在R中,如何用等于相同宽度的值的向量替换多列中的值? [英] In R, how to replace values in multiple columns with a vector of values equal to the same width?
问题描述
我试图用长度为2的向量替换2列中的每一行的值。更容易显示。
首先是一些数据。
set.seed(1234)
x< -data.frame(x = sample(c(0:3 ),10,replace = T))
x $ ab< -0 #column将被替换
x $ cd< -0 #column将被替换
数据如下所示:
x ab cd
1 0 0 0
2 2 0 0
3 2 0 0
4 2 0 0
5 3 0 0
6 2 0 0
7 0 0 0
8 0 0 0
9 2 0 0
10 2 0 0
每次x = 2或x = 3,我想要ab = 0和cd = 1。
我的尝试是这样:
x [with(x,which(x == 2 | x == 3)),c(2:3 )]< - c(0,1)
哪些没有预期的结果: p>
x ab cd
pre>
1 0 0 0
2 2 0 1
3 2 1 0
4 2 0 1
5 3 1 0
6 2 0 1
7 0 0 0
8 0 0 0
9 2 1 0
10 2 0 1
你能帮我吗?
解决方案因为R在列主要布局中存储矩阵和数组。而当您将较短的数组分配给较长的数组时,R会循环使用较短的数组。例如,如果您有
x< -rep(0,20)
x [1:10]< c(2,3)
然后你最终得到
[1] 2 3 2 3 2 3 2 3 2 3 0 0 0 0 0 0 0 0 0 0
pre>
您的情况发生的是,x等于2或3的子数组通过循环遍历向量
c(0,1)
。我不知道改变这种行为的任何简单的方法。
可能这里最简单的一件事是简单地填写一个列。或者你可以这样做:
索引< -with(x,which(x == 2 | x == 3))
x [indices,c(2,3)]< -rep(c(0,1),each = length(indices))
I am trying to replace every row's values in 2 columns with a vector of length 2. It is easier to show you.
First here is a some data.
set.seed(1234) x<-data.frame(x=sample(c(0:3), 10, replace=T)) x$ab<-0 #column that will be replaced x$cd<-0 #column that will be replaced
The data looks like this:
x ab cd 1 0 0 0 2 2 0 0 3 2 0 0 4 2 0 0 5 3 0 0 6 2 0 0 7 0 0 0 8 0 0 0 9 2 0 0 10 2 0 0
Every time x=2 or x=3, I want to ab=0 and cd=1.
My attempt is this:
x[with(x, which(x==2|x==3)), c(2:3)] <- c(0,1)
Which does not have the intended results:
x ab cd 1 0 0 0 2 2 0 1 3 2 1 0 4 2 0 1 5 3 1 0 6 2 0 1 7 0 0 0 8 0 0 0 9 2 1 0 10 2 0 1
Can you help me?
解决方案The reason it doesn't work as you want is because R stores matrices and arrays in column-major layout. And when you a assign a shorter array to a longer array, R cycles through the shorter array. For example if you have
x<-rep(0,20) x[1:10]<-c(2,3)
then you end up with
[1] 2 3 2 3 2 3 2 3 2 3 0 0 0 0 0 0 0 0 0 0
What is happening in your case is that the sub-array where x is equal to 2 or 3 is being filled in column-wise by cycling through the vector
c(0,1)
. I don't know of any simple way to change this behavior.Probably the easiest thing to do here is simply fill in the columns one at a time. Or, you could do something like this:
indices<-with(x, which(x==2|x==3)) x[indices,c(2,3)]<-rep(c(0,1),each=length(indices))
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