R中3个矩阵的元素中值
Elementwise Median of 3 matrices in R
我有 3 个矩阵，每个矩阵存储三次测量值（矩阵 1，测量值 1，矩阵 2 测量值 2，.....）
它们具有以下结构：
> a1
ACTIN 18S TET1 TET2 TET3
Control 25.943441 22.62984 <NA> 34.063107 34.034756
Sample1 24.48504 20.04858 <NA> 32.37173 32.341072
Sample2 25.265867 19.680647 28.086248 33.76187 33.41289
Sample3 24.441484 18.146513 <NA> 32.811428 31.22825
> a2
ACTIN 18S TET1 TET2 TET3
Control 25.980696 22.393877 <NA> 34.548923 33.7815
Sample1 24.263775 20.073978 27.23082 32.27775 32.343292
Sample2 25.25487 19.680494 27.214449 33.70534 33.48968
Sample3 24.26332 18.108198 <NA> 32.769787 31.19895
> a3
ACTIN 18S TET1 TET2 TET3
Control 25.937397 22.429556 30.020935 33.98415 33.858604
Sample1 24.44776 20.090088 28.328804 32.317287 32.291912
Sample2 25.148333 19.537455 <NA> 33.83607 33.3961
Sample3 24.242998 18.335524 <NA> 32.788536 31.147346
我想用 3 个测量值的中值创建一个新矩阵。
理想情况下，第一列保持不变。
如果没有值（未确定），则给出 NA 是首选
我想要一个带有中位数的矩阵，所以像这样：
median(a1[i,j], a2[i,j], a2[i,j])
我尝试了以下方法：
2 个 for 循环遍历数组：
med<-matrix(NA, nrow(a1), ncol(a1))
for(i in ncol(a1)){
for(j in nrow(a1)){
med[i,j]<-median(a1[i,j], a2[i,j], a2[i,j])
}
}
但这给我的值显然不是中位数，我觉得它过于复杂。
谢谢！
您可以先将 "Undetermined" 替换为 "NA"，您将自动获得 NA。我不想输入所有这些数字，所以我只使用了 1 到 5，但它应该适用于任何数字。
a1 <- data.frame(c("Control", "Sample1", "Sample2", "Sample3"), 1, 2, c("Undetermined", "Undetermined", 3, "Undetermined"), 4, 5)
a2 <- data.frame(c("Control", "Sample1", "Sample2", "Sample3"), 1, 2, c("Undetermined", 3, 3, "Undetermined"), 4, 5)
a3 <- data.frame(c("Control", "Sample1", "Sample2", "Sample3"), 1, 2, c(3, 3, "Undetermined", "Undetermined"), 4, 5)
names(a1) <- names(a2) <- names(a3) <- c("Sample", "CT ACTIN", "CT 18S", "CT TET1", "CT TET2", "CT TET3")
a1[a1 == "Undetermined"] <- NA
a2[a2 == "Undetermined"] <- NA
a3[a3 == "Undetermined"] <- NA
med <- matrix(NA, nrow = nrow(a1), ncol = ncol(a1))
for (i in 1:nrow(a1)) {
for (j in 1:ncol(a1)){
med[i, j] <- median(c(a1[i, j], a2[i, j], a3[i, j]))
}
}
med <- data.frame(a1[, 1], med)
names(med) <- c("Sample", "CT ACTIN", "CT 18S", "CT TET1", "CT TET2", "CT TET3")
您可以使用 mapply 并重塑生成的矩阵。假设您的数据最初是我从 <NA> 推断出的字符矩阵，可重现的解决方案如下：
dat <- mapply(function(...) median(as.numeric(c(...))), a1, a2, a3)
# this gives a warning message but you can ignore this which comes up when it converts the character `NA` to numeric `NA`;
matrix(dat, nrow(a1), ncol(a1), dimnames = dimnames(a1))
# ACTIN X18S TET1 TET2 TET3
# Control 25.94344 22.42956 NA 34.06311 33.85860
# Sample1 24.44776 20.07398 NA 32.31729 32.34107
# Sample2 25.25487 19.68049 NA 33.76187 33.41289
# Sample3 24.26332 18.14651 NA 32.78854 31.19895
数据:
a1 <- structure(c("25.94344", "24.48504", "25.26587", "24.44148", "22.62984",
"20.04858", "19.68065", "18.14651", "<NA>", "<NA>", "28.086248",
"<NA>", "34.06311", "32.37173", "33.76187", "32.81143", "34.03476",
"32.34107", "33.41289", "31.22825"), .Dim = 4:5, .Dimnames = list(
c("Control", "Sample1", "Sample2", "Sample3"), c("ACTIN",
"X18S", "TET1", "TET2", "TET3")))
a2 <- structure(c("25.98070", "24.26377", "25.25487", "24.26332", "22.39388",
"20.07398", "19.68049", "18.10820", "<NA>", "27.23082", "27.214449",
"<NA>", "34.54892", "32.27775", "33.70534", "32.76979", "33.78150",
