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# 计算x^2
x = c(1, 2, 3)
for (i in x) {
    print(i^2)
}

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## [1] 1
## [1] 4
## [1] 9

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# 九九乘法表
for (j in 1:9) {
    for (i in 1:j) {
        m = i * j
        cat(i, "*", j, "=", m, " ")
    }
    cat("\n")
}

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9

## 1 * 1 = 1  
## 1 * 2 = 2  2 * 2 = 4  
## 1 * 3 = 3  2 * 3 = 6  3 * 3 = 9  
## 1 * 4 = 4  2 * 4 = 8  3 * 4 = 12  4 * 4 = 16  
## 1 * 5 = 5  2 * 5 = 10  3 * 5 = 15  4 * 5 = 20  5 * 5 = 25  
## 1 * 6 = 6  2 * 6 = 12  3 * 6 = 18  4 * 6 = 24  5 * 6 = 30  6 * 6 = 36  
## 1 * 7 = 7  2 * 7 = 14  3 * 7 = 21  4 * 7 = 28  5 * 7 = 35  6 * 7 = 42  7 * 7 = 49  
## 1 * 8 = 8  2 * 8 = 16  3 * 8 = 24  4 * 8 = 32  5 * 8 = 40  6 * 8 = 48  7 * 8 = 56  8 * 8 = 64  
## 1 * 9 = 9  2 * 9 = 18  3 * 9 = 27  4 * 9 = 36  5 * 9 = 45  6 * 9 = 54  7 * 9 = 63  8 * 9 = 72  9 * 9 = 81

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# 分别对矩阵u和v执行线性回归
u = matrix(c(1, 2, 3, 1, 2, 4), ncol = 2)
v = matrix(c(8, 12, 20, 15, 10, 2), ncol = 2)
for (m in c("u", "v")) {
    z = get(m)
    fit = lm(z[, 2] ~ z[, 1])
    print(fit)
}

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## 
## Call:
## lm(formula = z[, 2] ~ z[, 1])
## 
## Coefficients:
## (Intercept)       z[, 1]  
##     -0.6667       1.5000  
## 
## 
## Call:
## lm(formula = z[, 2] ~ z[, 1])
## 
## Coefficients:
## (Intercept)       z[, 1]  
##      23.286       -1.071

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9

# 编程序计算1+2+3+4+6+7的值
x = 1:7
sum = 0
for (i in x) {
    if (i == 5) 
        next
    sum = sum + i
}
print(sum)

1

## [1] 23

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i = 1
while (i <= 10) {
    i = i + 4
}
print(i)

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## [1] 13

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i = 1
while (TRUE) {
    i = i + 4
    if (i > 10) 
        break
}
print(i)

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## [1] 13

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i = 1
repeat {
    i = i + 4
    if (i > 10) 
        break
}
print(i)

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## [1] 13

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# 如果x = 2，则y = x，否则y = x + 1
x = 3
if (x == 2) {
    y = x
} else {
    y = x + 1
}
print(y)

1

## [1] 4

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feelings = c("sad", "afraid")
for (i in feelings) {
  print(
    switch(i,
           happy = "I am glad you are happy",
           afraid = "There is nothing to fear",
           sad = "Cheer up",
           angry = "Calm down now"
           )
    )
  }

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## [1] "Cheer up"
## [1] "There is nothing to fear"

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# 编写函数(统计奇数个数)
oddcount = function(x) {
    k = 0
    for (i in x) {
        if (i%%2 == 1) 
            k = k + 1
    }
    return(k)
}

# 测试函数
x = 1:10
oddcount(x)

1

## [1] 5