R 循环以实现遗漏一个观察值并运行glm，每次一个变量

我有一个包含96个观察值和1106个变量的数据框架

• 我想对的观察结果进行逻辑回归，每次一个。（因此，对于第一组观测值，移除第一个观测值时总共有95个观测值，移除第二个观测值时总共有95个观测值，依此类推，这样就有95组观测值，每个观测值都有一个观测值遗漏。）

• 此外，我想一次只在一个变量上运行每组观察结果。在对一个变量进行95个观测值的回归后，我想提取p值（不包括截距p值）

• 我已经能够手动完成所有这些，一次一个。然而，这是非常乏味的96次这样做，我相信一定有一种方法来自动与一个或多个循环

下面是我如何在10次观察中手动执行此操作的演示

## Create 10 data frames by removing one observation from each ##
di.1 <- mainDF [-1,]
di.2 <- mainDF [-2,]
di.3 <- mainDF [-3,]
di.4 <- mainDF [-4,]
di.5 <- mainDF [-5,]
di.6 <- mainDF [-6,]
di.7 <- mainDF [-7,]
di.8 <- mainDF [-8,]
di.9 <- mainDF [-9,]
di.10 <- mainDF [-10,]

## Create data frames to put each p-value result in ## 
dt.1 <- data.frame(intercept=numeric(), gene=numeric(), stringsAsFactors=FALSE)
dt.2 <- data.frame(intercept=numeric(), gene=numeric(), stringsAsFactors=FALSE)
dt.3 <- data.frame(intercept=numeric(), gene=numeric(), stringsAsFactors=FALSE)
dt.4 <- data.frame(intercept=numeric(), gene=numeric(), stringsAsFactors=FALSE)
dt.5 <- data.frame(intercept=numeric(), gene=numeric(), stringsAsFactors=FALSE)
dt.6 <- data.frame(intercept=numeric(), gene=numeric(), stringsAsFactors=FALSE)
dt.7 <- data.frame(intercept=numeric(), gene=numeric(), stringsAsFactors=FALSE)
dt.8 <- data.frame(intercept=numeric(), gene=numeric(), stringsAsFactors=FALSE)
dt.9 <- data.frame(intercept=numeric(), gene=numeric(), stringsAsFactors=FALSE)
dt.10 <- data.frame(intercept=numeric(), gene=numeric(), stringsAsFactors=FALSE)

## Run logistic regression on each data frame with one one obs. left out ##
## GLM run on one variable at a time##
## Extract p-values and put in separate dfs ##

for (i in 2:1106)
{
  formulas <- glm(response ~ di.1[,i], data=di.1, family= "binomial")
  dt.1[i,] <- coef(summary(formulas))[,4]
}
for (i in 2:1106)
{
  formulas <- glm(response ~ di.2[,i], data=di.2, family= "binomial")
  dt.2[i,] <- coef(summary(formulas))[,4]
}
for (i in 2:1106)
{
  formulas <- glm(response ~ di.3[,i], data=di.3, family= "binomial")
  dt.3[i,] <- coef(summary(formulas))[,4]
}
for (i in 2:1106)
{
  formulas <- glm(response ~ di.4[,i], data=di.4, family= "binomial")
  dt.4[i,] <- coef(summary(formulas))[,4]
}
for (i in 2:1106)
{
  formulas <- glm(response ~ di.5[,i], data=di.5, family= "binomial")
  dt.5[i,] <- coef(summary(formulas))[,4]
}
for (i in 2:1106)
{
  formulas <- glm(response ~ di.6[,i], data=di.6, family= "binomial")
  dt.6[i,] <- coef(summary(formulas))[,4]
}
for (i in 2:1106)
{
  formulas <- glm(response ~ di.7[,i], data=di.7, family= "binomial")
  dt.7[i,] <- coef(summary(formulas))[,4]
}
for (i in 2:1106)
{
  formulas <- glm(response ~ di.8[,i], data=di.8, family= "binomial")
  dt.8[i,] <- coef(summary(formulas))[,4]
}
for (i in 2:1106)
{
  formulas <- glm(response ~ di.9[,i], data=di.9, family= "binomial")
  dt.9[i,] <- coef(summary(formulas))[,4]
}
for (i in 2:1106)
{
  formulas <- glm(response ~ di.10[,i], data=di.10, family= "binomial")
  dt.10[i,] <- coef(summary(formulas))[,4]
}

## Remove intercept p-values ##
dt.1<- dt.1[-c(1)]
dt.2<- dt.2[-c(1)]
dt.3<- dt.3[-c(1)]
dt.4<- dt.4[-c(1)]
dt.5<- dt.5[-c(1)]
dt.6<- dt.6[-c(1)]
dt.7<- dt.7[-c(1)]
dt.8<- dt.8[-c(1)]
dt.9<- dt.9[-c(1)]
dt.10<- dt.10[-c(1)]

