I was trying to calculate mean and SD per month of a variable from an environmental dataset (.nc file of Sea surface temp/day during 2 years) and the loop I used gives me the following error
Error in h(simpleError(msg, call)) :
error in evaluating the argument 'x' in selecting a method for function 'mean': recursive indexing failed at level 2
I have no idea where my error could be but if you are curious I was using the following .nc dataset just for SST for 2018-2019 from copernicus sstdata
Here is the script I used so far and the packages I'm using:
# Load required libraries (install the required libraries using the Packages tab, if necessary)
library(raster)
library(ncdf4)
#Opern the .nc file with the environmental data
ENV = nc_open("SST.nc")
ENV
#create an index of the month for every (daily) capture from 2018 to 2019 (in this dataset)
m_index = c()
for (y in 2018:2019) {
# if bisestile year (do not apply for this data but in case a larger year set is used)
if (y%%4==0) { m_index = c(m_index, rep(1:12 , times = c(31,29,31,30,31,30,31,31,30,31,30,31))) }
# if non-bisestile year
else { m_index = c(m_index, rep(1:12 , times = c(31,28,31,30,31,30,31,31,30,31,30,31))) }
}
length(m_index) # expected length (730)
table(m_index) # expected number of records assigned to each of the twelve months
# computing of monthly mean and standard deviation.
# We first create two empty raster stack...
SST_MM = stack() # this stack will contain the twelve average SST (one per month)
SST_MSD = stack() # this stack will contain the twelve SST st. dev. (one per month)
# We run the following loop (this can take a while)
for (m in 1:12) { # for every month
print(m) # print current month to track the progress of the loop...
sstMean = mean(ENV[[which(m_index==m)]], na.rm=T) # calculate the mean SST for all the records of the current month
sstSd = calc(ENV[[which(m_index==m)]], sd, na.rm=T) # calculate the st. dev. of SST for all the records of the current month
# add the monthly records to the stacks
SST_MM = stack(SST_MM, sstMean)
SST_MSD = stack(SST_MSD, sstSd)
}
And as mentioned, the output of the loop including the error:
SST_MM = stack() # this stack will contain the twelve average SST (one per month)
> SST_MSD = stack() # this stack will contain the twelve SST st. dev. (one per month)
> for (m in 1:12) { # for every month
+
+ print(m) # print current month to track the progress of the loop...
+
+ sstMean = mean(ENV[[which(m_index==m)]], na.rm=T) # calculate the mean SST for all the records of the current month
+ sstSd = calc(ENV[[which(m_index==m)]], sd, na.rm=T) # calculate the st. dev. of SST for all the records of the current month
+
+ # add the monthly records to the stacks
+
+ SST_MM = stack(SST_MM, sstMean)
+ SST_MSD = stack(SST_MSD, sstSd)
+
+ }
[1] 1
**Error in h(simpleError(msg, call)) :
error in evaluating the argument 'x' in selecting a method for function 'mean': recursive indexing failed at level 2**
It seems that you make things too complicated. I think the easiest way to do this is with terra::tapp like this:
library(terra)
x <- rast("SST.nc")
xmn <- tapp(x, "yearmonths", mean)
xsd <- tapp(x, "yearmonths", sd)
or more manually:
library(terra)
x <- rast("SST.nc")
y <- format(time(x),"%Y")
m <- format(time(x),"%m")
ym <- paste0(y, "_", m)
r <- tapp(x, ym, mean)
Related
I have a dataset (photos) with two column (one for the images ID and one with the images url).
