使用lapply和两个参数运行滞后回归 [英] Running lagged regressions with lapply and two arguments

问题描述

我正在运行多个单变量回归，例如在此可重现的示例中:

require(dynlm)
data(USeconomic)
US<-USeconomic
vars<-colnames(US)[-2]
a<-lapply(colnames(US),function(x) dynlm(log(GNP)~get(x),data=US))

a包含3个单变量回归的列表.现在假设我要进行3次滞后进行相同的回归:l<-c(0,1,4)其中0当然是我已经得到的情况.有没有办法像

lagged_vars <- paste("L(",rep(vars,each=3),",",l,")",sep="")

那样直接使用向量l

# this did not work for me, I obtain multivariate regressions including all lags at once
lapply(colnames(US),function(x) dynlm(log(GNP)~L(get(x),l),data=US),l=l)

此操作无效后，我尝试了另一种方法并将其添加到以下向量:

lagged_vars <- paste("L(",rep(vars,each=3),",",l,")",sep="")

获取:

[1] "L(log(M1),0)" "L(log(M1),1)" "L(log(M1),4)" "L(rs,0)"      "L(rs,1)"     
[6] "L(rs,4)"      "L(rl,0)"      "L(rl,1)"      "L(rl,4)"

不幸的是，我不能使用新的字符向量运行它，get()不能帮助您.我不明白为什么会导致它与vars一起使用，而不能同时与lagged_vars一起使用，这两个都是字符向量.

请注意，L()语法来自dynlm软件包. 附带问题:如果我只是打印回归结果中的系数，将其标记为get(x)–我该如何更改?

可能会出现i，j循环，但我宁愿使用lapply或该系列产品中的某些东西...

as.formula与dynlm中的L()不能一起使用.我收到此错误消息:

merge.zoo(log(GNP)，L(log(M1)，0)，retclass ="list"，all = FALSE)中的错误: 找不到功能"L"

找到了有趣的帖子bei Achim Zeileis 提到了这个问题.

解决方案

以下是使用plyr

的方法

library(plyr); library(dynlm); library(tseries)

# FUNCTION TO RUN A SINGLE REGRESSION
foo = function(x, l) dynlm(log(GNP) ~ L(get(as.character(x)), l), data = US)

# CREATE PARAMETER GRID
params = expand.grid(x = colnames(US)[-2], l = c(0, 1, 4))

# RUN REGRESSIONS
regressions = mlply(params, foo)

此列表的每个元素均包含有关单个回归的详细信息，您可以从中提取所需的输出

I am running multiple univariate regressions, like in this reproducible example:

require(dynlm)
data(USeconomic)
US<-USeconomic
vars<-colnames(US)[-2]
a<-lapply(colnames(US),function(x) dynlm(log(GNP)~get(x),data=US))

a contains a list of 3 univariate regressions. Assume now I´d like to run the same regressions with 3 lags: l<-c(0,1,4) where 0 is of course the case I already got. Is there a way to use the vector ldirectly, like

# this did not work for me, I obtain multivariate regressions including all lags at once
lapply(colnames(US),function(x) dynlm(log(GNP)~L(get(x),l),data=US),l=l)

After this did not work I tried another approach and added to following vector:

lagged_vars <- paste("L(",rep(vars,each=3),",",l,")",sep="")

to get:

[1] "L(log(M1),0)" "L(log(M1),1)" "L(log(M1),4)" "L(rs,0)"      "L(rs,1)"     
[6] "L(rs,4)"      "L(rl,0)"      "L(rl,1)"      "L(rl,4)"

Unfortunately, I can't run it with the new character vector, get() does not help. I can't understand why cause it works with vars but not lagged_vars which are both character vectors.

Note, that the L() syntax comes from the dynlm package. Side question: If I just print a the coefficients from the regression result remain labelled get(x) – how can I change that?

An i,j loop could be possible solution but I´d rather use lapply or something out of this family...

EDIT: as.formula does not work together with L() from dynlm. I get the this error message:

Error in merge.zoo(log(GNP), L(log(M1), 0), retclass = "list", all = FALSE) : could not find function "L"

EDIT: found an interesting post bei Achim Zeileis referring to this problem.

解决方案

Here is an approach using plyr

library(plyr); library(dynlm); library(tseries)

# FUNCTION TO RUN A SINGLE REGRESSION
foo = function(x, l) dynlm(log(GNP) ~ L(get(as.character(x)), l), data = US)

# CREATE PARAMETER GRID
params = expand.grid(x = colnames(US)[-2], l = c(0, 1, 4))

# RUN REGRESSIONS
regressions = mlply(params, foo)

Each element of this list contains details on a single regression from which you can extract your desired output

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