R 不同列表元素行的平均值（对应于正确的日期）
我在文章底部使用了以下数据： 这是一个列表，如下所示：R 不同列表元素行的平均值（对应于正确的日期）,r,xts,zoo,R,Xts,Zoo,我在文章底部使用了以下数据： 这是一个列表，如下所示： $ALB ALB.Close 2018-01-03 0.0019747387 2018-01-04 -0.0119843581 2018-01-05 0.0114531296 2018-01-08 0.0425817277 日期是行名称，数据有一列。dput是4个xts项目的列表。我正试图计算所有项目的平均每日回报率 名单 问题：我使用了一个lappy函数来执行此操作，但是第x列中的日期与第x列中的日期
$ALB
ALB.Close
2018-01-03 0.0019747387
2018-01-04 -0.0119843581
2018-01-05 0.0114531296
2018-01-08 0.0425817277
dputlappy data <- structure(list(NEM = structure(c(-0.00842110239640803, 0.0110381675545126,
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-0.0183538516399846, -0.0120168373425571, 0.0101974822735915,
-0.00530367719466085, -0.0306272745291913, -0.0112995552539328,
-0.0181639984003068, 0.0273939708477471, -0.031103806302724,
0.00629864977064276, 0.0068096148514476, 0.00283893227365883,
0.00456205179109137), .indexCLASS = "Date", tclass = "Date", .indexTZ = "UTC", tzone = "UTC", src = "yahoo", updated = structure(1530984643.37014, class = c("POSIXct",
"POSIXt")), class = c("xts", "zoo"), index = structure(c(1514937600,
1515024000, 1515110400, 1515369600, 1515456000, 1515542400, 1515628800,
1515715200, 1516060800, 1516147200, 1516233600, 1516320000, 1516579200,
1516665600, 1516752000, 1516838400, 1516924800, 1517184000, 1517270400,
1517356800, 1517443200, 1517529600, 1517788800, 1517875200, 1517961600,
1518048000, 1518134400, 1518393600, 1518480000, 1518566400, 1518652800,
1518739200, 1519084800, 1519171200, 1519257600, 1519344000, 1519603200,
1519689600, 1519776000, 1519862400, 1519948800, 1520208000, 1520294400,
1520380800, 1520467200, 1520553600, 1520812800, 1520899200, 1520985600,
1521072000, 1521158400, 1521417600, 1521504000, 1521590400, 1521676800,
1521763200, 1522022400, 1522108800, 1522195200, 1522281600, 1522627200,
1522713600, 1522800000, 1522886400, 1522972800, 1523232000, 1523318400,
1523404800, 1523491200, 1523577600, 1523836800, 1523923200, 1524009600,
1524096000, 1524182400, 1524441600, 1524528000, 1524614400, 1524700800,
1524787200, 1525046400, 1525132800, 1525219200, 1525305600, 1525392000,
1525651200, 1525737600, 1525824000, 1525910400, 1525996800, 1526256000,
1526342400, 1526428800, 1526515200, 1526601600, 1526860800, 1526947200,
1527033600, 1527120000, 1527206400, 1527552000, 1527638400, 1527724800,
1527811200, 1528070400, 1528156800, 1528243200, 1528329600, 1528416000,
1528675200, 1528761600, 1528848000, 1528934400, 1529020800, 1529280000,
1529366400, 1529452800, 1529539200, 1529625600, 1529884800, 1529971200,
1530057600, 1530144000, 1530230400), tzone = "UTC", tclass = "Date"), .Dim = c(124L,
1L), .Dimnames = list(NULL, "LYB.Close"))), .Names = c("NEM",
