如何使用“ metafor” R包一次對數千個基因進行薈萃分析

Question

我正在嘗試使用R包“ metafor”對許多基因進行薈萃分析，我知道一次只能做一個基因，但是對成千上萬個基因這樣做是很荒謬的。 有人可以幫我這個忙！ 感謝任何建議！

我分別獲得了名為“ se_summary”和“ HR_summary”的所有基因的se和HR的所有結果。 我需要使用來自五個研究“ ICGC，TCGA，G71，G62，G8”的這些基因的se和HR作為輸入來進行薈萃分析。

我用來對一個基因進行元分析的代碼（以基因AAK1為例）是：

library(metafor) 
se.AAK1 <- as.numeric(se_summary[rownames(se_summary) == 'AAK1',][,-1])
HR.AAK1 <- as.numeric(HR_summary[rownames(HR_summary) == 'AAK1',][,-1])
beta.AAK1 <- log(HR.AAK1)

####First I need to use the random model to see if the test for Heterogeneity is significant or not.

pool.AAK1 <- rma(beta.AAK1, sei=se.AAK1) 
summary(pool.AAK1)
#### and this gives the following output:

#>Random-Effects Model (k = 5; tau^2 estimator: REML)
#>  logLik  deviance       AIC       BIC      AICc 
#> -2.5686    5.1372    9.1372    7.9098   21.1372   

#>tau^2 (estimated amount of total heterogeneity): 0.0870 (SE = 0.1176)
#>tau (square root of estimated tau^2 value):      0.2950
#>I^2 (total heterogeneity / total variability):   53.67%
#>H^2 (total variability / sampling variability):  2.16

#>Test for Heterogeneity:
#>Q(df = 4) = 8.5490, p-val = 0.0734

#>Model Results:
#>estimate      se     zval    pval    ci.lb   ci.ub 
#> -0.3206  0.1832  -1.7500  0.0801  -0.6797  0.0385  . 
#>---
#>Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


####If the I^2 > 50%, we still use the Random-effect Model but if the I^2 <= 50%, we then use the Fixed-effect Model 

pool.AAK1 <- rma(beta.AAK1, sei=se.AAK1, method="FE") 
summary(pool.AAK1)
####this gives the following output:

#>Fixed-Effects Model (k = 5)
#> logLik  deviance       AIC       BIC      AICc 
#> -2.5793    8.5490    7.1587    6.7681    8.4920   

#>Test for Heterogeneity:
#>Q(df = 4) = 8.5490, p-val = 0.0734

#>Model Results:
#>estimate      se     zval    pval    ci.lb    ci.ub 
#> -0.2564  0.1191  -2.1524  0.0314  -0.4898  -0.0229  * 
#>---
#>Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

如果我只有一個基因，則此方法效果很好，但是我需要一次對所有這些基因進行所有操作，然后導出包括“異質性p-val”在內的輸出，並且所有模型結果“ estimate，se，zval， pval，ci.lb，ci.ub“到一個.txt文件，一個基因的每一行，輸出應如下所示：

Gene_symbol Heterogeneity_p-val estimate    se      zval    pval    ci.lb   ci.ub
AAK1        0.0734              -0.2564    0.1191   -2.1524 0.0314  -0.4898 -0.0229
A2M         0.9664               0.1688    0.1173    1.4388 0.1502  -0.0611 0.3987

如果需要，這里有一個樣例數據“ se_summary”

Gene_symbol ICGC_se TCGA_se G71_se  G62_se  G8_se
A1CF        0.312   0.21    0.219   0.292   0.381
A2M         0.305   0.21    0.219   0.292   0.387
A2ML1       0.314   0.211   0.222   0.289   0.389
A4GALT      0.305   0.21    0.225   0.288   0.388
A4GNT       0.306   0.211   0.222   0.288   0.385
AAAS        0.308   0.213   0.223   0.298   0.38
AACS        0.307   0.209   0.221   0.287   0.38
AADAC       0.302   0.212   0.221   0.293   0.404
AADAT       0.308   0.214   0.22    0.288   0.391
AAK1        0.304   0.209   0.22    0.303   0.438
AAMP        0.303   0.211   0.222   0.288   0.394

