确保R中的时间数据密度

Question

ISSUE ---------

I have thousands of time series files (.csv) that contain intermittent data spanning for between 20-50 years (see df). Each file contains the date_time and a metric (temperature). The data is hourly and where no measurement exists there is an 'NA'.

>df
date_time         temp 
01/05/1943 11:00  5.2
01/05/1943 12:00  5.2
01/05/1943 13:00  5.8
01/05/1943 14:00   NA
01/05/1943 15:00   NA
01/05/1943 16:00  5.8
01/05/1943 17:00  5.8
01/05/1943 18:00  6.3

I need to check these files to see if they have sufficient data density. Ie that the ratio of NA's to data values is not too high. To do this I have 3 criteria that must be checked for each file:

1. Ensure that no more than 10% of the hours in a day are NA's
2. Ensure that no more than 10% of the days in a month are NA's
3. Ensure that there are 3 continuous years of data with valid days and months.

Each criterion must be fulfilled sequentially and if the file does not meet the requirements then I must create a data frame (or any list) of the files that do not meet the criteria.

QUESTION--------

I wanted to ask the community how to go about this. I have considered the value of nested if loops, along with using sqldf, plyr, aggregate or even dplyr. But I do not know the simplest way to achieve this. Any example code or suggestions would be very much appreciated.

Answer 1

I think this will work for you. These will check every hour for NA's in the next day, month or 3 year period. Not tested because I don't care to make up data to test it. These functions should spit out the number of NA's in the respective time period. So for function checkdays if it returns a value greater than 2.4 then according to your 10% rule you'd have a problem. For months 72 and for 3 year periods you're hoping for values less than 2628. Again please check these functions. By the way the functions assume your NA data is in column 2. Cheers.

checkdays <- function(data){
countNA=NULL
for(i in 1:(length(data[,2])-23)){
nadata=data[i:(i+23),2]
countNA[i]=length(nadata[is.na(nadata)])}
return(countNA)
}

checkmonth <- function(data){
countNA=NULL
for(i in 1:(length(data[,2])-719)){
nadata=data[i:(i+719),2]
countNA[i]=length(nadata[is.na(nadata)])}
return(countNA)
}

check3years <- function(data){
countNA=NULL
for(i in 1:(length(data[,2])-26279)){
nadata=data[i:(i+26279),2]
countNA[i]=length(nadata[is.na(nadata)])}
return(countNA)
}

So I ended up testing these. They work for me. Here are system times for a dataset a year long. So I don't think you'll have problems.

> system.time(checkdays(RM_W1))
   user  system elapsed 
   0.38    0.00    0.37 
> system.time(checkmonth(RM_W1))
   user  system elapsed 
   0.62    0.00    0.62

Optimization: I took the time to run these functions with the data you posted above and it wasn't good. For loops are dangerous because they work well for small data sets but slow down exponentially as datasets get larger, that is if they're not constructed properly. I cannot report system times for the functions above with your data (it never finished) but I waited about 30 minutes. After reading this awesome post Speed up the loop operation in R I rewrote the functions to be much faster. By minimising the amount of things that happen in the loop and pre-allocating memory you can really speed things up. You need to call the function like checkdays(df[,2]) but its faster this way.

checkdays <- function(data){
countNA=numeric(length(data)-23)
for(i in 1:(length(data)-23)){
nadata=data[i:(i+23)]
countNA[i]=length(nadata[is.na(nadata)])}
return(countNA)
}
> system.time(checkdays(df[,2]))
   user  system elapsed 
   4.41    0.00    4.41 

I believe this should be sufficient for your needs. In regards to leap years you should be able to modify the optimized function as I mentioned in the comments. However make sure you specify a leap year dataset as second dataset rather than a second column.