用NA操纵矩阵

Question

我不想用零或其他东西填充矩阵。我想知道如何处理这些问题

data<- structure(c(79L, 106L, 156L, 194L, 248L, 248L, 248L, 266L, 272L, 
            79L, 106L, 125L, 156L, 156L, 156L, 156L, 156L, 194L, 79L, 156L, 
            156L, 156L, 156L, 156L, 156L, 156L, 156L, 79L, 248L, 393L, 674L, 
            2447L, NA, NA, NA, NA, 21L, 21L, 21L, 21L, 21L, 21L, 21L, 21L, 
            21L, NA, NA, NA, NA, NA, NA, NA, NA, NA), .Dim = c(9L, 6L), .Dimnames = list(
              NULL, c("a", "b", "c", "d", 
                      "e", "f"))) 

1-突出显示在一列中不止一次重复的那些

预期输出

      a        b       c    d     e     f
[1,]248(3)  156(3)   156(8)      21(9)
[2,]        248(2)            
[3,]                          
[4,]                          
[5,]                          
[6,]                          
[7,]                          
[8,]                         
[9,]                         

1-突出显示在行列中多次重复的那些

预期输出

[1,]  79(4)  
[2,] 106(2) 
[3,] 156(2) 
[4,] 156(2)
[5,] 156(2)
[6,] 156(2)
[7,] 156(2)
[8,] 156(2)
[9,] 

3-如何在不更改维度的情况下在每列中保留唯一元素？

预期产出

     a   b   c    d  e  f
[1,]  79  79  79   79 21 
[2,] 106 106 156  248    
[3,] 156 125      393    
[4,] 194 156      674    
[5,] 248 194     2447    
[6,] 266               
[7,] 272               
[8,]                   
[9,]        

4-如何根据行，排名

找出整个矩阵中出现的数字

预期产出

21(9) 156(8) 248(3) 156(3) 248(2) 

Answer 1

关于突出显示矩阵中的数字出现次数，不会：

table(data)

够了吗？ 对于多次出现，您可以这样做：

table(data)[table(data) > 1]

然后，如果您希望评估行和/或列的语句，您可以这样做：

lstRes <- list()
for (i in 1:dim(data)[1]) {
    lstRes[[i]] <-table(data[i,])[table(data[i,]) > 1]
}

到达data.frame：

lstRes <- list()
for (i in 1:dim(data)[1]) {
    lstRes[[i]] <- as.matrix(table(data[i,])[table(data[i,]) > 1])
}

Reduce(rbind, lstRes)

Answer 2

# this gives you min, median, mean, max of each column 
summary(data)
# this gives you which number are repeated 
data[duplicated(data),]
# gives you how many times each elemnt appears in the data 
as.data.frame(sort(table(data)))
# you can count how many unique values are in each columns and rows, respectively 
apply(data, 2 function(x)length(unique(x)))
apply(data, 1, function(x)length(unique(x)))
# this also give you a logical idea of duplicated elements 
apply(data,2,duplicated)
# if you want to see whether you have any duplicated row (it takes into acount all elements)
duplicated(data)