一秒钟内完成，但我不知道如何做到。我寻求一些建议，如果有人能告诉我怎么做，我将不胜感激。

基于basic的可复制示例：

代码语言：javascript

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start <- as.POSIXct("2022-08-01 00:00:00")#time when logger is installed stop <- as.POSIXct("2022-09-01 00:00:00")#time when retrieved dt <- seq.POSIXt(start,stop,by=3600)#generate datetime column, measured hourly #generate a bunch of values within my measured range co2 <- round(rnorm(length(dt),mean=600,sd=100)) #generate dummy dataframe dummy <- data.frame(dt,co2) #actual values used in characterization actual <- c(0,400,1000) #measured in the container by the instruments being characterized measured.pre <- c(105,520,1150) measured.post <- c(115,585,1250) diff.pre <- measured.pre-actual#diff at precharacterization diff.post <- measured.post-actual#diff at post #linear interpolation of how deviance from actual values change throughout the season #I assume that the temporal drift is linear diff.0 <- seq(diff.pre[1],diff.post[1],length.out=length(dummy$dt)) diff.400 <- seq(diff.pre[2],diff.post[2],length.out = length(dummy$dt)) diff.1000 <- seq(diff.pre[3],diff.post[3],length.out = length(dummy$dt)) #creates a data frame with the assumed drift at each increment throughout the season dummy <- data.frame(dummy,diff.0,diff.400,diff.1000) #this loop makes a 3-point calibration at each day in the dummy data set co2.corrected <- vector() for(i in 1:nrow(dummy)){ print(paste0("row: ",i))#to show the progress of the loop diff.0 <- dummy$diff.0[i]#get the differences at characterization increments diff.400 <- dummy$diff.400[i] diff.1000 <- dummy$diff.1000[i] #values below are only used for encompassing the range of measured values in the characterization #this is based on the interpolated difference at the given time point and the known concentrations used measured.0 <- diff.0+0 measured.400 <- diff.400+400 measured.1000 <- diff.1000+1000 #linear difference between calibration at 0 and 400 seg1 <- seq(diff.0,diff.400,length.out=measured.400-measured.0) #linear difference between calibration at 400 and 1000 seg2 <- seq(diff.400,diff.1000,length.out=measured.1000-measured.400) #bind them together to get one vector correction.ppm <- c(seg1,seg2) #the complete range of measured co2 in the characterization. #in reality it can not be below 0 and thus it can not be below the minimum measured in the range measured.co2.range <- round(seq(measured.0,measured.1000,length.out=length(correction.ppm))) #generate a table from which we can characterize the measured values from correction.table <- data.frame(measured.co2.range,correction.ppm) co2 <- dummy$co2[i] #measured co2 at the current row #find the measured value in the table and extract the difference diff <- correction.table$correction.ppm[match(co2,correction.table$measured.co2.range)] #correct the value and save it to vector co2.corrected[i] <- co2-diff } #generate column with calibrated values dummy$co2.corrected <- co2.corrected

Stack Overflow用户

回答已采纳

发布于 2022-11-04 07:24:16

这是我在查看代码之后所理解的。您有一系列的CO2浓度读数，但是它们需要根据时间序列开始和时间序列结束时的特性测量进行修正。这两套特征测量都使用了三种已知浓度: 0、400和1000。

您的代码似乎试图应用双线性插值(随时间和浓度)应用所需的校正。这很容易矢量化：

代码语言：javascript

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set.seed(1) start <- as.POSIXct("2022-08-01 00:00:00")#time when logger is installed stop <- as.POSIXct("2022-09-01 00:00:00")#time when retrieved dt <- seq.POSIXt(start,stop,by=3600)#generate datetime column, measured hourly #generate a bunch of values within my measured range co2 <- round(rnorm(length(dt),mean=600,sd=100)) #actual values used in characterization actual <- c(0,400,1000) #measured in the container by the instruments being characterized measured.pre <- c(105,520,1150) measured.post <- c(115,585,1250) # interpolate the reference concentrations over time cref <- mapply(seq, measured.pre, measured.post, length.out = length(dt)) #generate dummy dataframe with corrected values dummy <- data.frame( dt, co2, co2.corrected = ifelse( co2 < cref[,2], actual[1] + (co2 - cref[,1])*(actual[2] - actual[1])/(cref[,2] - cref[,1]), actual[2] + (co2 - cref[,2])*(actual[3] - actual[2])/(cref[,3] - cref[,2]) ) ) head(dummy) #> dt co2 co2.corrected #> 1 2022-08-01 00:00:00 537 416.1905 #> 2 2022-08-01 01:00:00 618 493.2432 #> 3 2022-08-01 02:00:00 516 395.9776 #> 4 2022-08-01 03:00:00 760 628.2707 #> 5 2022-08-01 04:00:00 633 507.2542 #> 6 2022-08-01 05:00:00 518 397.6533

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Stack Overflow用户

发布于 2022-11-04 04:54:40

我不知道你在计算什么(我觉得这样做可以不同)，但你可以通过以下方式提高速度：

remove print，这需要很多时间在循环删除data.frame创建在每次迭代，这是缓慢的，这里不需要

。

这个循环应该更快：

代码语言：javascript

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for(i in 1:nrow(dummy)){ diff.0 <- dummy$diff.0[i] diff.400 <- dummy$diff.400[i] diff.1000 <- dummy$diff.1000[i] measured.0 <- diff.0+0 measured.400 <- diff.400+400 measured.1000 <- diff.1000+1000 seg1 <- seq(diff.0,diff.400,length.out=measured.400-measured.0) seg2 <- seq(diff.400,diff.1000,length.out=measured.1000-measured.400) correction.ppm <- c(seg1,seg2) s <- seq(measured.0,measured.1000,length.out=length(correction.ppm)) measured.co2.range <- round(s) co2 <- dummy$co2[i] diff <- correction.ppm[match(co2, measured.co2.range)] co2.corrected[i] <- co2-diff }

附注：现在，我测试中最慢的部分是round(s)。也许可以删除或者重写..。

票数 1

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原文链接：

https://stackoverflow.com/questions/74300448

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