R，dplyr：获取矩阵列表中重复单元格的数量
R, dplyr: Get number of repeated cells in list of matrices
编辑
添加一个 4x4 矩阵的例子来测试
矩阵：
> mat1
[,1] [,2] [,3] [,4]
[1,] 0 0 1 0
[2,] 0 1 0 1
[3,] 1 1 1 1
[4,] 0 1 0 1
> mat2
[,1] [,2] [,3] [,4]
[1,] 1 1 1 0
[2,] 1 1 1 1
[3,] 1 1 1 1
[4,] 0 1 0 0
> mat3
[,1] [,2] [,3] [,4]
[1,] 1 1 0 1
[2,] 1 0 0 1
[3,] 0 0 0 1
[4,] 1 0 1 0
> mat4
[,1] [,2] [,3] [,4]
[1,] 0 0 1 0
[2,] 0 0 0 0
[3,] 0 0 1 1
[4,] 0 0 1 1
sample_list <- list(mat1, mat2, mat3, mat4)
m12 <- as.matrix(sample_list[[1]] + sample_list[[2]])
sum_row <- rowSums(m12*(m12 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m12 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame()%>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep2 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep2 = sum_row - num_rep) %>%
select(., id, sum_rep2)
m123 <- as.matrix(sample_list[[1]] + sample_list[[2]] + sample_list[[3]])
sum_row <- rowSums(m123*(m123 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m123 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep3 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep3 = sum_row - num_rep) %>%
select(., id, sum_rep3)
m1234 <- as.matrix(sample_list[[1]] + sample_list[[2]] + sample_list[[3]] + sample_list[[4]])
sum_row <- rowSums(m1234*(m1234 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m1234 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep4 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep4 = sum_row - num_rep) %>%
select(., id, sum_rep4)
all_reps <- full_join(rep2, rep3) %>%
full_join(., rep4)
最终输出：
id sum_rep2 sum_rep3 sum_rep4
1 1 3 4
2 2 4 4
3 4 5 7
4 1 1 3
原版POST
我有一个邻接矩阵列表。我正在尝试计算每次观察重复填充单元格的次数。我的目标是按顺序对矩阵执行此操作。所以，M1和M2的重复次数，然后是M1，M2，M3，等等
我正在尝试创建一个函数来使用列表中提供的尽可能多的矩阵来执行此操作。下面的代码是我一直用来一步一个脚印的代码。
示例矩阵：
set.seed(0)
mat1 <- matrix(sample(0:1, 10*10, replace=TRUE),10,10) %>%
replace(., col(.) == row(.), 0)
mat2 <- matrix(sample(0:1, 10*10, replace=TRUE),10,10) %>%
replace(., col(.) == row(.), 0)
mat3 <- matrix(sample(0:1, 10*10, replace=TRUE),10,10) %>%
replace(., col(.) == row(.), 0)
mat4 <- matrix(sample(0:1, 10*10, replace=TRUE),10,10) %>%
replace(., col(.) == row(.), 0)
sample_list <- list(mat1, mat2, mat3, mat4)
我用过的代码：
这里我按行计算单元格的总和，填充的单元格数大于1。我从总行总和中减去单元格数得到重复的总数。
m12 <- as.matrix(sample_list[[1]] + sample_list[[2]])
sum_row <- rowSums(m12*(m12 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m12 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame()%>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep2 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep2 = sum_row - num_rep) %>%
select(., id, sum_rep2)
m123 <- as.matrix(sample_list[[1]] + sample_list[[2]] + sample_list[[3]])
sum_row <- rowSums(m123*(m123 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m123 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep3 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep3 = sum_row - num_rep) %>%