和一塊樣本數據“ HR_summary”

Gene_symbol ICGC_HR TCGA_HR G71_HR  G62_HR  G8_HR
A1CF        1.689   1.427   0.864   1.884   1.133
A2M         1.234   1.102   1.11    1.369   1.338
A2ML1       0.563   0.747   0.535   1.002   0.752
A4GALT      0.969   0.891   0.613   0.985   0.882
A4GNT       1.486   0.764   1.051   1.317   1.465
AAAS        1.51    1.178   1.076   0.467   0.681
AACS        1.4     1.022   1.255   1.006   1.416
AADAC       0.979   0.642   1.236   1.581   1.234
AADAT       1.366   1.405   1.18    1.057   1.408
AAK1        1.04    0.923   0.881   0.469   0.329
AAMP        1.122   0.639   1.473   0.964   1.284

Answer 1

將具有效果大小和se數據的長格式數據並排放置，然后使用split並對每個參數應用rma 。 您可以僅針對rma對象制作自己的broom的tidy函數版本。

library(metafor) 
library(reshape)
se_summary<-read.table(text="
Gene_symbol ICGC_se TCGA_se G71_se  G62_se  G8_se
AADAT       0.308   0.214   0.22    0.288   0.391
AAK1        0.304   0.209   0.22    0.303   0.438
AAMP        0.303   0.211   0.222   0.288   0.394
",header=T)

HR_summary<-read.table(text="
Gene_symbol ICGC_HR TCGA_HR G71_HR  G62_HR  G8_HR
AADAT       0.308   0.214   0.22    0.288   0.391
AAK1        0.304   0.209   0.22    0.303   0.438
AAMP        0.303   0.211   0.222   0.288   0.394
                       ",header=T)

HR_summary<-melt(HR_summary,id.vars = "Gene_symbol")%>%
  mutate(.,variable=sapply(strsplit(as.character(variable), split='_', fixed=TRUE), function(x) (x[1])))%>%
  rename(gene=variable)
se_summary<-melt(se_summary,id.vars = "Gene_symbol")%>%
  mutate(.,variable=sapply(strsplit(as.character(variable), split='_', fixed=TRUE), function(x) (x[1])))%>%
  rename(gene=variable)
HR_summary<-merge(HR_summary,se_summary,by=c("Gene_symbol","gene"),suffixes=c(".HR",".se"))



tidy.rma<-function(x) {
  return(data.frame(estimate=x$b,se=x$se,zval=x$zval,ci.lb=x$ci.lb,ci.ub=x$ci.ub,k=x$k,Heterog_pv=x$QEp#the main stuff: overall ES, etc
                    #variance components( random effects stuff): nlvls is n sites
  )) #test for heterogeneity q value and p-value
}



rbindlist(lapply(split(HR_summary, droplevels(HR_summary$Gene_symbol)),
                 function(x)with(x, tidy.rma(rma(yi=value.HR, sei=value.se,method="FE")))),idcol = "Gene_symbol2")

Answer 2

要點1：如果您的數據是從不同人群中收集的，則不應使用固定效應模型。 因為人力資源在您的人群中可能有所不同。

要點2：如果將HR轉換為log（HR），則應該計算log（HR）的SE。

您的數據：

  se_summary=data.frame(
    Gene_symbol=c("A1CF","A2M","A2ML1","A4GALT","A4GNT","AAAS","AACS","AADAC","AADAT","AAK1","AAMP"),
    ICGC_se=c(0.312,0.305,0.314,0.305,0.306,0.308,0.307,0.302,0.308,0.304,0.303),
    TCGA_se=c(0.21,0.21,0.211,0.21,0.211,0.213,0.209,0.212,0.214,0.209,0.211),
    G71_se=c(0.219,0.219,0.222,0.225,0.222,0.223,0.221,0.221,0.22,0.22,0.222),
    G62_se=c(0.292,0.292,0.289,0.288,0.288,0.298,0.287,0.293,0.288,0.303,0.288),
    G8_se=c(0.381,0.387,0.389,0.388,0.385,0.38,0.38,0.404,0.391,0.438,0.394))

