3

我有三个数据帧，dfLON、dfMOS 和 dfATA。每个都具有相同的变量：y 是连续变量，a、b 和 c 是二元分类变量，还有一些NA.

我想建立单独的线性回归模型，每个数据集一个。

使用我当前的代码，我设法制作了一个数据帧列表并将其传递给 lm()。但是有没有比 eg 更简洁的方法来查看结果fitdfLON <- DfList[[1]]？我在这个例子中提供了三个数据框，但实际上我有大约 25 个，所以我必须输入 25 次！

任何帮助将非常感激。

起点（dfs）：

dfLON <- data.frame(y=c(1.23,2.32,3.21,2.43),a=c(1,NA,1,2),b=c(1,1,2,2),c=c(2,1,2,1))
dfMOS <- data.frame(y=c(4.56,6.54,4.43,5.78),a=c(2,1,2,1),b=c(2,1,1,2),c=c(1,2,1,2))
dfATA <- data.frame(y=c(1.22,6.54,3.23,4.23),a=c(2,2,2,1),b=c(1,2,1,2),c=c(1,NA,1,2))

当前代码：

Mylm <- function(df){
 fit <- lm(y ~ a + b + c, data=df)
  return(fit)
}
DfList <- lapply(list(dfLON, dfMOS, dfATA), Mylm)

fitdfLON <- DfList[[1]]
fitdfMOS <- DfList[[2]]
fitdfATA <- DfList[[3]]
4

3 回答 3

1

如果 data.frame 的名称具有共同的模式，您可以使用mget和的组合ls来提取它们并lm使用lapply

fit = lapply(mget(ls(pattern = "^df[A-Z]{3}")), function(x) lm(y ~ a + b + c, data = x))
fit$dfATA

#Call:
#lm(formula = y ~ a + b + c, data = x)

#Coefficients:
#(Intercept)            a            b            c  
#      6.235       -2.005           NA           NA

如果你只想要所有人的系数，你可以这样做

do.call(rbind,
        lapply(X = mget(ls(pattern = "^df[A-Z]{3}")),
               FUN = function(x) lm(formula = y ~ a + b + c, data = x)[[1]]))
#      (Intercept)      a      b  c
#dfATA      6.2350 -2.005     NA NA
#dfLON      0.0300 -0.780  1.980 NA
#dfMOS      8.2975 -1.665 -0.315 NA

而不是ls(pattern = "df[A-Z]{3}")，您也可以只提供一个包含所有 data.frame 名称的向量

于 2017-11-28T18:44:00.393 回答
1

每当您在许多不同的数据集上运行模型时，使用 broom 库整理它们是有意义的。这将为每个模型生成一个干净的数据框，然后您可以将其输出或用于下游分析。

最简单的例子：

library(broom)

Mylm <- function(df){
  fit <- lm(y ~ a + b + c, data=df)
  tidy(fit) # tidy the fit object
}

list(dfLON, dfMOS, dfATA) %>% lapply(Mylm)

#[[1]]
#         term estimate std.error statistic p.value
#1 (Intercept)     0.03       NaN       NaN     NaN
#2           a    -0.78       NaN       NaN     NaN
#3           b     1.98       NaN       NaN     NaN
#
#[[2]]
#         term estimate std.error  statistic    p.value
#1 (Intercept)   8.2975  0.969855  8.5554025 0.07407531
#2           a  -1.6650  0.445000 -3.7415730 0.16626155
#3           b  -0.3150  0.445000 -0.7078652 0.60785169
#
#[[3]]
#         term estimate std.error statistic   p.value
#1 (Intercept)    6.235  3.015000  2.067993 0.2867398
#2           a   -2.005  1.740711 -1.151828 0.4551559

现在，您可以将其与map_dfr()purrr 中的函数结合起来，将所有内容组合成一个组合数据框：

library(purrr)

# note the named list entries; these will go into the "model" column
# without them, you'd just get a model number
list("LON" = dfLON, "MOS" = dfMOS, "ATA" = dfATA) %>% 
  map_dfr(Mylm, .id = "model")

#  model        term estimate std.error  statistic    p.value
#1   LON (Intercept)   0.0300       NaN        NaN        NaN
#2   LON           a  -0.7800       NaN        NaN        NaN
#3   LON           b   1.9800       NaN        NaN        NaN
#4   MOS (Intercept)   8.2975  0.969855  8.5554025 0.07407531
#5   MOS           a  -1.6650  0.445000 -3.7415730 0.16626155
#6   MOS           b  -0.3150  0.445000 -0.7078652 0.60785169
#7   ATA (Intercept)   6.2350  3.015000  2.0679934 0.28673976
#8   ATA           a  -2.0050  1.740711 -1.1518281 0.45515586

为了使事情更紧凑，您可以在内部动态定义函数map_dfr。当您所做的只是拟合线性模型时，这似乎很合适。

list("LON" = dfLON, "MOS" = dfMOS, "ATA" = dfATA) %>% 
  map_dfr(~ tidy(lm(y ~ a + b + c, data = .)),
          .id = "model")

#  model        term estimate std.error  statistic    p.value
#1   LON (Intercept)   0.0300       NaN        NaN        NaN
#2   LON           a  -0.7800       NaN        NaN        NaN
#3   LON           b   1.9800       NaN        NaN        NaN
#4   MOS (Intercept)   8.2975  0.969855  8.5554025 0.07407531
#5   MOS           a  -1.6650  0.445000 -3.7415730 0.16626155
#6   MOS           b  -0.3150  0.445000 -0.7078652 0.60785169
#7   ATA (Intercept)   6.2350  3.015000  2.0679934 0.28673976
#8   ATA           a  -2.0050  1.740711 -1.1518281 0.45515586
于 2017-11-28T18:54:08.920 回答
0 
#make a list of all the dataframes
df = list(dfATA = dfATA, dfLON =dfLON, dfMOS = dfMOS)

#fitting the model
lmr = lapply(df, function(x){
  lmr = lm(x$y ~ x$a + x$b+ x$c, x)
})

#Get coefficients for each model
coefficients = lapply(lmr, function(x) x[["coefficients"]])
coefficients = unlist(coefficients)
于 2021-09-13T04:34:26.460 回答