使用metafor包在R中进行荟萃分析

在下面的小型荟萃分析的真实示例中，如何rma从metafor包中对函数进行语法化处理才能获得结果？（随机效应，摘要统计SMD）

study,      mean1, sd1,    n1,  mean2,  sd2,   n2

Foo2000,    0.78,  0.05,   20,  0.82,   0.07,  25    
Sun2003,    0.74,  0.08,   30,  0.72,   0.05,  19    
Pric2005,   0.75,  0.12,   20,  0.74,   0.09,  29    
Rota2008,   0.62,  0.05,   24,  0.66,   0.03,  24    
Pete2008,   0.68,  0.03,   10,  0.68,   0.02,  10

r meta-analysis

— 绝望的绝望
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创建一个适当的data.frame：

df <- structure(list(study = structure(c(1L, 5L, 3L, 4L, 2L), .Label = c("Foo2000", 
"Pete2008", "Pric2005", "Rota2008", "Sun2003"), class = "factor"), 
    mean1 = c(0.78, 0.74, 0.75, 0.62, 0.68), sd1 = c(0.05, 0.08, 
    0.12, 0.05, 0.03), n1 = c(20L, 30L, 20L, 24L, 10L), mean2 = c(0.82, 
    0.72, 0.74, 0.66, 0.68), sd2 = c(0.07, 0.05, 0.09, 0.03, 
    0.02), n2 = c(25L, 19L, 29L, 24L, 10L)), .Names = c("study", 
"mean1", "sd1", "n1", "mean2", "sd2", "n2"), class = "data.frame", row.names = c(NA, 
-5L))

运行rma-function：

library(metafor)
rma(measure = "SMD", m1i = mean1, m2i = mean2, 
    sd1i = sd1, sd2i = sd2, n1i = n1, n2i = n2, 
    method = "REML", data = df)

请注意rma假定(m1i-m2i)。这导致以下单变量随机效应模型的荟萃分析：

> rma(measure = "SMD", m1i = mean1, m2i = mean2, 
+     sd1i = sd1, sd2i = sd2, n1i = n1, n2i = n2, 
+     method = "REML", data = df)

Random-Effects Model (k = 5; tau^2 estimator: REML)

tau^2 (estimate of total amount of heterogeneity): 0.1951 (SE = 0.2127)
tau (sqrt of the estimate of total heterogeneity): 0.4416
I^2 (% of total variability due to heterogeneity): 65.61%
H^2 (total variability / within-study variance):   2.91

Test for Heterogeneity: 
Q(df = 4) = 11.8763, p-val = 0.0183

Model Results:

estimate       se     zval     pval    ci.lb    ci.ub          
 -0.2513   0.2456  -1.0233   0.3061  -0.7326   0.2300          

---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

您可能想要更改估算值method，例如method = "DL"（但我会坚持REML）。

— 伯恩德·魏斯
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