R 中的线性回归使每条记录成为回归变量
Linear Regression in R Making Every Record a Regressor
我正在尝试使用 R 中的 lm() 函数进行线性回归。当我 运行 它时，R 似乎正在获取每条记录并将其作为独立变量。
fit3 <- lm(no_outliers$strat_count ~ no_outliers$CDS_emerg + no_outliers$weighted_volume_avg)
summary(fit3)
这是输出样本（有 118 个自变量，但我不想把它全部放在 post 上）：
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -336.1017 1065.5530 -0.315 0.7652
no_outliers$weighted_volume_avg941931862 12.2047 31.2349 0.391 0.7121
no_outliers$weighted_volume_avg949989365.5 4.0453 33.3295 0.121 0.9081
no_outliers$weighted_volume_avg955100055.4 10.4469 31.3577 0.333 0.7525
no_outliers$weighted_volume_avg961033059.1 -17.2295 32.3160 -0.533 0.6168
no_outliers$weighted_volume_avg973785580 85.1891 40.0488 2.127 0.0867 .
no_outliers$weighted_volume_avg976666189.1 48.1133 39.9253 1.205 0.2821
no_outliers$weighted_volume_avg979529996 26.2521 31.2707 0.840 0.4395
no_outliers$weighted_volume_avg985701661.4 35.3185 35.8381 0.986 0.3696
no_outliers$weighted_volume_avg988019226.8 66.4781 31.3502 2.120 0.0875 .
no_outliers$weighted_volume_avg994324495.5 -13.4227 32.9220 -0.408 0.7004
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 29.2 on 5 degrees of freedom
Multiple R-squared: 0.9752, Adjusted R-squared: 0.3894
F-statistic: 1.665 on 118 and 5 DF, p-value: 0.3005
任何帮助或指点将不胜感激。
我想 weighted_volume_avg 是一个因素。尝试 class(no_outliers$weighted_volume_avg)。如果是应该先将其转换为数字。尝试：
no_outliers$weighted_volume_avg <- as.numeric(as.character(no_outliers$weighted_volume_avg))
然后再试一次 lm。
我正在尝试使用 R 中的 lm() 函数进行线性回归。当我 运行 它时，R 似乎正在获取每条记录并将其作为独立变量。
fit3 <- lm(no_outliers$strat_count ~ no_outliers$CDS_emerg + no_outliers$weighted_volume_avg)
summary(fit3)
这是输出样本（有 118 个自变量，但我不想把它全部放在 post 上）：
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -336.1017 1065.5530 -0.315 0.7652
no_outliers$weighted_volume_avg941931862 12.2047 31.2349 0.391 0.7121
no_outliers$weighted_volume_avg949989365.5 4.0453 33.3295 0.121 0.9081
no_outliers$weighted_volume_avg955100055.4 10.4469 31.3577 0.333 0.7525
no_outliers$weighted_volume_avg961033059.1 -17.2295 32.3160 -0.533 0.6168
no_outliers$weighted_volume_avg973785580 85.1891 40.0488 2.127 0.0867 .
no_outliers$weighted_volume_avg976666189.1 48.1133 39.9253 1.205 0.2821
no_outliers$weighted_volume_avg979529996 26.2521 31.2707 0.840 0.4395
no_outliers$weighted_volume_avg985701661.4 35.3185 35.8381 0.986 0.3696
no_outliers$weighted_volume_avg988019226.8 66.4781 31.3502 2.120 0.0875 .
no_outliers$weighted_volume_avg994324495.5 -13.4227 32.9220 -0.408 0.7004
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 29.2 on 5 degrees of freedom
Multiple R-squared: 0.9752, Adjusted R-squared: 0.3894
F-statistic: 1.665 on 118 and 5 DF, p-value: 0.3005
任何帮助或指点将不胜感激。
我想 weighted_volume_avg 是一个因素。尝试 class(no_outliers$weighted_volume_avg)。如果是应该先将其转换为数字。尝试：
no_outliers$weighted_volume_avg <- as.numeric(as.character(no_outliers$weighted_volume_avg))
然后再试一次 lm。