使用NA计算列中值

Question

我试图计算R中各列的中位数,然后用列中的每个值减去中值.我在这里遇到的问题是我在我的专栏中有N/A,我不想删除但只返回它们而不减去中位数.例如

ID <- c("A","B","C","D","E") 
Point_A <- c(1, NA, 3, NA, 5) 
Point_B <- c(NA, NA, 1, 3, 2)

df <- data.frame(ID,Point_A ,Point_B)

是否可以计算具有N/A的柱的中值？我的结果是

+----+---------+---------+
| ID | Point_A | Point_B |
+----+---------+---------+
| A  | -2      | NA      |
| B  | NA      | NA      |
| C  | 0       | -1      |
| D  | NA      | 1       |
| E  | 2       | 0       |
+----+---------+---------+

Answer 1

如果我们谈论真正的NA价值观(根据OP评论),人们可以做到

df[-1] <- lapply(df[-1], function(x) x - median(x, na.rm = TRUE))
df
#   ID Point_A Point_B
# 1  A      -2      NA
# 2  B      NA      NA
# 3  C       0      -1
# 4  D      NA       1
# 5  E       2       0

或使用matrixStats包

library(matrixStats)
df[-1] <- df[-1] - colMedians(as.matrix(df[-1]), na.rm = TRUE)

原来df的时候

df <- structure(list(ID = structure(1:5, .Label = c("A", "B", "C", 
"D", "E"), class = "factor"), Point_A = c(1, NA, 3, NA, 5), Point_B = c(NA, 
NA, 1, 3, 2)), .Names = c("ID", "Point_A", "Point_B"), row.names = c(NA, 
-5L), class = "data.frame")