在 R 中创建具有不均匀数据帧的二进制矩阵

Question

我有一个数据集，其中包含8个不同组中的microRNA。我需要使用 R 将此数据帧转换为二进制矩阵。组中的microRNA数量不同，我想将组放在行中，并将microRNA放在列上。以下是部分数据：

Group1    Group2    Group3   Group4
miR-133a  miR-133b  miR-456  miR777
miR-777   miR138    miR-564  miR-878
miR-878             miR-777  miR978
miR-878
miR-978

预期输出：

Groups  miR-133a  miR-133b  miR-456  miR-777.....
Group1  1             0      0        1
Group2  0             1      0        0
.
.
.

我尝试使用此代码：

im <- which(arr.ind=T,Dat!='');
u <- unique(Dat[im[order(im[,'row'],im[,'col']),]]);
res <- matrix(0L,nrow(Dat),length(u),dimnames=list(NULL,u));
res[cbind(im[,'row'],match(Dat[im],u))] <- 1L;
res

但它给了我很多行。谁能帮我？

Answer 1

这是一个带有tidyverse的选项。 改形为"长"格式，然后使用pivot_wider将其转换回"宽"格式

library(dplyr)
library(tidyr)
df1 %>%
pivot_longer(cols = everything(), names_to = 'Groups', 
values_drop_na = TRUE) %>%
distinct %>%
mutate(new =1) %>% 
pivot_wider(names_from =value, values_from = new,  
values_fill = list(new = 0))
#Groups `miR-133a` `miR-133b` `miR-456` miR777 `miR-777` miR138 `miR-564` `miR-878` miR978 `miR-978`
#  <chr>       <dbl>      <dbl>     <dbl>  <dbl>     <dbl>  <dbl>     <dbl>     <dbl>  <dbl>     <dbl>
#1 Group1          1          0         0      0         1      0         0         1      0         0
#2 Group2          0          1         0      0         0      1         0         0      0         0
#3 Group3          0          0         1      0         1      0         1         1      0         1
#4 Group4          0          0         0      1         0      0         0         1      1         0

---

或与tablebase R

table(names(df1)[col(df1)], unlist(df1))
#           miR-133a miR-133b miR-456 miR-564 miR-777 miR-878 miR-978 miR138 miR777 miR978
#  Group1        1        0       0       0       1       1       0      0      0      0
#  Group2        0        1       0       0       0       0       0      1      0      0
#  Group3        0        0       1       1       1       1       1      0      0      0
#  Group4        0        0       0       0       0       1       0      0      1      1

注意：在这里，我们假设空白为NA。如果是""，请先将其更改为NA，然后使用相同的代码

df1[df1 == ""] <- NA

数据

df1 <- structure(list(Group1 = c("miR-133a", "miR-777", "miR-878", NA, 
NA), Group2 = c("miR-133b", "miR138", NA, NA, NA), Group3 = c("miR-456", 
"miR-564", "miR-777", "miR-878", "miR-978"), Group4 = c("miR777", 
"miR-878", "miR978", NA, NA)), class = "data.frame", row.names = c(NA, 
-5L))

Answer 2

假设数据框中的空白为 "：

df = structure(list(Group1 = c("miR-133a", "miR-777", "miR-878", "", 
""), Group2 = c("miR-133b", "miR138", "", "", ""), Group3 = c("miR-456", 
"miR-564", "miR-777", "miR-878", "miR-978"), Group4 = c("miR777", 
"miR-878", "miR978", "", "")), row.names = c(NA, -5L), class = "data.frame")

然后，制作所有项目的母版集：

alla = setdiff(sort(unique(unlist(df))),"")
res = t(sapply(colnames(df),function(i)as.numeric(alla %in% df[,i])))
colnames(res) = alla
miR-133a miR-133b miR-456 miR-564 miR-777 miR-878 miR-978 miR138 miR777
Group1        1        0       0       0       1       1       0      0      0
Group2        0        1       0       0       0       0       0      1      0
Group3        0        0       1       1       1       1       1      0      0
Group4        0        0       0       0       0       1       0      0      1
miR978
Group1      0
Group2      0
Group3      0
Group4      1