[英]Using the ifelse statement in R
I would like to use the ifelse statement to create a new variable, say, z. 我想使用ifelse语句创建一个新变量,比如z。 However, one of the return values depends on the i-th column of a matrix. 但是,其中一个返回值取决于矩阵的第i列。 Here is a simple example 这是一个简单的例子
set.seed(1)
data <- data.frame(x = rnorm(10), y = rnorm(10), ind = rep(c(0, 1), 5))
m <- data.frame(matrix(rnorm(100), 10, 10))
z <- ifelse(data$ind == 1, data$x, sum(m[, i]))
I know the line with z won't run, but it illustrates what I would like to do. 我知道z行不会运行,但它说明了我想做的事情。 If a subject has the ind variable equal to 0, then I assign to z the sum of the 10 entries in m corresponding to subject i's column. 如果一个主题的ind变量等于0,那么我将z分配给m对应于主题i列的10个条目的总和。
Could I do this with ifelse, or would I need a for loop? 我可以用ifelse做这个,还是需要for循环? I'm trying to stay away from for loops, which is why I am trying ifelse in the first place. 我试图远离for循环,这就是为什么我首先尝试ifelse。
Here is what z should look like: 这是z应该是什么样子:
z
[1] -1.3367324 0.1836433 1.3413668 1.5952808 4.5120996 -0.8204684 1.2736029
[8] 0.7383247 3.4748021 -0.3053884
Thanks! 谢谢!
Yes you can do it with ifelse
and a one-liner, very close to what you wrote: 是的,你可以使用ifelse
和一个单行,与你写的非常接近:
z <- ifelse(data$ind == 0, colSums(m), data$x)
Here is what R does when it executes this statement: 以下是R执行此语句时的作用:
data$ind == 0
, and stores into memory the two numeric vectors colSums(m)
and data$x
它计算布尔矢量data$ind == 0
,并将两个数值向量colSums(m)
和data$x
存储到内存中 (data$ind == 0)
is True
, it outputs colSums(m)
; 其中(data$ind == 0)
为True
,输出colSums(m)
; where (data$ind == 0)
is False
, it outputs data$x
其中(data$ind == 0)
为False
,它输出data$x
Or we can use arithmetic 或者我们可以使用算术
colSums(m)*(data$ind==0) + (data$ind==1)*data$x
# X1 X2 X3 X4 X5 X6 X7
#-1.3367324 0.1836433 1.3413668 1.5952808 4.5120996 -0.8204684 1.2736029
# X8 X9 X10
# 0.7383247 3.4748021 -0.3053884
You can do it in a two-liner instead: 你可以用双线代替它:
z <- data$x
z[data$ind == 0] <- colSums(m[,data$ind == 0])
[1] -1.3367324 0.1836433 1.3413668 1.5952808 4.5120996 -0.8204684 1.2736029 0.7383247 3.4748021
[10] -0.3053884
more generally, you could use an apply
function. 更一般地说,您可以使用apply
函数。 This will in general be slower than a straight vectorised solution, like the above. 这通常比直接矢量化解决方案慢,如上所述。 Here's sapply: 这是开心的:
sapply(1:nrow(data), function(x){ifelse(data$ind[x] == 1, data$x[x], sum(m[, x]))})
[1] -1.3367324 0.1836433 1.3413668 1.5952808 4.5120996 -0.8204684 1.2736029 0.7383247 3.4748021
[10] -0.3053884
A benchmark: 基准:
microbenchmark::microbenchmark(
sapply = sapply(1:nrow(data), function(x){ifelse(data$ind[x] == 1, data$x[x], sum(m[, x]))}),
vectorised = {z <- data$x;
z[data$ind == 0] <- colSums(m[,data$ind == 0])})
Unit: microseconds
expr min lq mean median uq max neval cld
sapply 391.297 408.193 423.6525 412.4170 423.7450 853.249 100 b
vectorised 197.377 199.873 208.7701 202.5605 214.4645 284.545 100 a
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