select(., id, sum_rep3)
m1234 <- as.matrix(sample_list[[1]] + sample_list[[2]] + sample_list[[3]] + sample_list[[4]])
sum_row <- rowSums(m1234*(m1234 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m1234 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep4 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep4 = sum_row - num_rep) %>%
select(., id, sum_rep4)
all_reps <- full_join(rep2, rep3) %>%
full_join(., rep4)
最终输出 all_reps 为我提供了这个数据集（使用我在第一段代码中创建的随机矩阵）：
id sum_rep2 sum_rep3 sum_rep4
1 1 1 7 10
2 2 3 8 10
3 3 1 5 10
4 4 0 1 4
5 5 2 4 9
6 6 0 2 4
7 7 3 6 10
8 8 5 8 12
9 9 3 7 11
10 10 3 9 12
有没有一种方法可以使用循环或应用函数以更自动化的方式执行此操作，这也将获取更多矩阵的列表？
矩阵：
sample_list[[1]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 1 0 1 1 0 0 1 1 1
[2,] 0 0 0 1 1 1 0 1 1 0
[3,] 1 0 0 1 1 0 1 1 1 0
[4,] 0 0 0 0 0 1 1 0 0 1
[5,] 0 0 0 1 0 1 1 0 0 1
[6,] 1 0 0 0 1 0 1 0 0 0
[7,] 0 1 0 0 1 0 0 1 1 0
[8,] 0 1 1 1 1 1 1 0 0 0
[9,] 0 1 0 0 1 1 0 0 0 1
[10,] 1 1 0 1 0 1 1 0 0 0
sample_list[[2]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 1 0 0 0 1 1 0 0
[2,] 1 0 1 0 1 1 0 0 1 0
[3,] 1 0 0 0 0 0 0 0 0 1
[4,] 1 0 0 0 0 0 0 1 0 0
[5,] 0 1 0 0 0 0 1 0 0 1
[6,] 0 1 1 1 0 0 0 0 0 1
[7,] 0 1 0 0 1 1 0 1 0 1
[8,] 0 1 0 1 1 1 1 0 1 0
[9,] 1 0 1 1 1 1 0 1 0 1
[10,] 0 1 1 1 0 1 0 1 1 0
sample_list[[3]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 0 1 1 1 1 1 1 1
[2,] 1 0 1 0 1 0 1 1 1 0
[3,] 1 0 0 0 1 1 0 1 1 0
[4,] 0 0 1 0 0 0 0 0 0 1
[5,] 1 1 1 0 0 0 0 1 1 1
[6,] 1 0 1 0 0 0 0 0 1 0
[7,] 1 0 0 1 1 0 0 0 1 1
[8,] 0 0 0 1 0 1 0 0 1 1
[9,] 0 1 0 1 1 0 0 1 0 0
[10,] 1 1 0 1 0 0 1 1 1 0
sample_list[[4]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 1 1 0 0 0 0 0 0 1
[2,] 0 0 0 0 1 0 0 0 1 0
[3,] 0 0 0 1 1 1 1 0 1 0
[4,] 0 0 1 0 1 0 0 1 0 1
[5,] 0 0 0 1 0 1 1 1 0 1
[6,] 0 0 0 0 1 0 1 0 0 0
[7,] 0 0 1 1 0 1 0 1 1 0
[8,] 1 0 1 1 0 0 1 0 1 0
[9,] 1 0 1 0 1 1 1 0 0 0
[10,] 1 0 1 1 1 0 0 0 0 0
我认为这可以满足您的需求：
library(tidyverse)
accumulate(sample_list, `+`) %>%
tail(-1) %>%
map(~ rowSums(pmax(.x - 1, 0))) %>%
bind_cols(.name_repair = ~ paste0("sum_rep", seq_along(.x)+1)) %>%
rowid_to_column()
# A tibble: 4 x 4
rowid sum_rep2 sum_rep3 sum_rep4
<int> <dbl> <dbl> <dbl>
1 1 0 0 0
2 2 0 1 2
3 3 1 3 4
4 4 0 0 2
示例数据：
set.seed(0)
n <- 4
sample_list <- replicate(4, `diag<-`(matrix(sample(0:1, n*n, replace=TRUE),n), 0), simplify = FALSE)
编辑
添加一个 4x4 矩阵的例子来测试
矩阵：
> mat1
[,1] [,2] [,3] [,4]
[1,] 0 0 1 0
[2,] 0 1 0 1
[3,] 1 1 1 1
[4,] 0 1 0 1
> mat2
[,1] [,2] [,3] [,4]
[1,] 1 1 1 0
[2,] 1 1 1 1
[3,] 1 1 1 1
[4,] 0 1 0 0
> mat3
[,1] [,2] [,3] [,4]
[1,] 1 1 0 1
[2,] 1 0 0 1
[3,] 0 0 0 1
[4,] 1 0 1 0
> mat4
[,1] [,2] [,3] [,4]
[1,] 0 0 1 0
[2,] 0 0 0 0
[3,] 0 0 1 1
[4,] 0 0 1 1
sample_list <- list(mat1, mat2, mat3, mat4)
