如何编程比较R中的整行？ [英] How to programatically compare an entire row in R?
问题描述
我在R中有以下数据框：
data =
时间X1 X2 X3
1 1 0 0
2 1 1 1
3 0 0 1
4 1 1 1
5 0 0 0
6 0 1 1
7 1 1 1
8 0 0 0
9 1 1 1
10 0 0 0
I have the following dataframe in R:
data=
Time X1 X2 X3
1 1 0 0
2 1 1 1
3 0 0 1
4 1 1 1
5 0 0 0
6 0 1 1
7 1 1 1
8 0 0 0
9 1 1 1
10 0 0 0
有没有办法编程选择那些等于0,1,1）？我知道可以通过执行 data [data $ X1 == 0&数据$ X2 == 1&数据$ X3 == 1，] 但是，在我的场景中，（0,1,1）是变量中的列表。我的最终目标是确定等于（0,1,1）的行数，或列表变量可以容纳的任何其他组合。
Is there a way to programatically select those rows that are equal to (0,1,1)? I know it can be done by doing data[data$X1 == 0 & data$X2 == 1 & data$X3 == 1,] but, in my scenario, (0,1,1) is a list in a variable. My ultimate goal here is to determine the number of rows that are equal to (0,1,1), or any other combination that list variable can hold.
谢谢！
Mariano。
推荐答案
这里有几个选项， code> merge ：
Here's a couple of options using a merge:
merge(list(X1=0,X2=1,X3=1), dat)
#or
merge(setNames(list(0,1,1),c("X1","X2","X3")), dat)
甚至根据您要匹配的列使用位置索引：
Or even using positional indexes based on what columns you want matched up:
L <- list(0,1,1)
merge(L, dat, by.x=seq_along(L), by.y=2:4)
所有这些都返回：
# X1 X2 X3 Time
#1 0 1 1 6
如果您的匹配变量都是相同的类型，您还可以通过矩阵比较来安全地执行，如：
If your matching variables are all of the same type, you could also safely do it via matrix comparison like:
dat[colSums(t(dat[c("X1","X2","X3")]) == c(0,1,1)) == 3,]
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