In order to distribute a certain amount of TRUE (=1) values, depending on raster layer's values, I would like to do something like this:
为了分配一定数量的TRUE（= 1）值，取决于栅格图层的值，我想做这样的事情：
Let's say I have a 9x2 matrix looking like this:
假设我有一个9x2矩阵，如下所示：
Example <- matrix(c(0.8,0.6,0.5,0.3,0.2,0.4,0.9,0.7,
0.1,1,1,1,1,1,0,0,0,0),ncol=2)
> Example
[,1] [,2]
[1,] 0.8 1
[2,] 0.6 1
[3,] 0.5 1
[4,] 0.3 1
[5,] 0.2 1
[6,] 0.4 0
[7,] 0.9 0
[8,] 0.7 0
[9,] 0.1 0
Now I want to allocate the 1's from [,2] as a function of the highest values of [,1] without changing the order of [,1]. The result should look like this:
现在我想从[，2]中分配1作为[，1]的最高值的函数，而不改变[，1]的顺序。结果应如下所示：
> Example
[,1] [,2]
[1,] 0.8 1
[2,] 0.6 1
[3,] 0.5 1
[4,] 0.3 0
[5,] 0.2 0
[6,] 0.4 0
[7,] 0.9 1
[8,] 0.7 1
[9,] 0.1 0
Is there any elegant way to do this. I could't find any function to solve this problem. As I want to convert [,2] to a raster layer using the raster package a raster-based solution would work for me as well.
有没有优雅的方法来做到这一点。我找不到任何解决这个问题的功能。由于我想使用光栅包将[，2]转换为栅格图层，因此基于栅格的解决方案也适用于我。
2 个解决方案
#1
1
You can find how many 1's you have by summing column 2. You can then find your relative threshold of column 1 by checking if it is higher or lower than that value.
您可以通过对第2列求和来找到您有多少1个。然后，您可以通过检查第1列的相对阈值是高于还是低于该值来找到它的相对阈值。
how.many.ones <- sum(Example[, 2])
max.thresh <- sort(Example[, 1], decreasing=TRUE)[how.many.ones]
change.to.one <- Example[, 1] >= max.thresh
Example[, 2] <- ifelse(change.to.one, 1, 0)
RESULTS:
Example
[,1] [,2]
[1,] 0.8 1
[2,] 0.6 1
[3,] 0.5 1
[4,] 0.3 0
[5,] 0.2 0
[6,] 0.4 0
[7,] 0.9 1
[8,] 0.7 1
[9,] 0.1 0
Caveat:
There is no tie-handling in this process.
在此过程中没有绑定处理。
#2
0
I think what you want is something like this:
我想你想要的是这样的：
Example[,2]<-ifelse(Example[,1]>=0.5,1,0)
But, I'm only guessing that you want the column to be 1 if it is greater or equal to 0.5, and 0 if not.
但是，我只是猜测如果列大于或等于0.5，则希望列为1，否则为0。
But maybe you wanted the second column to be the order of the first one. It is a little confusing:
但也许你想要第二列是第一列的顺序。这有点令人困惑：
Example[,2]<-Example[order(Example[,1]),2]
#1
1
You can find how many 1's you have by summing column 2. You can then find your relative threshold of column 1 by checking if it is higher or lower than that value.
您可以通过对第2列求和来找到您有多少1个。然后，您可以通过检查第1列的相对阈值是高于还是低于该值来找到它的相对阈值。
how.many.ones <- sum(Example[, 2])
max.thresh <- sort(Example[, 1], decreasing=TRUE)[how.many.ones]
change.to.one <- Example[, 1] >= max.thresh
Example[, 2] <- ifelse(change.to.one, 1, 0)
RESULTS:
Example
[,1] [,2]
[1,] 0.8 1
[2,] 0.6 1
[3,] 0.5 1
[4,] 0.3 0
[5,] 0.2 0
[6,] 0.4 0
[7,] 0.9 1
[8,] 0.7 1
[9,] 0.1 0
Caveat:
There is no tie-handling in this process.
在此过程中没有绑定处理。
#2
0
I think what you want is something like this:
我想你想要的是这样的：
Example[,2]<-ifelse(Example[,1]>=0.5,1,0)
But, I'm only guessing that you want the column to be 1 if it is greater or equal to 0.5, and 0 if not.
但是，我只是猜测如果列大于或等于0.5，则希望列为1，否则为0。
But maybe you wanted the second column to be the order of the first one. It is a little confusing:
但也许你想要第二列是第一列的顺序。这有点令人困惑：
Example[,2]<-Example[order(Example[,1]),2]