我想对不同数据帧中的每一行取平均值。有谁知道使用 apply 语句更聪明的方法来做到这一点?抱歉,代码之墙。
您需要为每个 hiXXXX
文件提供一个 1000:1006 的向量,然后为列提供一个 2:13 的向量。我以前用过 mapply 来做一些像这样奇怪的事情,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)]))
}
最佳答案
由于只有 7 个数据集,我们可以将其用作 Map
的参数,然后cbind
它,并获取 rowMeans
Map(function(...) rowMeans(cbind(...)), hi1000, hi1001, hi1002, hi1003,
hi1004, hi1005, hi1006)
或者在获取列表
中的数据集后使用+
和Reduce
,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,最好使用第一个解决方案,因为 rowMeans
有 na.rm
参数。默认情况下它是FALSE
,但我们可以将其设置为TRUE
。
关于r - 使用 apply 来做到这一点的方法?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38421998/