和

 HR_summary=data.frame(
    Gene_symbol=c("A1CF","A2M","A2ML1","A4GALT","A4GNT","AAAS","AACS","AADAC","AADAT","AAK1","AAMP"),
    ICGC_HR=c(1.689,1.234,0.563,0.969,1.486,1.51,1.4,0.979,1.366,1.04,1.122),
    TCGA_HR=c(1.427,1.102,0.747,0.891,0.764,1.178,1.022,0.642,1.405,0.923,0.639),
    G71_HR=c(0.864,1.11,0.535,0.613,1.051,1.076,1.255,1.236,1.18,0.881,1.473),
    G62_HR=c(1.884,1.369,1.002,0.985,1.317,0.467,1.006,1.581,1.057,0.469,0.964),
    G8_HR=c(1.133,1.338,0.752,0.882,1.465,0.681,1.416,1.234,1.408,0.329,1.284))

1）合並數據

data=cbind(se_summary,log(HR_summary[,-1]))

2）計算meta-log HR的功能

met=function(x) {
y=rma(as.numeric(x[7:11]), sei=as.numeric(x[2:6]))
y=c(y$b,y$beta,y$se,y$zval,y$pval,y$ci.lb,y$ci.ub,y$tau2,y$I2)
y
}

3）對所有行執行功能

results=data.frame(t(apply(data,1,met)))
rownames(results)=rownames(data)
colnames(results)=c("b","beta","se","zval","pval","ci.lb","ci.ub","tau2","I2")

4）結果

> results
                 b        beta        se       zval        pval
A1CF    0.27683114  0.27683114 0.1538070  1.7998601 0.071882735
A2M     0.16877042  0.16877042 0.1172977  1.4388214 0.150201136
A2ML1  -0.37676308 -0.37676308 0.1182825 -3.1852811 0.001446134
A4GALT -0.18975044 -0.18975044 0.1179515 -1.6087159 0.107678477
A4GNT   0.09500277  0.09500277 0.1392486  0.6822528 0.495079085
AAAS   -0.07012629 -0.07012629 0.2000932 -0.3504680 0.725987468
AACS    0.15333550  0.15333550 0.1170061  1.3104915 0.190029610
AADAC   0.04902471  0.04902471 0.1738017  0.2820727 0.777887764
AADAT   0.23785528  0.23785528 0.1181503  2.0131593 0.044097875
AAK1   -0.32062727 -0.32062727 0.1832183 -1.7499744 0.080122725
AAMP    0.02722082  0.02722082 0.1724461  0.1578512 0.874574077
              ci.lb       ci.ub       tau2       I2
A1CF   -0.024625107  0.57828740 0.04413257 37.89339
A2M    -0.061128821  0.39866965 0.00000000  0.00000
A2ML1  -0.608592552 -0.14493360 0.00000000  0.00000
A4GALT -0.420931120  0.04143024 0.00000000  0.00000
A4GNT  -0.177919527  0.36792508 0.02455208 25.35146
AAAS   -0.462301836  0.32204926 0.12145183 62.23915
AACS   -0.075992239  0.38266324 0.00000000  0.00000
AADAC  -0.291620349  0.38966978 0.07385974 50.18761
AADAT   0.006285038  0.46942552 0.00000000  0.00000
AAK1   -0.679728455  0.03847392 0.08700387 53.66905
AAMP   -0.310767314  0.36520895 0.07266674 50.07330