假设我在一个数据框中有三列分别为A、B和C。每一列的行数都是10。现在，我想要对这些列进行两两组合的乘积运算。所以，我想要的列是A*B、A*C和B*C。我该如何在R中实现这个操作呢？另外，我想要将这些列A*B、A*C和B*C用于多次线性回归的多次迭代，如下所示：Y ~ A + B + C + A*B/A*C/B*C。模型的迭代需要通过一个for循环进行计算。
我已经尝试了以下代码：
```R
cc <- combn(data, 2, FUN = Reduce, f = `*`)
n = choose(ncol(data),2)
model <- list()
for (i in 1:n) {
model[[i]] <- lm(Y ~ A+B+C+cc[,i])
}

<details>
<summary>英文:</summary>
Suppose I have three columns in a data frame A, B and C respectively. The number of rows for each column are 10. Now, I want to multiply all possible combinations of the columns by taking 2 at a time. So, the columns that I want are A*B, A*C,B*C. How shall I get this using R. Also, I want to use these columns A*B, A*C, B*C into several iterations of a multiple linear regression as follows: Y ~ A + B + C + A*B/A*C/B*C. The model iterations has to be computed through a for loop. 
I have tried this code:
    cc &lt;- combn(data, 2, FUN = Reduce, f = `*`)
    n = choose(ncol(data),2)
    model &lt;- list()
    for (i in 1:n) {
    model[[i]] &lt;- lm(Y ~ A+B+C+cc[,i])
    }
Now, the problem is cc[,i] is a double containing all the possible combinations of the three columns and it does not have specific name to it. So, in the model summary the interaction variable is named as &quot;cc[,i]&quot; only. I want to change the variable name to either &quot;A*B&quot; or &quot;B*C&quot; or &quot;A*C&quot;. How shall I do it?
</details>
# 答案1
**得分**: 0
这是你的代码翻译结果：
```R
看看这个代码结构是否适合你：
set.seed(7)
df <- data.frame(A = rexp(10, 1), B = rexp(10, 2), C = rexp(10, 3))
Y <- rnorm(10)
df <- data.frame(cbind(Y, df))
m <- combn(3, 2)
mylist <- list()
for (i in 1:3) {
    new_col <- df[ , m[1, i] + 1] * df[ , m[2, i] + 1]
    df2 <- cbind(df, new_col)
    mylist[[i]] <- lm(Y ~ ., data = df2)
}
mylist
[[1]]
调用：
lm(formula = Y ~ ., data = df2)
系数：
(Intercept)            A            B            C      new_col  
    -1.9290       0.4046       2.0715       0.8587      -0.1156  
[[2]]
调用：
lm(formula = Y ~ ., data = df2)
系数：
(Intercept)            A            B            C      new_col  
   -1.81765      0.33800      1.89724      0.79561      0.03353  
[[3]]
调用：
lm(formula = Y ~ ., data = df2)
系数：
(Intercept)            A            B            C      new_col  
     -1.402        0.321        1.058       -1.445        5.114