"32.34329", "33.48968", "31.19895"), .Dim = 4:5, .Dimnames = list(
c("Control", "Sample1", "Sample2", "Sample3"), c("ACTIN",
"X18S", "TET1", "TET2", "TET3")))
a3 <- structure(c("25.93740", "24.44776", "25.14833", "24.24300", "22.42956",
"20.09009", "19.53746", "18.33552", "30.020935", "28.328804",
"<NA>", "<NA>", "33.98415", "32.31729", "33.83607", "32.78854",
"33.85860", "32.29191", "33.39610", "31.14735"), .Dim = 4:5, .Dimnames = list(
c("Control", "Sample1", "Sample2", "Sample3"), c("ACTIN",
"X18S", "TET1", "TET2", "TET3")))
假设您的数据集采用您在编辑前发布它们的形式：
> a1
# Sample CT ACTIN CT 18S CT TET1 CT TET2 CT TET3
#1: Control 25.94344 22.62984 Undetermined 34.06311 34.03476
#2: Sample1 24.48504 20.04858 Undetermined 32.37173 32.34107
#3: Sample2 25.26587 19.68065 28.086248 33.76187 33.41289
#4: Sample3 24.44148 18.14651 Undetermined 32.81143 31.22825
您可以使用 mget() 在您的环境中检索与 a[[:digit:]] 匹配的对象，并将它们 bind_rows() 一起检索：
library(dplyr)
dat <- bind_rows(mget(ls(pattern = "a[[:digit:]]")))
然后使用 na_if() 将 "Undetermined" 替换为 NA，将除 Sample 之外的所有列转换为数字，并使用 [=22 计算 median() =]
dat %>%
na_if("Undetermined") %>%
mutate_each(funs(as.numeric), -Sample) %>%
group_by(Sample) %>%
summarise_each(funs(median(., na.rm = TRUE)), -Sample)
给出：
# A tibble: 4 x 6
# Sample CT ACTIN CT 18S CT TET1 CT TET2 CT TET3
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 Control 25.94344 22.42956 30.02094 34.06311 33.85860
#2 Sample1 24.44776 20.07398 27.77981 32.31729 32.34107
#3 Sample2 25.25487 19.68049 27.65035 33.76187 33.41289
#4 Sample3 24.26332 18.14651 NA 32.78854 31.19895
我有 3 个矩阵，每个矩阵存储三次测量值（矩阵 1，测量值 1，矩阵 2 测量值 2，.....）
它们具有以下结构：
> a1
ACTIN 18S TET1 TET2 TET3
Control 25.943441 22.62984 <NA> 34.063107 34.034756
Sample1 24.48504 20.04858 <NA> 32.37173 32.341072
Sample2 25.265867 19.680647 28.086248 33.76187 33.41289
Sample3 24.441484 18.146513 <NA> 32.811428 31.22825
> a2
ACTIN 18S TET1 TET2 TET3
Control 25.980696 22.393877 <NA> 34.548923 33.7815
Sample1 24.263775 20.073978 27.23082 32.27775 32.343292
Sample2 25.25487 19.680494 27.214449 33.70534 33.48968
Sample3 24.26332 18.108198 <NA> 32.769787 31.19895
> a3
ACTIN 18S TET1 TET2 TET3
Control 25.937397 22.429556 30.020935 33.98415 33.858604
Sample1 24.44776 20.090088 28.328804 32.317287 32.291912
Sample2 25.148333 19.537455 <NA> 33.83607 33.3961
Sample3 24.242998 18.335524 <NA> 32.788536 31.147346
我想用 3 个测量值的中值创建一个新矩阵。
理想情况下，第一列保持不变。
如果没有值（未确定），则给出 NA 是首选
我想要一个带有中位数的矩阵，所以像这样：
median(a1[i,j], a2[i,j], a2[i,j])
我尝试了以下方法： 2 个 for 循环遍历数组：
med<-matrix(NA, nrow(a1), ncol(a1))
for(i in ncol(a1)){
for(j in nrow(a1)){
med[i,j]<-median(a1[i,j], a2[i,j], a2[i,j])
}
}
但这给我的值显然不是中位数，我觉得它过于复杂。
谢谢！
您可以先将 "Undetermined" 替换为 "NA"，您将自动获得 NA。我不想输入所有这些数字，所以我只使用了 1 到 5，但它应该适用于任何数字。
a1 <- data.frame(c("Control", "Sample1", "Sample2", "Sample3"), 1, 2, c("Undetermined", "Undetermined", 3, "Undetermined"), 4, 5)
a2 <- data.frame(c("Control", "Sample1", "Sample2", "Sample3"), 1, 2, c("Undetermined", 3, 3, "Undetermined"), 4, 5)
a3 <- data.frame(c("Control", "Sample1", "Sample2", "Sample3"), 1, 2, c(3, 3, "Undetermined", "Undetermined"), 4, 5)
