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使用summary()函数打印荟萃分析结果和访问荟萃分析对象中的元素时存在差异

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  • user115916  · 技术社区  · 2 年前

    我正在使用 meta 包,我正在尝试导出我的结果。然而,与直接从荟萃分析对象的元素中获取汇总效应相比,我在使用汇总函数时获得的每个亚组的汇总效应大小存在差异。

    以下是一个具有简化数据集的可复制示例:

    # Load the meta package
    library(meta)
    library(meta)
    
    # Create a data frame with meta-analysis data
    meta_data <- data.frame(
      subgroup = c("Subgroup1", "Subgroup1", "Subgroup2", "Subgroup2", "Subgroup1", "Subgroup2"),
      cor = c(0.5, 0.6, 0.7, 0.8, 0.7, 0.65),
      se = c(0.1, 0.2, 0.15, 0.25, 0.21, 0.18),
      n = c(50, 60, 70, 80, 90, 85)
    )
    
    # Perform the meta-analysis
    x <- metacor(
      cor = meta_data$cor,
      n = meta_data$n,
      data = meta_data,
      common = FALSE,
      random = TRUE,
      subgroup = subgroup
    )
    
    # Extract subgroup-specific results using 'summary'
       summary(x)
    
        summary(x)
         COR           95%-CI %W(random)  subgroup
    1 0.5000 [0.2575; 0.6833]       13.8 Subgroup1
    2 0.6000 [0.4083; 0.7410]       15.3 Subgroup1
    3 0.7000 [0.5566; 0.8029]       16.6 Subgroup2
    4 0.8000 [0.7040; 0.8673]       17.6 Subgroup2
    5 0.7000 [0.5765; 0.7922]       18.6 Subgroup1
    6 0.6500 [0.5071; 0.7581]       18.1 Subgroup2
    
    Number of studies: k = 6
    Number of observations: o = 435
    
                            COR           95%-CI     z  p-value
    Random effects model 0.6755 [0.5905; 0.7457] 11.30 < 0.0001
    
    Quantifying heterogeneity:
     tau^2 = 0.0169 [0.0000; 0.1912]; tau = 0.1302 [0.0000; 0.4373]
     I^2 = 53.6% [0.0%; 81.5%]; H = 1.47 [1.00; 2.32]
    
    Test of heterogeneity:
         Q d.f. p-value
     10.78    5  0.0558
    
    Results for subgroups (random effects model):
                           k    COR           95%-CI  tau^2    tau    Q   I^2
    subgroup = Subgroup1   3 0.6190 [0.4920; 0.7201] 0.0105 0.1022 3.24 38.3%
    subgroup = Subgroup2   3 0.7228 [0.6176; 0.8025] 0.0155 0.1247 4.35 54.0%
    
    Test for subgroup differences (random effects model):
                      Q d.f. p-value
    Between groups 1.95    1  0.1621
    
    Details on meta-analytical method:
    - Inverse variance method
    - Restricted maximum-likelihood estimator for tau^2
    - Q-Profile method for confidence interval of tau^2 and tau
    - Fisher's z transformation of correlations
    

    直接从荟萃分析对象的元素中获取亚组特异性效应大小

    cbind(k=results$k.study.w,
          COR=round(results$TE.random.w, 3), 
          `95%CI`= paste(round(results$lower.random.w, 3),
                         round(results$upper.random.w, 3),sep = "; "), 
          tau2 = round(results$tau2.w, 3),
          I2 = paste(round(results$I2.w*100, 2), "%", sep=""))
    

    结果显示

                k   COR     95%CI          tau2    I2      
    Subgroup1 "3" "0.723" "0.539; 0.908" "0.01"  "38.32%"
    Subgroup2 "3" "0.913" "0.721; 1.106" "0.016" "54%" 
    

    从摘要中查看Subgroup1和Subgroup2的汇总值,并将其与我排序的其他值的差异进行比较 cbind()
    根据摘要

                             k    COR           95%-CI  tau^2    tau    Q   I^2
       subgroup = Subgroup1   3 0.6190 [0.4920; 0.7201] 0.0105 0.1022 3.24 38.3%
        subgroup = Subgroup2   3 0.7228 [0.6176; 0.8025] 0.0155 0.1247 4.35 54.0%
    

    从元分析对象的元素访问它们

         k   COR     95%CI          tau2    I2      
    Subgroup1 "3" "0.723" "0.539; 0.908" "0.01"  "38.32%"
    Subgroup2 "3" "0.913" "0.721; 1.106" "0.016" "54%" 
    

    COR、下限值和上限值之间存在很大差异。I平方和Tau平方值相同。这看起来有bug。汇总功能是否应用了我不知道的任何额外处理或调整?或者我的分析中遗漏了什么?我们将非常感谢您的帮助。

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  •   Wolfgang    2 年前

    您提取的结果是r-to-z变换空间中的值(请注意,分析是用Fisher的z变换相关性完成的)。使用时 summary() ,将结果反转换为相关性。