R：如何将数据帧转换为具有空角的非对称矩阵

Question

我有以下数据框：

table <- data.frame(pop_1 = c("AL","AL","AL","AL","AL","AL","AL","ALT","ALT","ALT","ALT","ALT","ALT","BU","BU","BU","BU","BU","IRK","IRK","IRK","IRK","KK","KK","KK","KYA","KYA","TU"),
                    pop_2 = c("ALT","BU","IRK","KK","KYA","TU","ZAB","BU","IRK","KK","KYA","TU","ZAB","IRK","KK","KYA","TU","ZAB","KK","KYA","TU","ZAB","KYA","TU","ZAB","TU","ZAB","ZAB"),
                    value = c(0.43447,0.15267,0.25912,0.10435,0.19238,0.19186,0.18155,0.34969,0.07506,0.29206,0.13597,0.46354,0.17870,0.18658,0.02297,0.08851,0.18950,0.05176,0.12086,0.02690,0.29669,0.05551,0.04910,0.15779,0.03276,0.23422,0.00568,0.22181))

如何将其转换为具有空（或 NA 等）单元格的非对称矩阵，如下所示：

Answer 1

对您的数据框进行细微更改，在开头添加额外的“AL”、“AL”、“NA”组合。你会想在最后为一个额外的“ZAB”做同样的事情：

df<- data.frame(pop_1 = c("AL","AL","AL","AL","AL","AL","AL","AL","ALT","ALT","ALT","ALT","ALT","ALT","BU","BU","BU","BU","BU","IRK","IRK","IRK","IRK","KK","KK","KK","KYA","KYA","TU"),
              pop_2 = c("AL","ALT","BU","IRK","KK","KYA","TU","ZAB","BU","IRK","KK","KYA","TU","ZAB","IRK","KK","KYA","TU","ZAB","KK","KYA","TU","ZAB","KYA","TU","ZAB","TU","ZAB","ZAB"),
              value = c(NA,0.43447,0.15267,0.25912,0.10435,0.19238,0.19186,0.18155,0.34969,0.07506,0.29206,0.13597,0.46354,0.17870,0.18658,0.02297,0.08851,0.18950,0.05176,0.12086,0.02690,0.29669,0.05551,0.04910,0.15779,0.03276,0.23422,0.00568,0.22181))

library(tidyverse)
pivot_wider(df, names_from=pop_1, values_from=value)

 pop_2     AL     ALT      BU     IRK      KK      KYA     TU
  <chr>  <dbl>   <dbl>   <dbl>   <dbl>   <dbl>    <dbl>  <dbl>
1 AL    NA     NA      NA      NA      NA      NA       NA    
2 ALT    0.434 NA      NA      NA      NA      NA       NA    
3 BU     0.153  0.350  NA      NA      NA      NA       NA    
4 IRK    0.259  0.0751  0.187  NA      NA      NA       NA    
5 KK     0.104  0.292   0.0230  0.121  NA      NA       NA    
6 KYA    0.192  0.136   0.0885  0.0269  0.0491 NA       NA    
7 TU     0.192  0.464   0.190   0.297   0.158   0.234   NA    
8 ZAB    0.182  0.179   0.0518  0.0555  0.0328  0.00568  0.222

编辑：

df2<-df
names(df2)<-c("pop_2", "pop_1", "value")
rbind(df, df2) %>% pivot_wider(names_from=pop_1, values_from=value) %>% arrange(pop_2)

Answer 2

创建数据框 pop_1 和 pop_2 列中所有唯一值的向量。这将是矩阵的行和列的名称。

populations <- unique(c(table$pop_1, table$pop_2))

使用 matrix 函数创建一个空矩阵，其行数和列数与步骤 1 中的向量相同。使用 value 参数将矩阵的默认值设置为 NA。

matrix <- matrix(NA, nrow = length(populations), ncol = length(populations))

使用 rownames 和 colnames 函数将矩阵的行和列的名称设置为 populations 向量中的值。

rownames(matrix) <- populations
colnames(matrix) <- populations

使用 for 循环遍历数据框的行。对于每一行，使用 pop_1 和 pop_2 列在矩阵中找到相应的单元格，并使用 value 列设置这些单元格的值。

for (i in 1:nrow(table)) {
  row_name <- table[i, "pop_1"]
  col_name <- table[i, "pop_2"]
  value <- table[i, "value"]
  matrix[row_name, col_name] <- value
}

在这些步骤之后，矩阵应该是一个不对称矩阵，在适当的单元格中具有来自数据框的值，在所有其他单元格中具有 NA 。

当您查看矩阵的结果时：

       AL     ALT      BU     IRK      KK     KYA      TU     ZAB
AL  NA 0.43447 0.15267 0.25912 0.10435 0.19238 0.19186 0.18155
ALT NA      NA 0.34969 0.07506 0.29206 0.13597 0.46354 0.17870
BU  NA      NA      NA 0.18658 0.02297 0.08851 0.18950 0.05176
IRK NA      NA      NA      NA 0.12086 0.02690 0.29669 0.05551
KK  NA      NA      NA      NA      NA 0.04910 0.15779 0.03276
KYA NA      NA      NA      NA      NA      NA 0.23422 0.00568
TU  NA      NA      NA      NA      NA      NA      NA 0.22181
ZAB NA      NA      NA      NA      NA      NA      NA      NA