使用 apply 来做到这一点的方法？

Question

我想对不同数据帧的每一行取平均值。有谁知道使用 apply 语句执行此操作的更聪明的方法？对不起代码墙。

每个 hiXXXX 文件需要一个 1000:1006 的矢量，然后是列的 2:13 矢量。我以前用过 mapply 来做这样奇怪的事情，所以也许可以以某种方式做到这一点？

for (i in 1:nrow(subavg)) {
    subavg[i,c(2)] <- mean(c(hi1000[i,c(2)],hi1001[i,c(2)],hi1002[i,c(2)],hi1003[i,c(2)],hi1004[i,c(2)],hi1005[i,c(2)],hi1006[i,c(2)]))
    subavg[i,c(3)] <- mean(c(hi1000[i,c(3)],hi1001[i,c(3)],hi1002[i,c(3)],hi1003[i,c(3)],hi1004[i,c(3)],hi1005[i,c(3)],hi1006[i,c(3)]))
    subavg[i,c(4)] <- mean(c(hi1000[i,c(4)],hi1001[i,c(4)],hi1002[i,c(4)],hi1003[i,c(4)],hi1004[i,c(4)],hi1005[i,c(4)],hi1006[i,c(4)]))
    subavg[i,c(5)] <- mean(c(hi1000[i,c(5)],hi1001[i,c(5)],hi1002[i,c(5)],hi1003[i,c(5)],hi1004[i,c(5)],hi1005[i,c(5)],hi1006[i,c(5)]))
    subavg[i,c(6)] <- mean(c(hi1000[i,c(6)],hi1001[i,c(6)],hi1002[i,c(6)],hi1003[i,c(6)],hi1004[i,c(6)],hi1005[i,c(6)],hi1006[i,c(6)]))
    subavg[i,c(7)] <- mean(c(hi1000[i,c(7)],hi1001[i,c(7)],hi1002[i,c(7)],hi1003[i,c(7)],hi1004[i,c(7)],hi1005[i,c(7)],hi1006[i,c(7)]))
    subavg[i,c(8)] <- mean(c(hi1000[i,c(8)],hi1001[i,c(8)],hi1002[i,c(8)],hi1003[i,c(8)],hi1004[i,c(8)],hi1005[i,c(8)],hi1006[i,c(8)]))
    subavg[i,c(9)] <- mean(c(hi1000[i,c(9)],hi1001[i,c(9)],hi1002[i,c(9)],hi1003[i,c(9)],hi1004[i,c(9)],hi1005[i,c(9)],hi1006[i,c(9)]))
    subavg[i,c(10)] <- mean(c(hi1000[i,c(10)],hi1001[i,c(10)],hi1002[i,c(10)],hi1003[i,c(10)],hi1004[i,c(10)],hi1005[i,c(10)],hi1006[i,c(10)]))
    subavg[i,c(11)] <- mean(c(hi1000[i,c(11)],hi1001[i,c(11)],hi1002[i,c(11)],hi1003[i,c(11)],hi1004[i,c(11)],hi1005[i,c(11)],hi1006[i,c(11)]))
    subavg[i,c(12)] <- mean(c(hi1000[i,c(12)],hi1001[i,c(12)],hi1002[i,c(12)],hi1003[i,c(12)],hi1004[i,c(12)],hi1005[i,c(12)],hi1006[i,c(12)]))
    subavg[i,c(13)] <- mean(c(hi1000[i,c(13)],hi1001[i,c(13)],hi1002[i,c(13)],hi1003[i,c(13)],hi1004[i,c(13)],hi1005[i,c(13)],hi1006[i,c(13)]))
}

Answer 1

因为只有 7 个数据集，我们可以将其用作 Map 的参数，然后 cbind 它，并得到 rowMeans

Map(function(...) rowMeans(cbind(...)), hi1000, hi1001, hi1002, hi1003, 
                     hi1004, hi1005, hi1006)

或者用+和Reduce得到一个list的数据集，然后除以数据集总数，即7

Reduce(`+`, mget(paste0("hi", 1000:1006)))/7

第二个解决方案更紧凑，但如果数据集中有 NA，最好使用第一个，因为 rowMeans 有 na.rm 参数。默认是FALSE，但是我们可以设置成TRUE.

使用 apply 来做到这一点的方法？

Way to do this using apply?

r

lapply

sapply

mapply