如何更改因子级别的名称？

Question

训练集

trainSample <- cbind(data[1:980,1], data[1:980,2]) cl <-
factor(c(data[1:980,3]))

测试集

testSample <- data(data[981:1485,1], data[981:1485,2])
cl.test <- clknn

预测

k <- knn(trainSample, testSample, cl, k = 5)

产量

< k

  [1] 2 2 1 1 1 1 2 1 2 1 1 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 1 1 2 2 1 1 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2
 [60] 2 2 2 2 1 2 2 2 2 1 2 2 1 2 2 2 1 1 2 1 2 2 1 1 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 1 2 2 2 2 1 2 2 2 2 2 2
[119] 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 1 2 1 1 1 1 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 2 1 2 2 1 2 1 2 2 2 2
[178] 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 1 1 1 1 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 1
[237] 2 2 2 2 2 1 2 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 2 2 2 1 2 2 1 2 2 2 2 1 2 1 2 2 2 2 1 1 2 1 2 2 2 2 1 2 2 2
[296] 2 2 2 1 2 1 2 1 1 1 2 1 2 2 1 1 2 2 1 2 1 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 2 2 2 1 2 1 1 2 2 2 1 1 2
[355] 1 2 1 2 1 2 1 2 2 2 2 2 2 1 1 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2
[414] 2 2 1 2 2 2 2 2 2 2 2 2 1 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
[473] 2 2 2 2 2 1 1 2 2 2 2 2 1 2 2 1 1 2 2 1 2 2 1 2 1 2 2 1 2 2 2 2 2
Levels: 1 2

我想要"c"和"not-c"(就像在我原来的data.csv中),而不是1和2(我也不确定哪个数字应该代表哪个)

有人可以帮忙吗？

Answer 1

更改因子水平非常容易,也不会混淆哪个是:

示例数据:

> a <- factor(rep(c(1,2,1),50))
> a
  [1] 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2
 [75] 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1 2 1 1
[149] 2 1
Levels: 1 2

#this will help later as a verification
#this counts the instances for 1 and 2
> table(a)
a
  1   2 
100  50 

因此,正如您在上面看到的那样,级别的顺序是1第一个和2第二个.更改级别(下方)时,订单保持不变:

#the assignment function levels can be used to change the levels
#the order will remain the same i.e. 'c' for '1' and 'not-c' for '2'
levels(a) <- c('c', 'not-c')

> a
  [1] c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c    
 [25] c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c    
 [49] c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c    
 [73] c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c    
 [97] c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c    
[121] c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c     c     not-c c    
[145] c     not-c c     c     not-c c    
Levels: c not-c

这是验证:

> table(a)
a
    c not-c 
  100    50 

Answer 2

订阅作业也有效.例如,这是一个因素:

> a <- factor(sample(letters[1:5],100,replace=T))
> a
  [1] a d d d d a d d a b a b e a c d a c a a b e e d a e d e e a a c a a a b a
 [38] b b a a e b d b c a a a b e b c e d d b b c c a b a d c b c c d e b d e d
 [75] a a a b e e c b c b c c d d e e d a e e e b c e b e
Levels: a b c d e

现在,让我们给出几个级别的新名称:

> levels(a)[c(2,4)] <- c('y','z')
> a
  [1] a z z z z a z z a y a y e a c z a c a a y e e z a e z e e a a c a a a y a
 [38] y y a a e y z y c a a a y e y c e z z y y c c a y a z c y c c z e y z e z
 [75] a a a y e e c y c y c c z z e e z a e e e y c e y e
Levels: a y c z e