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我正在 R 中进行一项研究，我正在对不同的依赖项进行几个二项式逻辑回归。这些分析是反复进行的，只是进行了微小的更改，我正在与我的同事分享结果，最好是在漂亮的表格中，而不是凌乱的 R 结果。如果我只打算这样做几次，我可以将所有分析作为单一回归进行，然后使用 sjt.glm 制作漂亮的表格。尽管我一遍又一遍地进行这些类似的分析，但我正在使用 lapply 循环来加速和简化该过程。不幸的是，我无法让 lapply 和 sjt.glm 合作。理想情况下，我会从 lapply 循环中获取结果，并使用 sjt.glm 制作一个漂亮的水平对齐表。

请参阅示例（对于丑陋的编码感到抱歉）

    library(sjPlot)
    swiss$y1 <- ifelse(swiss$Fertility < median(swiss$Fertility), 0, 1)
    swiss$y2 <- ifelse(swiss$Infant.Mortality < median(swiss$Infant.Mortality), 0, 1)
    swiss$y3 <- ifelse(swiss$Agriculture < median(swiss$Agriculture), 0, 1)

    #Normal slow way would be
    fitOR1 <- glm(y1 ~ Education + Examination + Catholic, data = swiss,
         family = binomial(link = "logit"))
    fitOR2 <- glm(y2 ~ Education + Examination + Catholic, data = swiss,
          family = binomial(link = "logit"))
    fitOR3 <- glm(y3 ~ Education + Examination + Catholic, data = swiss,
          family = binomial(link = "logit"))

    #and then simply use summary and other formulas to look at the results
    summary(fitOR1);exp(cbind(OR = coef(fitOR1), confint(fitOR1)))

    #but with 20+ dependents, this would become tedious

    #Doing the same analysis as a laply loop, is relatively easy (and non-tedious)
    varlist <- names(swiss[c(7:9)])

    results <- lapply(varlist, function(x){
      glm(substitute(i ~ Education + Examination + Catholic, list(i=as.name(x))), 
  family =binomial, data = swiss)})

    for (i in 1:3) print(summary(results[[i]]))
    for (i in 1:3) print(exp(cbind(OR = coef(results[[i]]), confint(results[[i]]))))

    #Though here is the catch. To get the output/results into a nice table
    #I can easily use sjt.glm for the "standard" single logistic regressions.
    sjt.glm(fitOR1,fitOR2,fitOR3, file = "SwissFits.html")

    #Though I can't think of how I could do this for the loop-results.
    #The closest I have come is perhaps something like
    for(i in 1:3)(sjt.glm(results[[i]],file="LoopSwissFits.html"))

    #but then I only get the results from the last regression. 

    #One alternative is to do
    lapply(varlist,function(x){ sjt.glm(
      glm(substitute(i ~ Education + Examination + Catholic, list(i=as.name(x))), 
  family =binomial, data = swiss), file = paste0("SwissFits_",(i=as.name(x)),".html"))})

    #but then I get three separate files, when it would be preferable to
     #have the results in one horizontally oriented file

你们中有人对我的问题有简洁优雅的解决方案吗？

非常感谢您！

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2 回答 2

1

stargazer 软件包非常适合这个。使用 write 函数将结果存储在文本文件中。要在控制台中显示，只需使用 -

install.packages("stargazer")  
library(stargazer)

stargazer(results, align = TRUE, type = "text")

# To write to a word file
write(stargazer(results, align = TRUE, type = "text"), "results.txt")
于 2016-11-23T22:36:52.417 回答
1

只需将列表作为第一个参数：sjt.glm(results).

于 2016-11-24T09:03:33.430 回答