R: Creating and Labelling Groups by Increments

Question

I am working with the R programming language.

I have the following data:

set.seed(123)
my_data = data.frame(var1 =  rnorm(100,100,100))
min = min(my_data$var1)
max = max(my_data$var)

Here is what I am trying to do:

Starting from the smallest value of var1, I would like to create a variable that groups values of var1 by some "fixed increment" (e.g. by 10) until the maximum value of var1 is reached
Then, I would then like to create another variable which labels each of these groups by the min/max value of that group

Here is my attempt to do this:

# create a vector of increments
breaks <- seq(min(my_data$var1), max(my_data$var1), by = 10)

# initialize new variables
my_data$class <- NA
my_data$label <- NA

# get the number of breaks
n <- length(breaks)

# Loop 
for (i in 1:(n - 1)) {
    # find which "class" (i.e. break) each value of var1 is located within
    indices <- which(my_data$var1 > breaks[i] & my_data$var1 <= breaks[i + 1])
    
    # make assignment
    my_data$class[indices] <- i
    
    # create labels
    my_data$label[indices] <- paste(breaks[i], breaks[i + 1])
}

The code seems to have run, but I am not sure if this is correct (I don't think I have done this correctly because I see some NA's).

Can someone please tell show me how to do this correctly?

Thanks!

@ akrun: thank you for your reply! I will try to incorporate this into my code! — stats_noob, Jan 08 '23 at 20:58
@ akrun: thank you for your reply! If you have time, can you please show me what you mean? thank you so much! — stats_noob, Jan 08 '23 at 21:00
Related to `cut` is `findInterval`; `ix = findInterval(my_data$var1, breaks)`; `data.frame(lo = breaks[ix], hi = breaks[ix + 1])`. — Henrik, Jan 08 '23 at 21:26
Can you try with the update in the post i.e. `breaks <- unique(c(seq(min, max, by = 10), max))` — akrun, Jan 08 '23 at 21:28
Somewhat related: [Get lower and upper bounds from `cut` as numeric values](https://stackoverflow.com/questions/32356108/get-lower-and-upper-bounds-from-cut-as-numeric-values). See also example in `?cut` - "one way to extract the breakpoints". — Henrik, Jan 08 '23 at 21:34

akrun · Accepted Answer · 2023-01-08T21:28:04.960

This could be done with a non-equi join

library(data.table)
my_data1 <- copy(my_data)
setDT(my_data1)[data.table(start = breaks, end = shift(breaks, 
   type = "lead", fill = last(breaks))),  c("indices", "label") := .(.GRP, paste(start, end)), 
    on = .(var1 > start, var1 <= end), by = .EACHI]

-output

> head(my_data1)
        var1 indices                             label
1:  43.95244      18 39.0831124359188 49.0831124359188
2:  76.98225      21 69.0831124359188 79.0831124359188
3: 255.87083      39 249.083112435919 259.083112435919
4: 107.05084      24 99.0831124359188 109.083112435919
5: 112.92877      25 109.083112435919 119.083112435919
6: 271.50650      41 269.083112435919 279.083112435919

compare it with OP's for loop

> head(my_data)
       var1 class                             label
1  43.95244    18 39.0831124359188 49.0831124359188
2  76.98225    21 69.0831124359188 79.0831124359188
3 255.87083    39 249.083112435919 259.083112435919
4 107.05084    24 99.0831124359188 109.083112435919
5 112.92877    25 109.083112435919 119.083112435919
6 271.50650    41 269.083112435919 279.083112435919

Regarding the NAs in the output, it is a result of the seq output

> breaks
 [1] -130.9168876 -120.9168876 -110.9168876 -100.9168876  -90.9168876  -80.9168876  -70.9168876  -60.9168876  -50.9168876  -40.9168876  -30.9168876
[12]  -20.9168876  -10.9168876   -0.9168876    9.0831124   19.0831124   29.0831124   39.0831124   49.0831124   59.0831124   69.0831124   79.0831124
[23]   89.0831124   99.0831124  109.0831124  119.0831124  129.0831124  139.0831124  149.0831124  159.0831124  169.0831124  179.0831124  189.0831124
[34]  199.0831124  209.0831124  219.0831124  229.0831124  239.0831124  249.0831124  259.0831124  269.0831124  279.0831124  289.0831124  299.0831124
[45]  309.0831124

Note the max value is 309.083, and for the var1 > -130.9168876 would return FALSE for those values that are exactly same. Instead, it should be var1 >= -130.9168876. In order to correct this, we may need to concatenate with max at the end and then take the unique (in case there are duplicates)

breaks <- unique(c(seq(min, max, by = 10), max))

Now, we do the same

> setDT(my_data1)[data.table(start = breaks, end = shift(breaks, 
+    type = "lead", fill = last(breaks))),  c("indices", "label") := .(.GRP, paste(start, end)), 
+     on = .(var1 >= start, var1 <= end), by = .EACHI]
> 
> head(my_data1)
        var1 indices                             label
1:  43.95244      18 39.0831124359188 49.0831124359188
2:  76.98225      21 69.0831124359188 79.0831124359188
3: 255.87083      39 249.083112435919 259.083112435919
4: 107.05084      24 99.0831124359188 109.083112435919
5: 112.92877      25 109.083112435919 119.083112435919
6: 271.50650      41 269.083112435919 279.083112435919
> head(my_data)
       var1 class                             label
1  43.95244    18 39.0831124359188 49.0831124359188
2  76.98225    21 69.0831124359188 79.0831124359188
3 255.87083    39 249.083112435919 259.083112435919
4 107.05084    24 99.0831124359188 109.083112435919
5 112.92877    25 109.083112435919 119.083112435919
6 271.50650    41 269.083112435919 279.083112435919
> my_data1[is.na(indices)]
Empty data.table (0 rows and 3 cols): var1,indices,label

@ Akrun: thank you so much for your answer! Just to clarify : Your approach is NOT expected to produce NA's? — stats_noob, Jan 08 '23 at 21:33
@stats_noob If you use the `breaks` as I showed and then specify `my_data$var1 >= breaks[i]` in your code, it doesn't give NA — akrun, Jan 08 '23 at 21:35
@ Akrun: I have been trying to learn more and become more comfortable using different types of "cuts", "intervals", "breaks", "sequences" and "ranges" in R. I tend to get lost and find it very confusing at times. I had a question about problems involving overlapping ranges yesterday - if you had time, could you please take a look at it? https://stackoverflow.com/questions/75046097/r-overlapping-grouped-ntiles — stats_noob, Jan 08 '23 at 21:45

R: Creating and Labelling Groups by Increments

1 Answers1