我有一个如下所示的概率矩阵:
ID V1 V2 V3 V4
1 0.15 0.1 0.5 0.25
2. 0 0.1 0.3. 0.6
3. 0.2. 0.25. 0.2. 0.35
我想将其转换为1、0矩阵,将最高概率分配为1,其余概率分配为0:
ID V1 V2 V3 V4
1 0 0 1 0
2. 0 0. 0. 1
3. 0. 0 0 1
我该如何编写函数来完成任务?
答案 0 :(得分:4)
如果是按行,那么我们用max
得到pmax
的值,并与数据集进行比较
df1[-1] <- +(df1[-1] == do.call(pmax, df1[-1]))
df1
# ID V1 V2 V3 V4
#1 1 0 0 1 0
#2 2 0 0 0 1
#3 3 0 0 0 1
或与apply
df1[-1] <- t(apply(df1[-1], 1, function(x) +(x == max(x))))
df1 <- structure(list(ID = c(1, 2, 3), V1 = c("0.15", "0", "0.2."),
V2 = c("0.1", "0.1", "0.25."), V3 = c("0.5", "0.3.", "0.2."
), V4 = c(0.25, 0.6, 0.35)), class = "data.frame",
row.names = c(NA,
-3L))
答案 1 :(得分:1)
我们可以使用max.col
返回每一行的最大值索引。
#Create a copy of df
df2 <- df
#turn all values to 0
df2[-1] <- 0
#Get the max column number in each row and turn those values to 1
df2[cbind(1:nrow(df), max.col(df[-1]) + 1)] <- 1
df2
# ID V1 V2 V3 V4
#1 1 0 0 1 0
#2 2 0 0 0 1
#3 3 0 0 0 1
数据
df <- structure(list(ID = 1:3, V1 = c(0.15, 0, 0.2), V2 = c(0.1, 0.1,
0.25), V3 = c(0.5, 0.3, 0.2), V4 = c(0.25, 0.6, 0.35)),
class = "data.frame", row.names = c(NA, -3L))