"APD", "EMN", "LYB"))
data您可以merge（）
列表中的zoo
元素。在这之后，只需运行rowMeans（）
即可
library(zoo)
library(xts)
merged <- do.call(merge, data)
merged.avg <- merge(merged, avg=rowMeans(merged))
print(head(merged.avg), digits=3)
# NEM.Close APD.Close EMN.Close LYB.Close avg
# 2018-01-03 -0.00842 0.00542 0.011391 0.000267 0.002165
# 2018-01-04 0.01104 0.00384 0.012092 0.004799 0.007942
# 2018-01-05 0.00365 0.00835 0.006198 0.013822 0.008005
# 2018-01-08 -0.00104 0.00107 0.003803 -0.002276 0.000388
# 2018-01-09 -0.00654 0.00255 -0.002157 0.009073 0.000731
# 2018-01-10 0.00602 -0.00421 -0.000206 -0.001043 0.000140
图书馆（动物园）
图书馆（xts）
合并您可以merge（）
列表中的zoo
元素。在这之后，只需运行rowMeans（）
即可
library(zoo)
library(xts)
merged <- do.call(merge, data)
merged.avg <- merge(merged, avg=rowMeans(merged))
print(head(merged.avg), digits=3)
# NEM.Close APD.Close EMN.Close LYB.Close avg
# 2018-01-03 -0.00842 0.00542 0.011391 0.000267 0.002165
# 2018-01-04 0.01104 0.00384 0.012092 0.004799 0.007942
# 2018-01-05 0.00365 0.00835 0.006198 0.013822 0.008005
# 2018-01-08 -0.00104 0.00107 0.003803 -0.002276 0.000388
# 2018-01-09 -0.00654 0.00255 -0.002157 0.009073 0.000731
# 2018-01-10 0.00602 -0.00421 -0.000206 -0.001043 0.000140
图书馆（动物园）
图书馆（xts）
合并
lappy
返回与X
长度相同的列表，其中的每个元素都是将FUN
应用到X
的相应元素的结果
因此，当您对数据
列表进行平均值
评估时：
lapply(data , FUN = mean) # calculates mean for every stock and
# returns a length of 4
#$NEM
#[1] -9.566582e-05
#
#$APD
#[1] -0.0004911925
#
#$EMN
#[1] 0.0004726381
#
#$LYB
#[1] -0.0001728595
最好以适合您分析的方式简化数据。您可以使用apply
系列中的另一个函数来执行此操作，sapply
：
d <- data.frame(sapply(data, FUN = c))
d['avg'] <- rowMeans(d)
head(d, n = 3)
# NEM APD EMN LYB avg
#1 -0.0084211024 5.423061e-03 0.0113913375 2.672636e-04 2.165140e-03
#2 0.0110381676 3.838540e-03 0.0120923396 4.798738e-03 7.941946e-03
#3 0.0036525999 8.345765e-03 0.0061977163 1.382247e-02 8.004637e-03
d
lappy
返回与X
长度相同的列表，其中的每个元素都是将FUN
应用到X
的相应元素的结果
因此，当您对数据
列表进行平均值
评估时：
lapply(data , FUN = mean) # calculates mean for every stock and
# returns a length of 4
#$NEM
#[1] -9.566582e-05
#
#$APD
#[1] -0.0004911925
#
#$EMN
#[1] 0.0004726381
#
#$LYB
#[1] -0.0001728595
最好以适合您分析的方式简化数据。您可以使用apply
系列中的另一个函数来执行此操作，sapply
：
d <- data.frame(sapply(data, FUN = c))
d['avg'] <- rowMeans(d)
head(d, n = 3)
# NEM APD EMN LYB avg
#1 -0.0084211024 5.423061e-03 0.0113913375 2.672636e-04 2.165140e-03
#2 0.0110381676 3.838540e-03 0.0120923396 4.798738e-03 7.941946e-03
#3 0.0036525999 8.345765e-03 0.0061977163 1.382247e-02 8.004637e-03
d注意1：如果时间序列“没有完全相同的索引，这将出错；2：将剥离属性，而不返回zoo
对象。在这种特定情况下，1不是问题，2只是在您关心属性（如本例中的源和上次更新）的情况下才是问题，并且后续分析步骤不希望出现zoo
对象。注意1：如果时间序列“没有完全相同的索引，2：将剥离属性而不返回zoo
对象。在这种特定情况下，1不是问题，而2只是在您关心属性（如本例中的源和上次更新）并且后续分析步骤不希望出现zoo
对象的情况下才是问题。