## Export data frames, then manually copy and paste them into one CSV ##
write.csv(dt.1, file = "MyData.csv")
write.csv(dt.2, file = "MyData2.csv")
write.csv(dt.3, file = "MyData3.csv")
write.csv(dt.4, file = "MyData4.csv")
write.csv(dt.5, file = "MyData5.csv")
write.csv(dt.6, file = "MyData6.csv")
write.csv(dt.7, file = "MyData7.csv")
write.csv(dt.8, file = "MyData8.csv")
write.csv(dt.9, file = "MyData9.csv")
write.csv(dt.10, file = "MyData10.csv")

---

非常感谢您抽出时间

正如我在前面的评论中所说，我不会使用

glm

和

summary.glm

，因为这对于您的任务来说太慢了，因为您将适应

96*1106

glm。我将使用

glm.fit

，自己计算回归系数的p值。下面的函数

f

执行此操作。它将一个一维向量

x

作为协变量（不允许

NA

使用），另一个一维向量

y

作为响应（不允许

NA

使用）。由于进行了逻辑回归，因此要求

y

是两个级别的因子（或0-1二进制值）

所以p值匹配

---

我们现在需要另一个函数

g

来组织您计划作为双嵌套循环执行的操作。外部循环控制“省略一个”，而内部循环通过

lappy

排列，以循环数据帧列。在外循环的每次迭代结束时，p值的结果数据帧被写入“.csv”文件

---

g每个变量都是一个基因，我正在计算每个基因在所有观察数据集中被发现显著的次数。这是我用来确定哪些基因对我的研究有重要意义的不同方法之一。是的，不要这样做。使用正则化方法，如弹性网。你可以使用glmnet软件包来实现这一点。因为LOO是CV的一种形式，所以也有标记。
  Response  X1  X2  X3  X4  X5  X6  X7  X8  X9  X10

P1  N       1   1   1   0   1   0   1   0   2    2
P2  N       2   1   1   0   2   2   1   2   2    2
P3  N       2   1   2   1   1   0   1   1   0    1
P4  Y       1   1   2   0   1   0   0   1   1    1
P5  N       2   2   1   1   1   0   0   0   1    1
P6  N       2   1   2   1   1   0   0   0   2    1
P7  Y       2   1   1   0   2   0   0   0   2    0
P8  Y       2   1   1   0   2   0   0   1   0    2
P9  N       1   1   1   0   2   0   0   0   1    0
P10 N       2   1   2   1   1   0   1   0   0    2

f <- function (x, y) {
  ## call `glm.fit`
  fit <- glm.fit(cbind(1,x), y, family = binomial())
  ## estimated regression coefficients
  beta <- unname(fit$coefficients)
  ## since there are only two coefficients, I don't bother using `chol2inv`
  ## then extract square root of diagonals for standard errors
  se <- sqrt(diag(chol2inv(fit$qr$qr, size = fit$qr$rank)))
  ## deal with possible rank-deficient case
  if (length(se) < 2L) se <- c(se, NA_real_)
  ## z-score
  z <- beta / se
  ## p-value (0.05 significance level)
  2 * pnorm(-abs(z))
  }

dat <- 
structure(list(Response = structure(c(1L, 1L, 1L, 2L, 1L, 1L, 
2L, 2L, 1L, 1L), .Label = c("N", "Y"), class = "factor"), X1 = c(1L, 
2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L), X2 = c(1L, 1L, 1L, 1L, 2L, 
1L, 1L, 1L, 1L, 1L), X3 = c(1L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 
2L), X4 = c(0L, 0L, 1L, 0L, 1L, 1L, 0L, 0L, 0L, 1L), X5 = c(1L, 
2L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 1L), X6 = c(0L, 2L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, 0L), X7 = c(1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 
1L), X8 = c(0L, 2L, 1L, 1L, 0L, 0L, 0L, 1L, 0L, 0L), X9 = c(2L, 
2L, 0L, 1L, 1L, 2L, 2L, 0L, 1L, 0L), X10 = c(2L, 2L, 1L, 1L, 
1L, 1L, 0L, 2L, 0L, 2L)), .Names = c("Response", "X1", "X2", 
"X3", "X4", "X5", "X6", "X7", "X8", "X9", "X10"), row.names = c("P1", 
"P2", "P3", "P4", "P5", "P6", "P7", "P8", "P9", "P10"), class = "data.frame")

## code response into factor
dat[[1]] <- factor(dat[[1]])

## call `f`
f(dat[[2]], dat[[1]])
# [1] 0.8559137 0.8804148

## call `glm` + `summary.glm`
coef(summary(glm(Response ~ X1, data = dat, family = binomial())))
#              Estimate Std. Error    z value  Pr(>|z|)
#(Intercept) -0.4700036   2.588435 -0.1815783 0.8559137
#X1          -0.2231436   1.483239 -0.1504434 0.8804148

g <- function (dat) {
  ## convert response to factor (if it is not readily is)
  y <- as.factor(dat[[1]])
  ## leave-one-out
  for (i in 1:nrow(dat)) {
    ## covariates data frame
    covariates <- dat[-i, -1]
    ## response vector
    response <- y[-i]
    ## call `f` to get a data frame of p-values
    result <- as.data.frame(lapply(covariates, f, y = response))
    ## write data frame to file
    write.csv(result, file = paste0(i,".csv"), row.names = FALSE)
    }
  }