R将值替换为垃圾箱

Question

I have a df with integer values. For purposes of classification, I'd like to replace this df with a simpler one that has pre-determined intervals instead of integers. How do I do this efficiently? An example is below:

df:

   1   2   3
1  5   3   0 
2  1   10  12
3  3   0   10

transforms into:

   1      2      3
1  [3-5]  [3-5]  [0-2]
2  [0-2]  [10-12][10-12]
3  [3-5]  [0-2]  [10-12]

Answer 1

Is df a data frame or a matrix? The name suggests the former, but the way you describe it suggests the latter.

If it's a matrix:

df2 <- cut(df, c(0, 2, 5, 9 12))
dim(df2) <- dim(df)

If it's a data frame:

df[] <- lapply(df, cut, c(0, 2, 5, 9, 12))

Answer 2

In addition to Hong, who proposes a good solution, I found something quite useful in ggplot2:

cut_interval - make n groups with equal range

cut_number - make n groups with approximately equal observations

cut_width - make n groups of equal width

In my opinion these functions offer more flexibility and are easier to understand than the base cut function. Note that the functions return factors instead of a matrix.

You could use something like this:

df <- matrix(c(5,3,0,1,10,12,3,0,10), nrow=3)
m.df <- melt(df)
m.df$value <- cut_width(m.df$value, width=2, boundary=0)

This will return

   Var1 Var2   value
1    1    1   (4,6]
2    2    1   (2,4]
3    3    1   [0,2]
4    1    2   [0,2]
5    2    2  (8,10]
6    3    2 (10,12]
7    1    3   (2,4]
8    2    3   [0,2]
9    3    3  (8,10]

If needed, you can cast it back to a square matrix:

df.bins <- acast(m.df, Var1~Var2)

Finally giving:

  1     2       3     
1 (4,6] [0,2]   (2,4] 
2 (2,4] (8,10]  [0,2] 
3 [0,2] (10,12] (8,10]
Levels: [0,2] (2,4] (4,6] (6,8] (8,10] (10,12]