R：使用成对组合更新邻接矩阵/数据帧 [英] R: Update adjacency matrix/data frame using pairwise combinations
问题描述
问题
假设我有此数据框：
# mock data set
df.size = 10
cluster.id<- sample(c(1:5), df.size, replace = TRUE)
letters <- sample(LETTERS[1:5], df.size, replace = TRUE)
test.set <- data.frame(cluster.id, letters)
将会是这样的东西：
cluster.id letters
<int> <fctr>
1 5 A
2 4 B
3 4 B
4 3 A
5 3 E
6 3 D
7 3 C
8 2 A
9 2 E
10 1 A
现在我想对每个cluster.id进行分组，看看在集群中可以找到哪种字母，例如，集群3 包含字母 A ，E，D，C 。然后我想获得所有唯一的成对组合（但不是与自身的组合，因此没有 A，A ）： A，E; A，D，A，C等。然后，我想更新邻接矩阵/数据帧中这些组合的成对距离。
Now I want to group these per cluster.id and see what kind of letters I can find within a cluster, so for example cluster 3 contains the letters A,E,D,C. Then I want to get all unique pairwise combinations (but not combinations with itself so no A,A e.g.): A,E ; A,D, A,C etc. Then I want to update the pairwise distance for these combination in an adjacency matrix/data frame.
想法
# group by cluster.id
# per group get all (unique) pairwise combinations for the letters (excluding pairwise combinations with itself, e.g. A,A)
# update adjacency for each pairwise combinations
我尝试过的事情
# empty adjacency df
possible <- LETTERS
adj.df <- data.frame(matrix(0, ncol = length(possible), nrow = length(possible)))
colnames(adj.df) <- rownames(adj.df) <- possible
# what I tried
update.adj <- function( data ) {
for (comb in combn(data$letters,2)) {
# stucked
}
}
test.set %>% group_by(cluster.id) %>% update.adj(.)
可能很容易y之所以这样做，是因为我一直都在查看邻接矩阵，但我无法弄清楚。.请让我知道是否不清楚
Probably there is an easy way to do this because I see adjacency matrices all the time, but I'm not able to figure it out.. Please let me know if it's not clear
回答评论
@Manuel Bickel的答案：
对于我作为示例给出的数据（就像）：
对于整个数据集，此矩阵将为A-> Z，请记住这一点。
Answer to comment
Answer to @Manuel Bickel:
For the data I gave as example (the table under "will be something like"):
This matrix will be A-->Z for the full dataset, keep that in mind.
A B C D E
A 0 0 1 1 2
B 0 0 0 0 0
C 1 0 0 1 1
D 1 0 1 0 1
E 2 0 1 1 0
我将解释我做了什么：
cluster.id letters
<int> <fctr>
1 5 A
2 4 B
3 4 B
4 3 A
5 3 E
6 3 D
7 3 C
8 2 A
9 2 E
10 1 A
仅包含更多> 1个唯一字母的聚类是相关的（因为我们不希望与自身组合，例如，聚类1仅包含字母B，因此将导致组合 B，B ，因此不相关）：
4 3 A
5 3 E
6 3 D
7 3 C
8 2 A
9 2 E
现在我为每个群集寻找可以配对的组合：
Now I look for each cluster what pairwise combinations I can make:
群集3 ：
A,E
A,D
A,C
E,D
E,C
D,C
更新这些组合在邻接矩阵中：
Update these combination in the adjacency matrix:
A B C D E
A 0 0 1 1 1
B 0 0 0 0 0
C 1 0 0 1 1
D 1 0 1 0 1
E 2 0 1 1 0
然后转到下一个集群
Then go to the next cluster
集群2
A，E
再次更新邻接矩阵：
A B C D E
A 0 0 1 1 2 <-- note the 2 now
B 0 0 0 0 0
C 1 0 0 1 1
D 1 0 1 0 1
E 2 0 1 1 0
对庞大数据集的反应
As reaction to the huge dataset
library(reshape2)
test.set <- read.table(text = "
cluster.id letters
1 5 A
2 4 B
3 4 B
4 3 A
5 3 E
6 3 D
7 3 C
8 2 A
9 2 E
10 1 A", header = T, stringsAsFactors = F)
x1 <- reshape2::dcast(test.set, cluster.id ~ letters)
x1
#cluster.id A B C D E
#1 1 1 0 0 0 0
#2 2 1 0 0 0 1
#3 3 1 0 1 1 1
#4 4 0 2 0 0 0
#5 5 1 0 0 0 0
x2 <- table(test.set)
x2
# letters
#cluster.id A B C D E
# 1 1 0 0 0 0
# 2 1 0 0 0 1
# 3 1 0 1 1 1
# 4 0 2 0 0 0
# 5 1 0 0 0 0
x1.c <- crossprod(x1)
#Error in crossprod(x, y) :
# requires numeric/complex matrix/vector arguments
x2.c <- crossprod(x2)
#works fine
推荐答案
关注g在上面的注释中，此处是泰勒·林克（Tyler Rinker）的代码与您的数据一起使用。我希望这就是您想要的。
Following above comment, here the code of Tyler Rinker used with your data. I hope this is what you want.
更新：在下面的评论之后，我使用了 reshape2软件包添加了一个解决方案。 ，以便能够处理大量数据。
UPDATE: Following below comments, I added a solution using the package reshape2 in order to be able to handle larger amounts of data.
test.set <- read.table(text = "
cluster.id letters
1 5 A
2 4 B
3 4 B
4 3 A
5 3 E
6 3 D
7 3 C
8 2 A
9 2 E
10 1 A", header = T, stringsAsFactors = F)
x <- table(test.set)
x
letters
#cluster.id A B C D E
# 1 1 0 0 0 0
# 2 1 0 0 0 1
# 3 1 0 1 1 1
# 4 0 2 0 0 0
# 5 1 0 0 0 0
#base approach, based on answer by Tyler Rinker
x <- crossprod(x)
diag(x) <- 0 #this is to set matches such as AA, BB, etc. to zero
x
# letters
# letters
# A B C D E
# A 0 0 1 1 2
# B 0 0 0 0 0
# C 1 0 0 1 1
# D 1 0 1 0 1
# E 2 0 1 1 0
#reshape2 approach
x <- acast(test.set, cluster.id ~ letters)
x <- crossprod(x)
diag(x) <- 0
x
# A B C D E
# A 0 0 1 1 2
# B 0 0 0 0 0
# C 1 0 0 1 1
# D 1 0 1 0 1
# E 2 0 1 1 0
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