将多个直方图频率计数转换为R中的数组

Question

For each row in the matrix "result" shown below

            A   B   C   D   E   F   G   H   I   J      
       1    4   6   3   5   9   9   9   3   4   4
       2    5   7   5   5   8   8   8   7   4   5
       3    7   5   4   4   7   9   7   4   4   5
       4    6   6   6   6   8   9   8   6   3   6
       5    4   5   5   5   8   8   7   4   3   7
       6    7   9   7   6   7   8   8   5   7   6
       7    5   6   6   5   8   8   7   3   3   5
       8    6   7   4   5   8   9   8   4   6   5
       9    6   8   8   6   7   7   7   7   6   6

I would like to plot a histogram for each row with 3 bins as shown below:

samp<-result[1,]
hist(samp, breaks = 3, col="lightblue", border="pink")

Now what is needed is to convert the histogram frequency counts into an array as follows If I have say 4 bins and say first bin has count=5 and second bin has a count=2 and fourth bin=3. Now I want a vector of all values in each of these bins, coming from data result(for every row) in a vector as my output.

       row1  5 2 0 3

For hundreds of rows I would like to do it in an automated way and hence posted this question.

In the end the matrix should look like

             bin 2-4 bin 4-6 bin6-8 bin8-10
      row 1   5       2       0     3
      row 2
      row 3
      row 4
      row 5
      row 6
      row 7
      row 8
      row 9

Answer 1

DF <- read.table(text="A   B   C   D   E   F   G   H   I   J      
1    4   6   3   5   9   9   9   3   4   4
2    5   7   5   5   8   8   8   7   4   5
3    7   5   4   4   7   9   7   4   4   5
4    6   6   6   6   8   9   8   6   3   6
5    4   5   5   5   8   8   7   4   3   7
6    7   9   7   6   7   8   8   5   7   6
7    5   6   6   5   8   8   7   3   3   5
8    6   7   4   5   8   9   8   4   6   5
9    6   8   8   6   7   7   7   7   6   6", header=TRUE)

m <- as.matrix(DF)

apply(m,1,function(x) hist(x,breaks = 3)$count)
# $`1`
# [1] 5 2 0 3
# 
# $`2`
# [1] 5 0 2 3
# 
# $`3`
# [1] 6 3 1
# 
# $`4`
# [1] 1 6 2 1
# 
# $`5`
# [1] 3 3 4
# 
# $`6`
# [1] 3 4 2 1
# 
# $`7`
# [1] 2 5 3
# 
# $`8`
# [1] 6 3 1
# 
# $`9`
# [1] 4 4 0 2

Note that according to the documentation the number of breaks is only a suggestion. If you want to have the same number of breaks in all rows, you should do the binning outside of hist :

breaks <- 1:5*2
t(apply(m,1,function(x) table(cut(x,breaks,include.lowest = TRUE))))
#   [2,4] (4,6] (6,8] (8,10]
# 1     5     2     0      3
# 2     1     4     5      0
# 3     4     2     3      1
# 4     1     6     2      1
# 5     3     3     4      0
# 6     0     3     6      1
# 7     2     5     3      0
# 8     2     4     3      1
# 9     0     4     6      0

Answer 2

您可以访问hist返回的counts向量（有关详细信息，请参见?hist ）：

counts <- hist(samp, breaks = 3, col="lightblue", border="pink")$counts