Suren
2020-12-10 12:30:52
If you are using cut, then the the first break has to be smaller than the min(data). Otherwise, there will be a NA in the intervals.
For example using quantiles,
cut(data, breaks= c(min(data) - diff(range(data)) / 1000,
quantile(data)[2:5]))
Also, you already know the two middle intervals (25.8 - 50 and 50 - 74.6), its then just finding the min and max of the data
cut(data, breaks= c(min(data) - diff(range(data)) / 1000,
25.8, 50, 74.6, max(data)))
It might be better to scale it slightly, which is what
cut.defaultdoes - e.g.min(data) * 0.99rather than having a fixed offset.@thelatemail I initially changed to what you suggested. Then I realized that if
min(data)is zero, there is a problem. I thinkmin(data) - diff(range(data)) / 1000is whatcut.defaultdo.