How to sort vector into bins in R? - r

Question

I have a vector that consists of numbers that can take on any value between 1 and 100.
I want to sort that vector into bins of a certain size.
My logic:
1.) Divide the range (in this case, 1:100) into the amount of bins you want (lets say 10 for this example)
Result: (1, 10.9], 10.9,20.8], (20.8,30.7], (30.7,40.6], (40.6,50.5], (50.5,60.4], (60.4,70.3], (70.3,80.2], (80.2,90.1], (90.1,100]
2.) Then sort my vector
I found a handy function that almost does all this in one fell swoop: cut(). Here is my code:
> table(cut(vector, breaks = 10))
(0.959,10.9] (10.9,20.8] (20.8,30.7] (30.7,40.5] (40.5,50.4] (50.4,60.3] (60.3,70.1] (70.1,80] (80,89.9] (89.9,99.8]
175 171 117 103 82 67 54 46 39 31
Unfortunately, the intervals are different than the bins we calculated from the possible range (1:100). So I tried fixing this by adding in that range into the vector:
> table(cut(c(1,100,vector), breaks = 10))
(0.901,10.9] (10.9,20.8] (20.8,30.7] (30.7,40.6] (40.6,50.5] (50.5,60.4] (60.4,70.3] (70.3,80.2] (80.2,90.1] (90.1,100]
176 171 117 104 82 66 54 48 38 31
This almost worked perfectly except the left-most interval which starts from 0.901 for some reason.
My questions:
1.) Is there a way to do this (using cut or another function/package) without having to insert artificial data points to get the specified bin ranges?
2.) If not, why does the lower bin start from 0.901 and not 1?

Answer 1

Based on your response to #Allan Cameron, I understand taht you want to divide your vector in 10 bins of the same size. But when you define this number of breaks in the cut() function, the size of the intervals calculated by the function, are different accros the groups. As #akrun sad, this occurs because of the method of calculus that the function uses on this case you define only the number's of breaks.
I do not know if there is a way to avoid this in the function. But I think it will be easier if you define the bins as you want as #Gregor Thomas suggested. Here is an example of how I would approach your desire:
vec <- sample(1:100, size = 500, replace = T)
# Here I suppose that you want to divide the data in
# intervals of the same length
breaks <- seq(min(vec), max(vec), by = 9.9)
cut(vec, breaks = breaks)
Other option, would be the cut_interval() function from ggplot2 package, that cut's the vector in n groups with the same length.
library(ggplot2)
cut_interval(vec, n = 10)

Answer 2

why does the lower bin start from 0.901 and not 1?
The answer is the first bit of the Details section of the ?cut help page:
When breaks is specified as a single number, the range of the data is divided into breaks pieces of equal length, and then the outer limits are moved away by 0.1% of the range to ensure that the extreme values both fall within the break intervals.
That .1% adjustment is the reason your lower bound is 0.901 --- the upper bound isn't adjusted because it is a closed, ], not open ) interval on that end.
If you'd like to use other breaks, you can specify exact breaks however you want. Perhaps this:
my_breaks = seq(1, 100, length.out = 11) ## for n bins, you need n+1 breaks
my_breaks
# [1] 1.0 10.9 20.8 30.7 40.6 50.5 60.4 70.3 80.2 90.1 100.0
cut(vector, breaks = my_breaks, include.lowest = TRUE)
But I actually think Allan's suggestion of 0:10 * 10 might be what you really want. I wouldn't dismiss it too quickly:
table(cut(1:100, breaks = 0:10*10))
# (0,10] (10,20] (20,30] (30,40] (40,50] (50,60] (60,70] (70,80] (80,90] (90,100]
# 10 10 10 10 10 10 10 10 10 10