After I plugged-in my Google Cloud Platform credentials, I ran the following code to generate keywords:
require(RoogleVision)
#add extra columns for 10 x 3 rows of data (keyword, probability score, and topicality score)
photos[,3:32] <- NA
##Loop##
for(i in 1:length(photos$url)){
te <- getGoogleVisionResponse(photos$url[i], feature="LABEL_DETECTION", numResults = 10)
#If not successful, return NA matrix
if(length(te)==1){ te <- matrix(NA, 10,4)}
if (is.null(te)){ te <- matrix(NA, 10,4)}
te <- te[,2:4]
#if successful but no. of keywords <10, put NAs in remaining rows
if(length(te[,1])<10){
te[(length(te[,1])+1):10,] <- NA}
#Append all data!
photos[i, 3:12] <- te[,1] #keywords
photos[i, 13:22] <- te[,2] #probability scores
photos[i, 23:32] <- te[,3] #topicality scores
cat("<row", i, "/", length(photos[,1]), "> ")
}
I got the following error
Error: Assigned data `te[, 1]` must be compatible with row subscript `i`.
x 1 row must be assigned.
x Assigned data has 10 rows.
ℹ Row updates require a list value. Do you need `list()` or `as.list()`?
Run `rlang::last_error()` to see where the error occurred.
Any help will be very much appreciated!
I have 5 variables for one questionnaire about social support. I want to define the group with low vs. high support. According to the authors low support is defined as a sum score <= 18 AND two items scoring <= 3.
It would be great to get a dummy variable which shows which people are low vs high in support.
How can I do this in the syntax?
Thanks ;)
Assuming your variables are named Var1, Var2 .... Var5, and that they are consecutive in the dataset, this should work:
recode Var1 to Var5 (1 2 3=1)(4 thr hi=0) into L1 to L5.
compute LowSupport = sum(Var1 to Var5) <= 18 and sum(L1 to L5)>=2.
execute.
New variable LowSupport will have value 1 for rows that have the parameters you defined and 0 for other rows.
Note: If your variables are not consecutive you'll have to list all of them instead of using Var1 to var5.
I'm looking to generate a table of random values, but want to make sure that none of those values are repeated within the table.
So my basic table generation looks like this:
numbers = {}
for i = 1, 5 do
table.insert(numbers, math.random(20))
end
So that will work in populating a table with 5 random values between 1-20. However, it's the making sure none of those values repeat is where I'm stuck.
One approach would be to shuffle an array of numbers and then take the first n numbers. The wrong way to go about shuffling an array is to maintain a list of previously generated random numbers, checking against that with each newly generated random number before adding it to the final array. Such a solution is O(n^2) in time complexity when iterating over the array during the check; this will be painful for large arrays, or for small arrays when many must be created. Lua has constant time array access since tables are really hash tables, so you could get away with this, except: sometimes many random numbers will need to be tried before a suitable one (that has not already been used) is found. This can be a real problem near the end of an array of many random numbers, i.e., when you want 1000 random numbers and have filled all but the last slot, how many random tries (and how many iterations of the 999 numbers already selected) will it take to find the only number (42, of course) that is still available?
The right way to go about shuffling is to use a shuffling algorithm. The Fisher-Yates shuffle is a common solution to this problem. The idea is that you start at one end of an array, and swap each element with a random element that occurs later in the list until the entire array has been shuffled. This solution is O(n) in time complexity, thus much less wasteful of computational resources.
