从R中的单个数据帧运行多个线性回归 [英] Running several linear regressions from a single dataframe in R

问题描述

我有一个具有21列的单个国家/地区的出口贸易数据的数据集.第一列表示年份(1962-2014年)，而其他20个是贸易伙伴.我正在尝试对Years列和其他各列进行线性回归.我已经尝试过这里推荐的方法:运行多个简单的线性回归从R 的数据帧(需要使用

I have a dataset of export trade data for a single country with 21 columns. The first column indicates the years (1962-2014) while the other 20 are trading partners. I am trying to run linear regressions for the years column and each other column. I have tried the method recommended here: Running multiple, simple linear regressions from dataframe in R that entails using

combn(names(DF), 2, function(x){lm(DF[, x])}, simplify = FALSE)

但是，这仅产生了每一对的截距，对我而言，它比回归的斜率不那么重要.

However this only yields the intercept for each pair which is less important to me than the slope of the regressions.

此外，我尝试将数据集用作时间序列，但是当我尝试运行

Additionally I have tried to use my dataset as a time series, however when I try to run

lm(dimnames~., brazilts, na.action=na.exclude)

(其中brazilts是我的数据集，是从"1962"到"2014"的时间序列)，它返回:

(where brazilts is my dataset as a time series from "1962" to "2014") it returns:

Error in model.frame.default(formula = dimnames ~ ., data = brazilts,  : 
  object is not a matrix.

因此，我尝试了使用矩阵的相同方法，但随后返回了错误:

I therefore tried the same method with a matrix but then it returned the error:

Error in model.frame.default(formula = . ~ YEAR, data = brazilmatrix,  : 
  'data' must be a data.frame, not a matrix or an array

(其中brazilmatrix是我的数据集，作为data.matrix，其中包含多年的列).

(where brazilmatrix is my dataset as a data.matrix which includes a column for years).

真的，我现在甚至都不精通R.最终目标是创建一个循环，我可以使用该循环对28个国家/地区的每年更大的按国家/地区划分的总出口数据集进行回归分析.也许我以完全错误的方式对此进行了攻击，因此欢迎您提供任何帮助或批评.切记，实际上，年份(1962-2014)是我的解释变量，而总出口值是我的因变量，在上面的示例中，这可能与我的语法不符.预先感谢！

Really I am not even proficient in R and at this point. The ultimate goal is to create a loop that I can use to get take regressions for a significantly larger dataset of gross exports by country-pair per year for 28 countries. Perhaps I am attacking this in entirely the wrong way, so any help or criticism is welcome. Bare in mind that the years (1962-2014) are in effect my explanatory variable and the value of gross export is my dependent variable, which may be throwing off my syntax in the above examples. Thanks in advance!

推荐答案

只需添加一个替代方法，我建议沿着这条路线走:

Just to add an alternative, I would propose going down this route:

library(reshape2)
library(dplyr)
library(broom)

df <- melt(data.frame(x = 1962:2014, 
                      y1 = rnorm(53), 
                      y2 = rnorm(53), 
                      y3 = rnorm(53)), 
          id.vars = "x")

df %>% group_by(variable) %>% do(tidy(lm(value ~ x, data=.)))

在这里，我只是融化数据，以便所有相关的列都由行组给出，以便能够使用dplyr的分组操作.这给出了以下数据帧作为输出:

Here, I just melt the data so that all relevant columns are given by groups of rows, to be able to use dplyr's grouped actions. This gives the following dataframe as output:

Source: local data frame [6 x 6]
Groups: variable [3]

  variable        term     estimate    std.error  statistic   p.value
    (fctr)       (chr)        (dbl)        (dbl)      (dbl)     (dbl)
1       y1 (Intercept) -3.646666114 18.988154862 -0.1920495 0.8484661
2       y1           x  0.001891627  0.009551103  0.1980533 0.8437907
3       y2 (Intercept) -8.939784046 16.206935047 -0.5516024 0.5836297
4       y2           x  0.004545156  0.008152140  0.5575415 0.5795966
5       y3 (Intercept) 21.699503502 16.785586452  1.2927462 0.2019249
6       y3           x -0.010879271  0.008443204 -1.2885240 0.2033785

这是继续使用系数的非常方便的形式.所需要做的就是熔化数据框，使所有列都是数据集中的行，然后使用dplyr的group_by对所有子集进行回归. broom::tidy将回归输出放入一个不错的数据框.有关更多信息，请参见?broom.

This is a pretty convenient form to continue working with the coefficients. All that is required is to melt the dataframe so that all columns are rows in the dataset, and then to use dplyr's group_by to carry out the regression in all subsets. broom::tidy puts the regression output into a nice dataframe. See ?broom for more information.

如果需要保留模型以进行某种调整(针对lm对象实现)，则还可以执行以下操作:

In case you need to keep the models to do adjustments of some sort (which are implemented for lm objects), then you can also do the following:

df %>% group_by(variable) %>% do(mod = lm(value ~ x, data=.))

Source: local data frame [3 x 2]
Groups: <by row>

# A tibble: 3 x 2
  variable      mod
*   <fctr>   <list>
1       y1 <S3: lm>
2       y2 <S3: lm>
3       y3 <S3: lm>

在这里，对于每个变量，lm对象都存储在数据框中.因此，如果您想获得第一个模型的输出，则可以像访问任何常规数据框一样访问它，例如

Here, for each variable, the lm object is stored in the dataframe. So, if you want to get the model output for the first, you can just access it as you would access any normal dataframe, e.g.

tmp <- df %>% group_by(variable) %>% do(mod = lm(value ~ x, data=.))
tmp[tmp$variable == "y1",]$mod
[[1]]

Call:
lm(formula = value ~ x, data = .)

Coefficients:
(Intercept)            x  
  -1.807255     0.001019  

如果要对所有lm对象应用某些方法，这很方便，因为您可以使用tmp$mod为您提供它们的列表这一事实，这使得传递给例如lapply.

This is convenient if you want to apply some methods to all lm objects since you can use the fact that tmp$mod gives you a list of them, which makes it easy to pass to e.g. lapply.

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