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its super hard to explain in text but ill try. it works up until where it says 'good until here'. so i have a data set with years that range from 5-9 year intervals associated with a number of scales. the first for loop on line 31 makes it so all of the intervals are one years and now i am trying to assign the number of scales for all of the years within the interval. I cannot get the nested for loops to work for that part

I have tried multiple variations of the for loops but am unsure how to do it in r. I have a successful matlab code that worked

remove(years)
remove(yearly_data, year_new_data)

num_intervals = nrow(Sard1301)                     #counts the numbers of intervals in the dataset
number_intervals = data.frame(interval = 1:num_intervals)

first_year = (Sard1301[num_intervals,1])           #Specifies the youngest/earliest year in the dataset (on the bottom of the dataset)

last_year_first_interval = (Sard1301[1, 1])        #Specified the first interval of the oldest year (top of the dataset)

end_samp_interval = Sard1301[1,1]-Sard1301[2,1]    #Specifies the interval between the most recent and second most recent intervals

last_year = round(Sard1301[1,1]+end_samp_interval)        #Specifies the second interval in the last sampling interval (i.e. the most recent year plus the interval)

total_years = (last_year-first_year)
num_years = as.numeric(last_year-first_year)            #Specifies the total number of years and converts it to a numeric value (from a data.frame)


yearly_data = round(matrix(data = 0, num_years, ncol=2))     #This creates a blank matrix for the data to be added to
years=yearly_data[ ,1]

##################################
#yearly_data1 = rbind(yearly_data[1,1], last_year)     ##########
#yearly_data[1,1] = as.numeric(last_year)                    #This makes the first row in the first column the most recent year (second part of the interval)
years[1] = as.numeric(round(last_year))  

#for (i in total_years) {
#yearly_data[i+1,1]=yearly_data[i,1]-1  
#}

num_steps=length(years)-1
counter=1
for(i in 1:num_steps) {
  years[counter+1]= years[counter]-1
  counter=counter+1
}
yearly_data[,1]=years

#yearly_data1 = rbind(yearly_data[1,1], last_year)    #good up until here


num.int = length(num_intervals)

  for(j in 1:num_int){
    for(k in 1:num_steps){
      year_new_data = yearly_data#[k, 1]

      if (j==1){
        if(year_new_data>Sard1301){#[j,1]) {
          yearly_data=Sard1301#[j,2]
          }}
        else((j>1) & (j<num_intervals-1))
          if(year_new_data>=round(Sard1301) & year_new_data<round(Sard1301)) {
            yearly_data=Sard1301
        }
       if(j==num.int) {
           if(year_new_data<=Sard1301) {
             yearly_data[2] = Sard1301[2]
           }}
     }
 }

my data is as follows column 1: c(1922, 1913, 1905, 1896, 1888, 1879, 1871, 1862, 1855, 1847, 1840) column 2: c(1, 6, 3, 0, 6, 0,2,0,8, 5, 1.333)

I hope to get a matrix reading (the first number is column 1 - second number s column 2) 1922 - 1, 1921, 1, 1920 - 1, 1919 - 1, 1918 -1, 1917 - 1, 1916 - 1, 1915 - 1, 1914 - 1, 1913 - 6, 1912 - 6, 1911 - 6, 1910 - 6, 1909 - 6, 1908 - 6, 1907 - 6, 1906 - 6, 1905 - 3, 1904 - 3, etc.

I expect to get a matrix with 1 year intervals that have the same number of scales for each of the intervals. i.e 1922-1915 = 6 scales---> output = 1922-6 scales, 1921 - 6 scales, 1920 - 6 scales, etc.

Alex Filardo
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  • Do you need help writing for loops, or do you need help solving your problem? Loops have their place in R code, but more often than not the optimal solution is either vectorized or uses the implicit looping abstracted in functions in the `apply` family. – John Coleman Jul 14 '19 at 19:55
  • It'd be easier to help if you just share a small subset of your data, briefly describe the calculation, and show expected output. – Shree Jul 14 '19 at 19:55
  • John- what does the apply function do? – Alex Filardo Jul 14 '19 at 20:06
  • Shree- I added what my dataset looks like and what i expect to see – Alex Filardo Jul 14 '19 at 20:34
  • See [How to make a great R reproducible example](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) and update your question accordingly. It is very hard to understand in it's current form. – Shree Jul 15 '19 at 02:07

1 Answers1

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I'm not sure to understand your problem but you can try this:

# 2 columns
x1<-c(1922, 1913, 1905, 1896, 1888, 1879, 1871, 1862, 1855, 1847, 1840)
x2<-c(1, 6, 3, 0, 6, 0,2,0,8, 5, 1.333) 



# order data
a1<-order(x1)
x1<-x1[a1]
x2<-x2[a1]

# create two vectors
sc<-fromto<-min(x1):max(x1)
for(i in 1:length(x1)){
  if(i == 1){
   d<-which(fromto<=x1[i])
   sc[d]<-x2[i]
  }else{
    d<-which(fromto<=x1[i] & fromto>=x1[i-1])
    sc[d]<-x2[i]
  }
}
finalmatrix<-cbind(fromto,sc)

# matrix
print(finalmatrix)
GreenLantern
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