Here is an implementation in Lua:
function shuffle (arr)
for i = 1, #arr - 1 do
local j = math.random(i, #arr)
arr[i], arr[j] = arr[j], arr[i]
end
end
Testing in the REPL:
> t = { 1, 2, 3, 4, 5, 6 }
> table.inspect(t)
1 = 1
2 = 2
3 = 3
4 = 4
5 = 5
6 = 6
> shuffle(t)
> table.inspect(t)
1 = 4
2 = 5
3 = 1
4 = 6
5 = 2
6 = 3
This can easily be extended to create lists of random numbers:
function shuffled_numbers (n)
local numbers = {}
for i = 1, n do
numbers[i] = i
end
shuffle(numbers)
return numbers
end
REPL interaction:
> s = shuffled_numbers(10)
> table.inspect(s)
1 = 9
2 = 5
3 = 3
4 = 4
5 = 7
6 = 6
7 = 2
8 = 10
9 = 8
10 = 1
If you want to see what is happening during the shuffle, add a print statement in the shuffle function:
function shuffle (arr)
for i = 1, #arr - 1 do
local j = math.random(i, #arr)
print(string.format("%d (%d) <--> %d (select %d)", i, arr[i], j, arr[j]))
arr[i], arr[j] = arr[j], arr[i]
end
end
Now you can see the swaps as they occur if you recall that in the above implementation of shuffled_numbers the array { 1, 2, ..., n } is the starting point of the shuffle. Note that sometimes a number is swapped with itself, which is to say that the number in the current unselected position is a valid choice, too. Also note that the last number is automatically the correct selection, since it is the only number that has not yet been randomly selected:
> s = shuffled_numbers(10)
1 (1) <--> 5 (select 5)
2 (2) <--> 10 (select 10)
3 (3) <--> 5 (select 1)
4 (4) <--> 9 (select 9)
5 (3) <--> 8 (select 8)
6 (6) <--> 9 (select 4)
7 (7) <--> 8 (select 3)
8 (7) <--> 10 (select 2)
9 (6) <--> 9 (select 6)
> table.inspect(s)
1 = 5
2 = 10
3 = 1
4 = 9
5 = 8
6 = 4
7 = 3
8 = 2
9 = 6
10 = 7
Obtaining a selection of 5 random numbers between 1 and 20 is easy enough to accomplish using the shuffle function; one of the virtues of this approach is that the shuffling operation has been abstracted to an O(n) procedure which can shuffle any array, numeric or otherwise. The function that calls shuffle is responsible for supplying the input and returning the results.
A simple solution for more flexibility in the range of random numbers returned:
-- Take the first N numbers from a shuffled range [A, B].
function shuffled_range_take (n, a, b)
local numbers = {}
for i = a, b do
numbers[i] = i
end
shuffle(numbers)
return { table.unpack(numbers, 1, n) }
-- table.unpack won't work for very large ranges, e.g. [1, 1000000]
-- You could instead use this for arbitrarily large ranges:
-- local take = {}
-- for i= 1, n do
-- take[i] = numbers[i]
-- end
-- return take
end
REPL interaction creating a table containing 5 random values between 1 and 20:
> s = shuffled_range_take(5, 1, 20)
> table.inspect(s)
1 = 1
2 = 10
3 = 4
4 = 8
5 = 20
But, there is a disadvantage to the shuffle method in some circumstances. When the number of elements needed is small compared with the number of available elements, the above solution must shuffle a large array to obtain comparatively few random elements. The shuffle is O(n) in the number of elements available, while the memoization method is roughly O(n) in the number of elements chosen. A memoization method like that of #AlexanderMashin performs poorly when the goal is to create an array of 20 random numbers between 1 and 20, because the final numbers chosen may need to be chosen many times before suitable numbers are found. But when only 5 random numbers between 1 and 20 are needed, this problem with duplicate choices is less of an issue. This approach seems to perform better than the shuffle, up to about 10 numbers needed from 20 random numbers. When more than 10 numbers are needed from 20, the shuffle begins to perform better. This break-even point is different for larger numbers of elements to choose from; for 1000 available elements, parity is reached at about 700 chosen. When performance is critical, testing is the only way to determine the best solution.
numbers = {}
local i = 1;
while i<=5 do
n = 0
local rand = math.random(20)
for x=1,#numbers do
if numbers[x] == rand then
n = n + 1
end
end
if n == 0 then
table.insert(numbers, rand)
i = i + 1
end
n = 0
end
the method I used for this process was to use a for to scan each of the elements in the table and increase the variable n if one of them was equal to the random value given, so if x was different from 0, the value would not be inserted in the table and would not increment the variable i (I had to use the while to work with i)
if you want to print each of the elements in the table to check the values you can use this:
for i=1,#numbers do
print(numbers[i])
end
I suggest an alternative method based on the fact that it is easy to make sets in Lua: they are just tables with true values.