m12 <- as.matrix(sample_list[[1]] + sample_list[[2]])
sum_row <- rowSums(m12*(m12 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m12 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame()%>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep2 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep2 = sum_row - num_rep) %>%
select(., id, sum_rep2)
m123 <- as.matrix(sample_list[[1]] + sample_list[[2]] + sample_list[[3]])
sum_row <- rowSums(m123*(m123 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m123 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep3 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep3 = sum_row - num_rep) %>%
select(., id, sum_rep3)
m1234 <- as.matrix(sample_list[[1]] + sample_list[[2]] + sample_list[[3]] + sample_list[[4]])
sum_row <- rowSums(m1234*(m1234 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m1234 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep4 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep4 = sum_row - num_rep) %>%
select(., id, sum_rep4)
all_reps <- full_join(rep2, rep3) %>%
full_join(., rep4)
最终输出：
id sum_rep2 sum_rep3 sum_rep4
1 1 3 4
2 2 4 4
3 4 5 7
4 1 1 3
原版POST
我有一个邻接矩阵列表。我正在尝试计算每次观察重复填充单元格的次数。我的目标是按顺序对矩阵执行此操作。所以，M1和M2的重复次数，然后是M1，M2，M3，等等
我正在尝试创建一个函数来使用列表中提供的尽可能多的矩阵来执行此操作。下面的代码是我一直用来一步一个脚印的代码。
示例矩阵：
set.seed(0)
mat1 <- matrix(sample(0:1, 10*10, replace=TRUE),10,10) %>%
replace(., col(.) == row(.), 0)
mat2 <- matrix(sample(0:1, 10*10, replace=TRUE),10,10) %>%
replace(., col(.) == row(.), 0)
mat3 <- matrix(sample(0:1, 10*10, replace=TRUE),10,10) %>%
replace(., col(.) == row(.), 0)
mat4 <- matrix(sample(0:1, 10*10, replace=TRUE),10,10) %>%
replace(., col(.) == row(.), 0)
sample_list <- list(mat1, mat2, mat3, mat4)
我用过的代码：
这里我按行计算单元格的总和，填充的单元格数大于1。我从总行总和中减去单元格数得到重复的总数。
m12 <- as.matrix(sample_list[[1]] + sample_list[[2]])
sum_row <- rowSums(m12*(m12 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m12 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame()%>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep2 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep2 = sum_row - num_rep) %>%
select(., id, sum_rep2)
m123 <- as.matrix(sample_list[[1]] + sample_list[[2]] + sample_list[[3]])
sum_row <- rowSums(m123*(m123 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m123 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep3 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep3 = sum_row - num_rep) %>%
select(., id, sum_rep3)
m1234 <- as.matrix(sample_list[[1]] + sample_list[[2]] + sample_list[[3]] + sample_list[[4]])
sum_row <- rowSums(m1234*(m1234 > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "sum_row" = ".")
num_rep <- m1234 %>%
apply(.,1,function(x) sum(x > 1)) %>% as.data.frame() %>%
rowid_to_column(., var = "id") %>%
rename(., "num_rep" = ".")
