Format and export the output of Mann-Kendall test in R to excel from Rstudio

Format and export the output of Mann-Kendall test in R to excel from Rstudio

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Format and export the output of Mann-Kendall test in R to excel from Rstudio
Tag : r , By : user98986
Date : January 11 2021, 05:14 PM

may help you . I've just started using R and would like to use the trend package to perform Mann-Kendall tests on yearly groundwater level data of multiple wells. Here is a sample from my dataframe pri.csv: , An option would be tidy from broom
map_dfr(results_list[!bad], tidy)
out <- do.call(rbind, lapply(results_list[!bad], function(x) 
  cbind(data.frame(p.value =  x$p.value, statistic  = x$statistic), 
row.names(out) <- NULL

#    p.value statistic  S     varS        tau
#1 1.0000000  0.000000  1 3.666667  0.3333333
#2 1.0000000  0.000000  1 3.666667  0.3333333
#3 0.2962699 -1.044466 -3 3.666667 -1.0000000
#4 1.0000000  0.000000 -1 3.666667 -0.3333333

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Split vector and apply Mann Kendall test

Tag : r , By : user186435
Date : March 29 2020, 07:55 AM
I hope this helps you . I have a long vector of financial time series information, price and volume. I expand the price vector based on the volume vector so I essentially get a price per share traded, and so only one vector is analyzed. I would like to split this vector into lengths of 100 (could be any number) each, and apply the Mann-Kendall trend analysis on each of these new vectors. Below is the code I have for this. , Using lapply for example, here I do it only for the first 5 elements.
WARNING: Error exit, tauk2. IFAULT =  12
WARNING: Error exit, tauk2. IFAULT =  12
WARNING: Error exit, tauk2. IFAULT =  12
WARNING: Error exit, tauk2. IFAULT =  12
WARNING: Error exit, tauk2. IFAULT =  12
tau = 1, 2-sided pvalue =1

tau = 1, 2-sided pvalue =1

tau = 1, 2-sided pvalue =1

tau = 1, 2-sided pvalue =1

tau = 1, 2-sided pvalue =1

  tau sl S D varS
1   1  1 0 0    0
2   1  1 0 0    0
3   1  1 0 0    0
4   1  1 0 0    0
5   1  1 0 0    0

Autocorrelation and Mann-Kendall Trend test for multiple timeseries

Tag : r , By : user183442
Date : March 29 2020, 07:55 AM
I hope this helps you . My above comments aside I think this will get you what you're looking for:
stationList <- unique(df$stn_num)
resultsList <- vector("list", length(stationList))
for(i in stationList){
  tempDF <- df[df$stn_num == i, ]
  t <- dim(c$acf)
  tempDF$prewhit1<-c$acf[[t[1], t[2], t[3]]]*tempDF$value
  prewhitseries<-data.frame(with(tempDF, (tempDF$value[-1] - prewhit1[-length(prewhit1)])))
  resultsList[[grep(i, stationList)]] <- MannKendall(autocordata[,5])
names(resultsList) <- stationList

Speed-up a parallel process calculating a mann-kendall test over a huge dataset in R

Tag : r , By : oiyto
Date : March 29 2020, 07:55 AM
help you fix your problem You note that you have already fixed your problem. Is obtainable using one of the following steps:
1: Copy the necessary objects to the foreach loops using .packages and .export. This ensures that each instance will not clash when trying to access the same memory.
data<- data.frame(cbind(runif(10000,-180,180), runif(10000,-90,90))
                  , replicate(1200, runif(10000,0,150)))

coords<-data[,1:2] #get the coordinates out of the initial dataset
data_t<- as.data.frame(t(data[,3:1202])) #each column is now the time series associated to a point
data_t$month<-rep(seq(1,12,1),100) # month index as last column of the data frame
# start the parallel processing

cores=detectCores() #count cores
cl <- makeCluster(cores[1]-1) #take all the cores minus 1 not to overload the pc

#user  system elapsed 
#17.80   35.12   98.72
  test <- data_t[,parLapply(cl, 
                            .SD, function(x){
                            ), by = month] #Perform the calculations across each month
  #create a column that indicates what each row is measuring
  rows <- rep(c("p.value","statistic.z","estimates.S","estimates.var","estimates.tau"),12)

  final_tests <- dcast( #Cast the melted structure to a nice form
                      melt(cbind(test,rowname = rows), #Melt the data for a better structure
                        id.vars = c("rowname","month"), #Grouping variables
                        measure.vars = paste0("V",seq.int(1,10000))), #variable names
                      month + variable ~ rowname, #LHS groups the data along rows, RHS decides the value columns
                      value.var = "value", #Which column contain values? 
                      drop = TRUE) #should we drop unused columns? (doesnt matter here)
  #rename the columns as desired
  names(final_tests) <- c("month","variable","S","tau","var","p.value","z_stat")
  #finally add the coordinates
  final_tests <- cbind(final_form,coords) 

How do you perform Mann-kendall test on multiple stations using loops?

Tag : r , By : TheMoo
Date : March 29 2020, 07:55 AM
To fix the issue you can do Something like this should do. split() splits along the well column, creating a list with a vector for each well. Only vectors of length 3 or more are kept. MannKendall() is then run on each of the remaining vectors using lapply()

tt <- read.table(text="
well    year    month   value
684     1994    Jan     8.53
684     1995    Jan     8.74
684     1996    Jan     8.88
684     1997    Jan     8.24
1001    2000    Jan     9.1
1001    2001    Jan     9.2
1001    2002    Jan     9.54
1001    2003    Jan     9.68
2003    1981    Jan     55.2
2003    1982    Jan     55.8
2003    1983    Jan     56.4
2003    1984    Jan     53.2
2004    1984    Jan     53.2", header=TRUE)

tt.wells <- split(tt$value, tt$well)
tt.wells <- tt.wells[lengths(tt.wells) >= 3]

lapply(tt.wells, MannKendall)

# $`684`
# tau = 0, 2-sided pvalue =1

# $`1001`
# tau = 1, 2-sided pvalue =0.089429

# $`2003`
# tau = 0, 2-sided pvalue =1

How to perform mann kendall trend test on multiple levels

Tag : r , By : Jeskl
Date : March 29 2020, 07:55 AM
it should still fix some issue I am working with station precipitation data. Each station has precipitation data for 60 years and there are 30 stations. I want to perform a Mann Kendall trend test on each station to see if there is a significant trend for precipitation. , @A.Suliman, you're right.
This seems to work:

mk<-df %>%
  as.data.frame() %>% 
  group_by(ID) %>%
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