在 data.table 中使用 `:=` 对 R 中两列的值求和,忽略 NA [英] Using `:=` in data.table to sum the values of two columns in R, ignoring NAs
问题描述
我有一个我认为与使用 data.table 和 :=
函数有关的非常简单的问题.我认为我不太了解 :=
的行为,并且经常遇到类似的问题.
I have what I think is a very simple question related to the use of data.table and the :=
function. I don't think I quite understand the behaviour of :=
and often I run into similar problems.
这是一些示例数据
mat <- structure(list(
col1 = c(NA, 0, -0.015038, 0.003817, -0.011407),
col2 = c(0.003745, 0.007463, -0.007407, -0.003731, -0.007491)),
.Names = c("col1", "col2"),
row.names = c(NA, 10L),
class = c("data.table", "data.frame"))
给了
> mat
col1 col2
1: NA 0.003745
2: 0.000000 0.007463
3: -0.015038 -0.007407
4: 0.003817 -0.003731
5: -0.011407 -0.007491
我想创建一个名为 col3 的列,它给出 col1 和 col2 的总和.如果我使用
I want to create a column called col3 which gives the sum of col1 and col2. If I use
mat[,col3 := col1 + col2]
# col1 col2 col3
#1: NA 0.003745 NA
#2: 0.000000 0.007463 0.007463
#3: -0.015038 -0.007407 -0.022445
#4: 0.003817 -0.003731 0.000086
#5: -0.011407 -0.007491 -0.018898
然后我得到第一行的 NA,但我希望忽略 NA.所以我尝试了
then I get an NA for the first row, but I want NAs to be ignored. So I tried instead
mat[,col3 := sum(col1,col2,na.rm=TRUE)]
# col1 col2 col3
#1: NA 0.003745 -0.030049
#2: 0.000000 0.007463 -0.030049
#3: -0.015038 -0.007407 -0.030049
#4: 0.003817 -0.003731 -0.030049
#5: -0.011407 -0.007491 -0.030049
这不是我想要的,因为它给了我 col1 和 col2 的所有元素的总和.我想我不太明白 :=
...如何在忽略 NA 值的情况下得到 col1 和 col2 元素的总和?
which is not what I am after, since it is giving me the sum of all elements of col1 and col2. I think I don't quite get :=
... How can I get the sum of the element of col1 and col2 ignoring NA values?
不确定这是否相关,但这是我的 sessionInfo
Not sure this is relevant, but here is my sessionInfo
> sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)
locale:
[1] en_AU.UTF-8/en_AU.UTF-8/en_AU.UTF-8/C/en_AU.UTF-8/en_AU.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] data.table_1.8.3
推荐答案
这不是缺乏对 data.table 的理解,而是对 R 中的向量化函数的理解.您可以定义一个与+"行为不同的二元运算符" 关于缺失值的运算符:
It's not a lack of understanding of data.table but rather one regarding vectorized functions in R. You can define a dyadic operator that will behave differently than the "+" operator with regard to missing values:
`%+na%` <- function(x,y) {ifelse( is.na(x), y, ifelse( is.na(y), x, x+y) )}
mat[ , col3:= col1 %+na% col2]
#-------------------------------
col1 col2 col3
1: NA 0.003745 0.003745
2: 0.000000 0.007463 0.007463
3: -0.015038 -0.007407 -0.022445
4: 0.003817 -0.003731 0.000086
5: -0.011407 -0.007491 -0.018898
您可以使用 mrdwad 的注释来做到这一点 sum(... , na.rm=TRUE
):
You can use mrdwad's comment to do it with sum(... , na.rm=TRUE
):
mat[ , col4 := sum(col1, col2, na.rm=TRUE), by=1:NROW(mat)]
这篇关于在 data.table 中使用 `:=` 对 R 中两列的值求和,忽略 NA的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!