如何使用 R 中的 lapply 多次 运行 具有来自不同数据帧的变量的模型
How to run a model with variables from different dataframes multiple times with lapply in R
我有 2 个数据帧
#dummy df for examples:
set.seed(1)
df1 <- data.frame(t = (1:16),
A = sample(20, 16),
B = sample(30, 16),
C = sample(30, 16))
df2 <- data.frame(t = (1:16),
A = sample(20, 16),
B = sample(30, 16),
C = sample(30, 16))
我想对两个数据框中的每一列都执行此操作（t 列除外）：
model <- lm(df2$A ~ df1$A, data = NULL)
我试过这样的方法：
model <- function(yvar, xvar){
lm(df1$as.name(yvar) ~ df2$as.name(xvar), data = NULL)
}
lapply(names(data), model)
但显然不行。我做错了什么？
最后，我真正想要的是从模型中获取系数和其他内容。但是阻止我的是如何多次 运行 一个包含来自不同数据帧的变量的线性模型。
我想要的输出我猜它应该看起来像这样：
# [[1]]
# Call:
# lm(df1$as.name(yvar) ~ df2$as.name(xvar), data = NULL)
#
# Residuals:
# Min 1Q Median 3Q Max
# -0.8809 -0.2318 0.1657 0.3787 0.5533
#
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) -0.013981 0.169805 -0.082 0.936
# predmodex[, 2] 1.000143 0.002357 424.351 <2e-16 ***
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.4584 on 14 degrees of freedom
# Multiple R-squared: 0.9999, Adjusted R-squared: 0.9999
# F-statistic: 1.801e+05 on 1 and 14 DF, p-value: < 2.2e-16
#
# [[2]]
# Call:
# lm(df1$as.name(yvar) ~ df2$as.name(xvar), data = NULL)
#
# Residuals:
# Min 1Q Median 3Q Max
# -0.8809 -0.2318 0.1657 0.3787 0.5533
#
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) -0.013981 0.169805 -0.082 0.936
# predmodex[, 2] 1.000143 0.002357 424.351 <2e-16 ***
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.4584 on 14 degrees of freedom
# Multiple R-squared: 0.9999, Adjusted R-squared: 0.9999
# F-statistic: 1.801e+05 on 1 and 14 DF, p-value: < 2.2e-16
#
# [[3]]
# Call:
# lm(df1$as.name(yvar) ~ df2$as.name(xvar), data = NULL)
#
# Residuals:
# Min 1Q Median 3Q Max
# -0.8809 -0.2318 0.1657 0.3787 0.5533
#
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) -0.013981 0.169805 -0.082 0.936
# predmodex[, 2] 1.000143 0.002357 424.351 <2e-16 ***
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.4584 on 14 degrees of freedom
# Multiple R-squared: 0.9999, Adjusted R-squared: 0.9999
# F-statistic: 1.801e+05 on 1 and 14 DF, p-value: < 2.2e-16
由于 df1 和 df2 具有相同的名称，您可以这样做：
model <- function(var){
lm(df1[[var]] ~ df2[[var]])
}
result <- lapply(names(df1)[-1], model)
result
#[[1]]
#Call:
#lm(formula = df1[[var]] ~ df2[[var]])
#Coefficients:
#(Intercept) df2[[var]]
# 15.1504 -0.4763
#[[2]]
#Call:
#lm(formula = df1[[var]] ~ df2[[var]])
#Coefficients:
#(Intercept) df2[[var]]
# 3.0227 0.6374
#[[3]]
#Call:
#lm(formula = df1[[var]] ~ df2[[var]])
#Coefficients:
#(Intercept) df2[[var]]
# 15.4240 0.2411
要从模型中获取汇总统计信息，您可以使用 broom::tidy :
purrr::map_df(result, broom::tidy, .id = 'model_num')
# model_num term estimate std.error statistic p.value
# <chr> <chr> <dbl> <dbl> <dbl> <dbl>
#1 1 (Intercept) 15.2 3.03 5.00 0.000194
#2 1 df2[[var]] -0.476 0.248 -1.92 0.0754
#3 2 (Intercept) 3.02 4.09 0.739 0.472
#4 2 df2[[var]] 0.637 0.227 2.81 0.0139