rep4 <- full_join(sum_row, num_rep) %>%
mutate(sum_rep4 = sum_row - num_rep) %>%
select(., id, sum_rep4)
all_reps <- full_join(rep2, rep3) %>%
full_join(., rep4)
最终输出 all_reps 为我提供了这个数据集（使用我在第一段代码中创建的随机矩阵）：
id sum_rep2 sum_rep3 sum_rep4
1 1 1 7 10
2 2 3 8 10
3 3 1 5 10
4 4 0 1 4
5 5 2 4 9
6 6 0 2 4
7 7 3 6 10
8 8 5 8 12
9 9 3 7 11
10 10 3 9 12
有没有一种方法可以使用循环或应用函数以更自动化的方式执行此操作，这也将获取更多矩阵的列表？
矩阵：
sample_list[[1]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 1 0 1 1 0 0 1 1 1
[2,] 0 0 0 1 1 1 0 1 1 0
[3,] 1 0 0 1 1 0 1 1 1 0
[4,] 0 0 0 0 0 1 1 0 0 1
[5,] 0 0 0 1 0 1 1 0 0 1
[6,] 1 0 0 0 1 0 1 0 0 0
[7,] 0 1 0 0 1 0 0 1 1 0
[8,] 0 1 1 1 1 1 1 0 0 0
[9,] 0 1 0 0 1 1 0 0 0 1
[10,] 1 1 0 1 0 1 1 0 0 0
sample_list[[2]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 1 0 0 0 1 1 0 0
[2,] 1 0 1 0 1 1 0 0 1 0
[3,] 1 0 0 0 0 0 0 0 0 1
[4,] 1 0 0 0 0 0 0 1 0 0
[5,] 0 1 0 0 0 0 1 0 0 1
[6,] 0 1 1 1 0 0 0 0 0 1
[7,] 0 1 0 0 1 1 0 1 0 1
[8,] 0 1 0 1 1 1 1 0 1 0
[9,] 1 0 1 1 1 1 0 1 0 1
[10,] 0 1 1 1 0 1 0 1 1 0
sample_list[[3]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 0 1 1 1 1 1 1 1
[2,] 1 0 1 0 1 0 1 1 1 0
[3,] 1 0 0 0 1 1 0 1 1 0
[4,] 0 0 1 0 0 0 0 0 0 1
[5,] 1 1 1 0 0 0 0 1 1 1
[6,] 1 0 1 0 0 0 0 0 1 0
[7,] 1 0 0 1 1 0 0 0 1 1
[8,] 0 0 0 1 0 1 0 0 1 1
[9,] 0 1 0 1 1 0 0 1 0 0
[10,] 1 1 0 1 0 0 1 1 1 0
sample_list[[4]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 1 1 0 0 0 0 0 0 1
[2,] 0 0 0 0 1 0 0 0 1 0
[3,] 0 0 0 1 1 1 1 0 1 0
[4,] 0 0 1 0 1 0 0 1 0 1
[5,] 0 0 0 1 0 1 1 1 0 1
[6,] 0 0 0 0 1 0 1 0 0 0
[7,] 0 0 1 1 0 1 0 1 1 0
[8,] 1 0 1 1 0 0 1 0 1 0
[9,] 1 0 1 0 1 1 1 0 0 0
[10,] 1 0 1 1 1 0 0 0 0 0
我认为这可以满足您的需求：
library(tidyverse)
accumulate(sample_list, `+`) %>%
tail(-1) %>%
map(~ rowSums(pmax(.x - 1, 0))) %>%
bind_cols(.name_repair = ~ paste0("sum_rep", seq_along(.x)+1)) %>%
rowid_to_column()
# A tibble: 4 x 4
rowid sum_rep2 sum_rep3 sum_rep4
<int> <dbl> <dbl> <dbl>
1 1 0 0 0
2 2 0 1 2
3 3 1 3 4
4 4 0 0 2
示例数据：
set.seed(0)
n <- 4
sample_list <- replicate(4, `diag<-`(matrix(sample(0:1, n*n, replace=TRUE),n), 0), simplify = FALSE)