我有这两个列表。
library(tidyquant)
standin1<-cbind(c("AAPL","JPM"),c("MSFT","FB"))
wantedstocks<-unique(c(standin1))
eeee<-tq_get(wantedstocks,"stock.prices", from = "2015-01-01", to = "2015-01-27")
prices_tbl1<-split(eeee,eeee$symbol)
prices_tbl2<-lapply(prices_tbl1,function(x) xts(x$adjusted,x$date))
prices_tbl3<-do.call(merge,prices_tbl2) #combine xts objects
split_prices_into_years<-split.xts(prices_tbl3,f="weeks",k=1)
jjj<-split_prices_into_years[2:4]
tt<-lapply(jjj, function(x) x[1,])
jjj
[[1]]
FB JPM MSFT AAPL
2015-01-05 77.19 55.15255 42.39050 99.43689
2015-01-06 76.15 53.72248 41.76831 99.44625
2015-01-07 76.15 53.80447 42.29900 100.84071
2015-01-08 78.18 55.00679 43.54335 104.71526
2015-01-09 77.74 54.05040 43.17737 104.82755
[[2]]
FB JPM MSFT AAPL
2015-01-12 76.72 53.58586 42.63753 102.24452
2015-01-13 76.45 53.59496 42.41794 103.15231
2015-01-14 76.28 51.74593 42.05195 102.75926
2015-01-15 74.05 50.08816 41.61278 99.97034
2015-01-16 75.18 50.94437 42.30815 99.19356
[[3]]
FB JPM MSFT AAPL
2015-01-20 76.24 50.74397 42.44539 101.7485
2015-01-21 76.74 50.90794 42.01535 102.5253
2015-01-22 77.65 52.45639 43.12246 105.1925
2015-01-23 77.83 51.62751 43.16822 105.7353
> tt
[[1]]
FB JPM MSFT AAPL
2015-01-05 77.19 55.15255 42.3905 99.43689
[[2]]
FB JPM MSFT AAPL
2015-01-12 76.72 53.58586 42.63753 102.2445
[[3]]
FB JPM MSFT AAPL
2015-01-20 76.24 50.74397 42.44539 101.7485
我想将 jjj 列表中的每列与 tt 列表中的相应元素相乘。因此
FB
2015-01-05 77.19 2015-01-05 5958.296
2015-01-06 76.15 FB 2015-01-06 5878.019
2015-01-07 76.15 (X) 77.19 = 2015-01-07 5878.019
2015-01-08 78.18 2015-01-08 6034.714
2015-01-09 77.74 2015-01-09 6000.751
FB
2015-01-12 76.72 2015-01-12 5958.296
2015-01-13 76.45 FB 2015-01-13 5865.244
2015-01-14 76.28 (X) 76.72 = 2015-01-14 5852.202
2015-01-15 74.05 2015-01-15 5681.116
2015-01-16 75.18 2015-01-16 5767.81
等等。所以我希望输出与 jjj 的形式相同，只是 tt 中的元素相乘，如上所示。任何帮助真的非常感谢。
不是很漂亮，但通过循环完成工作：
result_list <- list(length=length(jjj))
for(i in 1:length(jjj)) {
result_list[[i]] <- as.matrix(jjj[[i]]) * matrix(rep(tt[[i]], nrow(jjj[[i]])), ncol=ncol(tt[[i]]), byrow=T)
}
或带有sapply，但不是更漂亮的：
sapply(1:length(jjj), function(i) as.matrix(jjj[[i]]) * matrix(rep(tt[[i]], nrow(jjj[[i]])), ncol=ncol(tt[[i]]), byrow=T))
如果我们将每个matrix或data.frame的维度从jjj规范化为tt的维度，乘积将简单地将列表中的每个元素乘以另一个元素。
首先，我们整理了一个更简单的数据示例
list1 <- list( data.frame(COL1 = c(1,2,3,4),
COL2 = c(5,6,7,8)),
data.frame(COL1 = c(2,2,2,2),
COL2 = c(3,3,3,3)))
list2 <- list( data.frame(COL1 = c(2),
COL2 = c(3)),
data.frame(COL1 = c(3),
COL2 = c(2))
)
> list1
[[1]]
COL1 COL2
1 1 5
2 2 6
3 3 7
4 4 8
[[2]]
COL1 COL2
1 2 3
2 2 3
3 2 3
4 2 3
> list2
[[1]]
COL1 COL2
1 2 3
[[2]]
COL1 COL2
1 3 2
multm函数将 data.framedf2规范化为df1的维度，并返回两者之间的乘积。
multm <- function(df1, df2) {
m1 <- as.matrix(df1)
m2 <- as.matrix(df2)
m2 <- matrix(rep(as.vector(m2), nrow(m1)), ncol=ncol(m1), byrow=T)
as.data.frame(m1 * m2)
}
最后，我们将函数应用于两个列表的每个元素
lapply(1:length(list1), FUN=function(x){multm(list1[[x]], list2[[x]])})
[[1]]
COL1 COL2
1 2 15
2 4 18
3 6 21
4 8 24
[[2]]
COL1 COL2
1 6 6
2 6 6
3 6 6
4 6 6
你可以试试：
lapply(1:length(jjj),
function(k) mapply("*", as.data.frame(jjj[[k]]), jjj[[k]][1,]))
#returns
[[1]]
AAPL FB JPM MSFT
[1,] 9887.695 5958.296 3041.803 1796.954
[2,] 9888.626 5878.019 2962.932 1770.580
[3,] 10027.287 5878.019 2967.454 1793.076
[4,] 10412.559 6034.714 3033.765 1845.824
[5,] 10423.726 6000.751 2981.017 1830.310
[[2]]
AAPL FB JPM MSFT
[1,] 10453.94 5885.959 2871.444 1817.959
[2,] 10546.76 5865.244 2871.932 1808.596
[3,] 10506.57 5852.202 2772.850 1792.991
[4,] 10221.42 5681.116 2684.017 1774.266
[5,] 10142.00 5767.810 2729.898 1803.915
[[3]]
AAPL FB JPM MSFT
[1,] 10352.76 5812.537 2574.951 1801.611
[2,] 10431.80 5850.657 2583.271 1783.358
[3,] 10703.18 5920.036 2661.846 1830.350
[4,] 10758.41 5933.759 2619.785 1832.292