#5 3 (Intercept) 15.4 4.40 3.50 0.00351
#6 3 df2[[var]] 0.241 0.272 0.888 0.390
我有 2 个数据帧
#dummy df for examples:
set.seed(1)
df1 <- data.frame(t = (1:16),
A = sample(20, 16),
B = sample(30, 16),
C = sample(30, 16))
df2 <- data.frame(t = (1:16),
A = sample(20, 16),
B = sample(30, 16),
C = sample(30, 16))
我想对两个数据框中的每一列都执行此操作（t 列除外）：
model <- lm(df2$A ~ df1$A, data = NULL)
我试过这样的方法：
model <- function(yvar, xvar){
lm(df1$as.name(yvar) ~ df2$as.name(xvar), data = NULL)
}
lapply(names(data), model)
但显然不行。我做错了什么？
最后，我真正想要的是从模型中获取系数和其他内容。但是阻止我的是如何多次 运行 一个包含来自不同数据帧的变量的线性模型。
我想要的输出我猜它应该看起来像这样：
# [[1]]
# Call:
# lm(df1$as.name(yvar) ~ df2$as.name(xvar), data = NULL)
#
# Residuals:
# Min 1Q Median 3Q Max
# -0.8809 -0.2318 0.1657 0.3787 0.5533
#
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) -0.013981 0.169805 -0.082 0.936
# predmodex[, 2] 1.000143 0.002357 424.351 <2e-16 ***
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.4584 on 14 degrees of freedom
# Multiple R-squared: 0.9999, Adjusted R-squared: 0.9999
# F-statistic: 1.801e+05 on 1 and 14 DF, p-value: < 2.2e-16
#
# [[2]]
# Call:
# lm(df1$as.name(yvar) ~ df2$as.name(xvar), data = NULL)
#
# Residuals:
# Min 1Q Median 3Q Max
# -0.8809 -0.2318 0.1657 0.3787 0.5533
#
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) -0.013981 0.169805 -0.082 0.936
# predmodex[, 2] 1.000143 0.002357 424.351 <2e-16 ***
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.4584 on 14 degrees of freedom
# Multiple R-squared: 0.9999, Adjusted R-squared: 0.9999
# F-statistic: 1.801e+05 on 1 and 14 DF, p-value: < 2.2e-16
#
# [[3]]
# Call:
# lm(df1$as.name(yvar) ~ df2$as.name(xvar), data = NULL)
#
# Residuals:
# Min 1Q Median 3Q Max
# -0.8809 -0.2318 0.1657 0.3787 0.5533
#
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) -0.013981 0.169805 -0.082 0.936
# predmodex[, 2] 1.000143 0.002357 424.351 <2e-16 ***
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.4584 on 14 degrees of freedom
# Multiple R-squared: 0.9999, Adjusted R-squared: 0.9999
# F-statistic: 1.801e+05 on 1 and 14 DF, p-value: < 2.2e-16
由于 df1 和 df2 具有相同的名称，您可以这样做：
model <- function(var){
lm(df1[[var]] ~ df2[[var]])
}
result <- lapply(names(df1)[-1], model)
result
#[[1]]
#Call:
#lm(formula = df1[[var]] ~ df2[[var]])
#Coefficients:
#(Intercept) df2[[var]]
# 15.1504 -0.4763
#[[2]]
#Call:
#lm(formula = df1[[var]] ~ df2[[var]])
#Coefficients:
#(Intercept) df2[[var]]
# 3.0227 0.6374
#[[3]]
#Call:
#lm(formula = df1[[var]] ~ df2[[var]])
#Coefficients:
#(Intercept) df2[[var]]
# 15.4240 0.2411
要从模型中获取汇总统计信息，您可以使用 broom::tidy :
purrr::map_df(result, broom::tidy, .id = 'model_num')
# model_num term estimate std.error statistic p.value
# <chr> <chr> <dbl> <dbl> <dbl> <dbl>
#1 1 (Intercept) 15.2 3.03 5.00 0.000194
#2 1 df2[[var]] -0.476 0.248 -1.92 0.0754
#3 2 (Intercept) 3.02 4.09 0.739 0.472
#4 2 df2[[var]] 0.637 0.227 2.81 0.0139
#5 3 (Intercept) 15.4 4.40 3.50 0.00351
#6 3 df2[[var]] 0.241 0.272 0.888 0.390