从字符名称矩阵移动到这些名称的向量(对于fMRI数据)

Question

我有一个下三角矩阵的fMRI网络连接性之和(1:235)，所以有27730的值。我有这些值，但是，我想要绑定另一个向量，它有这些感兴趣区域的名称( ROIs )，但是我不知道如何从这些ROIs的236个向量移到填充的27730向量。

所以连接应该是这样的: SN1-SN2，SN1-SN3…。SN1-CB4，SN2-SN3…。SN2-CB4，SN3-SN4…SN3-CB4等。如果采用所有的唯一连接，那么236 ROIs中的第一个有235个连接，第二个ROI有234个连接，第三个ROI有233个连接等等。因此，唯一连接总数为求和(1:235)= 27730。

不过，根据注释，我已经将向量更改为只包含其中的7个值。

因此，我还将连接性更改为有和(1:8)值。

非常感谢！

roi <- c("SN2", "SN3", "SN4", "SN5", "CON1", "CON2", "CB4")

connectivities <- rnorm(1:28)

Answer 1

以下是一种方法：

m <- outer(roi, roi, paste, sep = "-")
m
#      [,1]       [,2]       [,3]       [,4]       [,5]        [,6]        [,7]      
# [1,] "SN2-SN2"  "SN2-SN3"  "SN2-SN4"  "SN2-SN5"  "SN2-CON1"  "SN2-CON2"  "SN2-CB4" 
# [2,] "SN3-SN2"  "SN3-SN3"  "SN3-SN4"  "SN3-SN5"  "SN3-CON1"  "SN3-CON2"  "SN3-CB4" 
# [3,] "SN4-SN2"  "SN4-SN3"  "SN4-SN4"  "SN4-SN5"  "SN4-CON1"  "SN4-CON2"  "SN4-CB4" 
# [4,] "SN5-SN2"  "SN5-SN3"  "SN5-SN4"  "SN5-SN5"  "SN5-CON1"  "SN5-CON2"  "SN5-CB4" 
# [5,] "CON1-SN2" "CON1-SN3" "CON1-SN4" "CON1-SN5" "CON1-CON1" "CON1-CON2" "CON1-CB4"
# [6,] "CON2-SN2" "CON2-SN3" "CON2-SN4" "CON2-SN5" "CON2-CON1" "CON2-CON2" "CON2-CB4"
# [7,] "CB4-SN2"  "CB4-SN3"  "CB4-SN4"  "CB4-SN5"  "CB4-CON1"  "CB4-CON2"  "CB4-CB4" 

m[upper.tri(m)]
#  [1] "SN2-SN3"   "SN2-SN4"   "SN3-SN4"   "SN2-SN5"   "SN3-SN5"   "SN4-SN5"   "SN2-CON1"  "SN3-CON1"  "SN4-CON1" 
# [10] "SN5-CON1"  "SN2-CON2"  "SN3-CON2"  "SN4-CON2"  "SN5-CON2"  "CON1-CON2" "SN2-CB4"   "SN3-CB4"   "SN4-CB4"  
# [19] "SN5-CB4"   "CON1-CB4"  "CON2-CB4" 

由于roi中有7，第一个元素("SN2")有六个连接，第二个元素("SN3")有五个，等等。共产生21个连接。

另一种方式，使用(并改进)Ben对combn的使用

apply(combn(roi,2), 2, paste, collapse = "-")
#  [1] "SN2-SN3"   "SN2-SN4"   "SN2-SN5"   "SN2-CON1"  "SN2-CON2"  "SN2-CB4"   "SN3-SN4"   "SN3-SN5"   "SN3-CON1" 
# [10] "SN3-CON2"  "SN3-CB4"   "SN4-SN5"   "SN4-CON1"  "SN4-CON2"  "SN4-CB4"   "SN5-CON1"  "SN5-CON2"  "SN5-CB4"  
# [19] "CON1-CON2" "CON1-CB4"  "CON2-CB4" 

Answer 2

下面是一个较小的值集(7)的示例。对于7个值，有21个组合：6 + 5 + 4 + 3 + 2 + 1 = 45。

roi <- c("SN2", "SN3", "SN4", "SN5", "CON1", "CON2", "CB4")

combn()函数以矩阵形式生成所需的输出：

     [,1]  [,2]  [,3]  [,4]   [,5]   [,6]  [,7]  [,8]  [,9]   [,10]  [,11]
[1,] "SN2" "SN2" "SN2" "SN2"  "SN2"  "SN2" "SN3" "SN3" "SN3"  "SN3"  "SN3"
[2,] "SN3" "SN4" "SN5" "CON1" "CON2" "CB4" "SN4" "SN5" "CON1" "CON2" "CB4"
     [,12] [,13]  [,14]  [,15] [,16]  [,17]  [,18] [,19]  [,20]  [,21] 
[1,] "SN4" "SN4"  "SN4"  "SN4" "SN5"  "SN5"  "SN5" "CON1" "CON1" "CON2"
[2,] "SN5" "CON1" "CON2" "CB4" "CON1" "CON2" "CB4" "CON2" "CB4"  "CB4" 

要获得您最终想要的输出，请转置矩阵，转换为data.frame，并使用tidyr中的unite()将两个roi值缝合在一起。

library(dplyr) # for the piper %>%
library(tidy)
combn(roi, 2) %>%
  t() %>% as.data.frame() %>%
  unite(col = "combination", sep = "-") 

    combination
1      SN2-SN3
2      SN2-SN4
3      SN2-SN5
4     SN2-CON1
5     SN2-CON2
6      SN2-CB4
7      SN3-SN4
8      SN3-SN5
9     SN3-CON1
10    SN3-CON2
11     SN3-CB4
12     SN4-SN5
13    SN4-CON1
14    SN4-CON2
15     SN4-CB4
16    SN5-CON1
17    SN5-CON2
18     SN5-CB4
19   CON1-CON2
20    CON1-CB4
21    CON2-CB4