线性模型解决学前班问题 2012-04-30

以上这个据说是学前班同学更擅长的题目,微博上有人表示逻辑性强的头脑确实不能解出来,其实未必。我们试着用线性模型来解决这个问题:等式右边的数字一定是根据左边的数字来的,那么假设每个数字对右侧的贡献是有规律的,即右侧数字~第1位+第2位+…+第4位数字。最后注意这里的x实际上是因子,因为它只能有0~9总计10个水平。

训练数据文件train.csv如下:

one two three four answer
8 8 0 9 6
7 1 1 1 0
2 1 7 2 0
6 6 6 6 4
1 1 1 1 0
3 2 1 3 0
7 6 6 2 2
9 3 1 3 1
0 0 0 0 4
2 2 2 2 0
3 3 3 3 0
5 5 5 5 0
8 1 9 3 3
8 0 9 6 5
7 7 7 7 0
9 9 9 9 4
7 7 5 6 1
6 8 5 5 3
9 8 8 1 5
5 5 3 1 0

线性模型的代码及说明如下:

## 读取训练数据
train <- read.csv(file="D:\\Workplace\\train.csv")

## 把数据变为适用于线性模型的形式
freqTable <- as.data.frame( t(apply(train[,1:4], 1, function(X) table(c(X, 0:9))-1)) )
names(freqTable) <- c("zero","one","two","three","four","five","six","seven","eight","nine")
freqTable$y <- train[,5]

## 用无截距的线性模型拟合之
myModel <- lm(y ~ 0 + zero + one + two + three + four + five + six + seven + eight + nine, data=freqTable)

## 测试集相当于是2,8,5,1
test <- data.frame(t(c(0,1,1,0,0,1,0,0,1,0)))
names(test) <- c("zero","one","two","three","four","five","six","seven","eight","nine")
predict.lm(myModel,newdata=test)
# 1 --这只是行号
# 2 --这才是结果
# 警告信息: --无视警告
#   用秩缺乏拟合来进行预测的结果很可能不可靠

## 看看为什么这么准(训练数据没有4所以NA了)
round(myModel$coefficients)
# zero   one   two three  four  five   six seven eight  nine
# 1     0     0     0    NA     0     1     0     2     1

看到这里,是不是发现只用线性模型,就把这个数圈圈的秘密给训练出来了:-)



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