names(a1) <- names(a2) <- names(a3) <- c("Sample", "CT ACTIN", "CT 18S", "CT TET1", "CT TET2", "CT TET3")
a1[a1 == "Undetermined"] <- NA
a2[a2 == "Undetermined"] <- NA
a3[a3 == "Undetermined"] <- NA
med <- matrix(NA, nrow = nrow(a1), ncol = ncol(a1))
for (i in 1:nrow(a1)) {
for (j in 1:ncol(a1)){
med[i, j] <- median(c(a1[i, j], a2[i, j], a3[i, j]))
}
}
med <- data.frame(a1[, 1], med)
names(med) <- c("Sample", "CT ACTIN", "CT 18S", "CT TET1", "CT TET2", "CT TET3")
您可以使用 mapply 并重塑生成的矩阵。假设您的数据最初是我从 <NA> 推断出的字符矩阵，可重现的解决方案如下：
dat <- mapply(function(...) median(as.numeric(c(...))), a1, a2, a3)
# this gives a warning message but you can ignore this which comes up when it converts the character `NA` to numeric `NA`;
matrix(dat, nrow(a1), ncol(a1), dimnames = dimnames(a1))
# ACTIN X18S TET1 TET2 TET3
# Control 25.94344 22.42956 NA 34.06311 33.85860
# Sample1 24.44776 20.07398 NA 32.31729 32.34107
# Sample2 25.25487 19.68049 NA 33.76187 33.41289
# Sample3 24.26332 18.14651 NA 32.78854 31.19895
数据:
a1 <- structure(c("25.94344", "24.48504", "25.26587", "24.44148", "22.62984",
"20.04858", "19.68065", "18.14651", "<NA>", "<NA>", "28.086248",
"<NA>", "34.06311", "32.37173", "33.76187", "32.81143", "34.03476",
"32.34107", "33.41289", "31.22825"), .Dim = 4:5, .Dimnames = list(
c("Control", "Sample1", "Sample2", "Sample3"), c("ACTIN",
"X18S", "TET1", "TET2", "TET3")))
a2 <- structure(c("25.98070", "24.26377", "25.25487", "24.26332", "22.39388",
"20.07398", "19.68049", "18.10820", "<NA>", "27.23082", "27.214449",
"<NA>", "34.54892", "32.27775", "33.70534", "32.76979", "33.78150",
"32.34329", "33.48968", "31.19895"), .Dim = 4:5, .Dimnames = list(
c("Control", "Sample1", "Sample2", "Sample3"), c("ACTIN",
"X18S", "TET1", "TET2", "TET3")))
a3 <- structure(c("25.93740", "24.44776", "25.14833", "24.24300", "22.42956",
"20.09009", "19.53746", "18.33552", "30.020935", "28.328804",
"<NA>", "<NA>", "33.98415", "32.31729", "33.83607", "32.78854",
"33.85860", "32.29191", "33.39610", "31.14735"), .Dim = 4:5, .Dimnames = list(
c("Control", "Sample1", "Sample2", "Sample3"), c("ACTIN",
"X18S", "TET1", "TET2", "TET3")))
假设您的数据集采用您在编辑前发布它们的形式：
> a1
# Sample CT ACTIN CT 18S CT TET1 CT TET2 CT TET3
#1: Control 25.94344 22.62984 Undetermined 34.06311 34.03476
#2: Sample1 24.48504 20.04858 Undetermined 32.37173 32.34107
#3: Sample2 25.26587 19.68065 28.086248 33.76187 33.41289
#4: Sample3 24.44148 18.14651 Undetermined 32.81143 31.22825
您可以使用 mget() 在您的环境中检索与 a[[:digit:]] 匹配的对象，并将它们 bind_rows() 一起检索：
library(dplyr)
dat <- bind_rows(mget(ls(pattern = "a[[:digit:]]")))
然后使用 na_if() 将 "Undetermined" 替换为 NA，将除 Sample 之外的所有列转换为数字，并使用 [=22 计算 median() =]
dat %>%
na_if("Undetermined") %>%
mutate_each(funs(as.numeric), -Sample) %>%
group_by(Sample) %>%
summarise_each(funs(median(., na.rm = TRUE)), -Sample)
给出：
# A tibble: 4 x 6
# Sample CT ACTIN CT 18S CT TET1 CT TET2 CT TET3
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 Control 25.94344 22.42956 30.02094 34.06311 33.85860
#2 Sample1 24.44776 20.07398 27.77981 32.31729 32.34107
#3 Sample2 25.25487 19.68049 27.65035 33.76187 33.41289
#4 Sample3 24.26332 18.14651 NA 32.78854 31.19895