-- needed is how many random numbers in the table are needed,
-- maximum is the maximum value of a random non-negtive integer.
local function fill_table( needed, maximum )
math.randomseed ( os.time () ) -- reseed the random numbers generator
local numbers = {}
local used = {} -- which numbers are already used
for i = 1, needed do
local random
repeat
random = math.random( maximum )
until not used[random]
used[random] = true
numbers[i] = random
end
return numbers
end
Making a table with 20 keys (use for/do/end) and then do your desired times
rand_number=table.remove(tablename, math.random(1,#tablename))
EDIT: Corrected - See first comment
And rand_number never holds the same value. I use this as a simulation for a "Lottozahlengenerator" (german, sorry) or random video/music clips playing where duplicates are unwanted.
I'm trying to forecast hourly data for 30 days for a process.
I have used the following code:
# The packages required for projection are loaded
library("forecast")
library("zoo")
# Data Preparation steps
# There is an assumption that we have all the data for all
# the 24 hours of the month of May
time_index <- seq(from = as.POSIXct("2014-05-01 07:00"),
to = as.POSIXct("2014-05-31 18:00"), by = "hour")
value <- round(runif(n = length(time_index),100,500))
# Using zoo function , we merge data with the date and hour
# to create an extensible time series object
eventdata <- zoo(value, order.by = time_index)
# As forecast package requires all the objects to be time series objects,
# the below command is used
eventdata <- ts(value, order.by = time_index)
# For forecasting the values for the next 30 days, the below command is used
z<-hw(t,h=30)
plot(z)
I feel the output of this code, is not working fine.
The forecasted line looks wrong and the dates are not getting correctly projected on the chart.
I'm not sure the fault lies in the data preparation, or if the output is as expected. Any ideas?
I'm learning Lua from a book and this is the exact question I'm stuck on:
Given that you need to sum the numbers 1 through 100, write a loop to complete the operation.
I've tried various things, but my most recent attempt following:
n = 1
while (n < 100) do
n = n + 1
print (n)
end
As mentioned earlier, you need at least two variables: one to hold sum and second to count to 100.
Fixed steps calculations is better to do with for loop.
local sum = 0
for i = 1, 100 do
sum = sum + i
end
print(sum)
P.S. Where is the question? Add not only broken code, but some words about what is wrong with it please.
It looks like you need to do something like this:
local n = 1
local sum = 0
while (n <= 100) do
sum = sum + n
n = n + 1
end
print(sum)
It should help if you keep your sum and counter in separate variables.
You need another variable to hold the sum :)
I believe this should do it:
i=0
n=0
while i <= 100 do
n = i + n
i = i + 1
end
print(n)
Variables are visible after they their first assignment. So you need one variable declared outside the loop to hold the sum as it is updated inside the loop, like this:
n = 0
sum = 0
while (n < 100) do
n = n + 1 -- n variable output is 1,2,3,4,5,...100
sum = sum + n -- sum variable remembers its value from previous iteration
print (sum)
end
When you do sum = sum + n, the interpreter takes the current value of sum, adds n to it, and puts the result into sum. At next iteration, sum still has that most recent value. Compare, if you had done
while (n < 100) do
n = n + 1 -- n variable output is 1,2,3,4,5,...100
local sum = sum + n -- sum is "new" at every iteration so fails
print (sum)
end
This sum variable is local to loop so every time through loop, a new sum is created. Only problem is,
local sum = sum + n
that statement tries to get value of "sum" and add it to n, but sum is being created on that line so it doesn't exist yet so interpreter will throw error about attempt to do arithmetic on global "sum" (the sum that appears on right hand side is not know to compiler so it thinks it is a global since it hasn't created the local sum yet).
All previous answers ignore that sum can be calculated using a single equation;
Assume largest number is "N"
Sum of integers from 1 to N is; ( N x ( N + 1 )) / 2