英文:
"variable lengths differ" error while running regressions in a loop
问题
我正在尝试运行一个基于我在以前的回答中找到的代码的回归循环（https://stackoverflow.com/q/27952653/21208453），但我一直收到一个错误。我的因变量（依赖变量）有940个变量（代谢物），我的自变量（独立变量）是"bmi"、"Age"、"sex"、"lpa2c"和"smoking"，其中BMI和Age是连续变量。BMI是平均暴露，对其他变量我在控制。所以我正在测试BMI对940个代谢物的影响。此外，我想知道如何提取BMI的系数、p值、标准误差和置信区间，仅当它显著时。
这是我使用的代码：
y <- c(1653:2592) # 响应变量x1 <- c("bmi","Age", "sex","lpa2c", "smoking") # 预测变量for (i in x1){model <- lm(paste("y ~", i), data = QBB_clean)print(summary(model))}
这是错误：
Error in model.frame.default(formula = paste("y ~", i), data = QBB_clean) :
variable lengths differ (found for 'bmi').
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英文:
I am trying to run a regression loop based on code that I have found in a previous answer (https://stackoverflow.com/q/27952653/21208453) but I keep getting an error. My outcomes (dependent) are 940 variables (metabolites) and my exposure (independent) are "bmi","Age", "sex","lpa2c", and "smoking". where BMI and Age are continuous. BMI is the mean exposure, and for others, I am controlling for them.
So I'm testing the effect of BMI on 940 metabolites.
Also, I would like to know how I can extract coefficient, p-value, standard error, and confidence interval for BMI only and when it is significant.
This is the code I have used:
y<- c(1653:2592) # responsex1<- c("bmi","Age", "sex","lpa2c", "smoking") # predictorfor (i in x1){model <- lm(paste("y ~", i[[1]]), data= QBB_clean)print(summary(model))}
And this is the error:
> Error in model.frame.default(formula = paste("y ~", i[[1]]), data = QBB_clean, :
variable lengths differ (found for 'bmi').
>
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-html -->
y1 y2 y3 y4 bmi age sex lpa2c smoking1 0.2875775201 0.59998896 0.238726027 0.784575267 24 18 1 0.470681834 12 0.7883051354 0.33282354 0.962358936 0.009429905 12 20 0 0.365845473 13 0.4089769218 0.48861303 0.601365726 0.779065883 18 15 0 0.121272054 04 0.8830174040 0.95447383 0.515029727 0.729390652 16 21 0 0.046993681 05 0.9404672843 0.48290240 0.402573342 0.630131853 18 28 1 0.262796304 16 0.0455564994 0.89035022 0.880246541 0.480910830 13 13 0 0.968641168 17 0.5281054880 0.91443819 0.364091865 0.156636851 11 12 0 0.488495482 18 0.8924190444 0.60873498 0.288239281 0.008215520 21 23 0 0.477822030 09 0.5514350145 0.41068978 0.170645235 0.452458394 18 17 1 0.748792881 010 0.4566147353 0.14709469 0.172171746 0.492293329 20 15 1 0.667640231 1
<!-- end snippet -->
答案1
得分: 1
如果您想循环遍历响应变量，您可能需要类似以下的代码：
respvars <- names(QBB_clean[1653:2592])predvars <- c("bmi","Age", "sex","lpa2c", "smoking")results <- list()for (v in respvars) {form <- reformulate(predvars, response = v)results[[v]] <- lm(form, data = QBB_clean)}
然后，您可以使用类似 lapply(results, summary) 的方式打印结果，提取系数等等。（我稍微有点难以理解只是打印 940 次回归结果会有多大用处...您真的打算检查它们吗？）
如果您想要BMI的系数等信息，我 认为 以下代码应该可以工作（未经测试）：
t(sapply(results, function(m) coef(summary(m))["bmi",]))
或者要获取系数的区间：
t(sapply(results, function(m) confint(m)["bmi",]))
英文:
If you want to loop over responses you will want something like this:
respvars <- names(QBB_clean[1653:2592])predvars <- c("bmi","Age", "sex","lpa2c", "smoking")results <- list()for (v in respvars) {form <- reformulate(predvars, response = v)results[[v]] <- lm(form, data = QBB_clean)}
You can then print the results with something like lapply(results, summary), extract coefficients, etc.. (I have a little trouble seeing how it's going to be useful to just print the results of 940 regressions ... are you really going to inspect them all?
If you want coefficients etc. for BMI, I think this should work (not tested):
t(sapply(results, function(m) coef(summary(m))["bmi",]))
Or for coefficients:
t(sapply(results, function(m) confint(m)["bmi",]))
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