如果条件为真，则将一个矩阵的值除以另一个矩阵
Divide the values from a matrix by another matrix if the conditions are true
我有三个带值的数据框。它们彼此相关并且具有相同的列名但显示不同的内容。
dat1 = cbind(a = c(2,1,3),
b = c(2,2,3),
c = c(2,2,2))
a b c
[1,] 2 2 2
[2,] 1 2 2
[3,] 3 3 2
dat2 = cbind(a = c(20, 20, 20),
b = c(30, 30, 30),
c = c(50, 50, 50))
a b c
[1,] 20 30 50
[2,] 20 30 50
[3,] 20 30 50
dat3 = cbind("1" = c(100,100,100),
"2" = c(200, 200, 200),
"3" = c(300, 300, 300))
1 2 3
[1,] 100 200 300
[2,] 100 200 300
[3,] 100 200 300
忽略 dat2 和 dat3 中值的模式。这只是为了说明。我的数据集有更多的随机值。
获得result
a b c
[1,] 0.100 0.150 0.25
[2,] 0.200 0.150 0.25
[3,] 0.067 0.067 0.25
这基本上是
1 2 3
[1,] "20/200" "30/200" "50/200"
[2,] "20/100" "30/200" "50/200"
[3,] "20/300" "30/300" "50/200"
基本原理是，如果 dat1 中的值具有值 1，则从 dat2 中取相应的值并除以 [=18= 中的相应值和列].同样，如果dat1中的值在dat1中的值为2，则从dat2中取对应的值（同"position"）除以dat3 中相应的值和列。 3.
的值也是如此
例如，让我们看一下 dat1 和 1 的值。我们只有一个数据点，值为 1，它是 dat2[2,1]，值为 20。取 20 除以 dat3[2,1]。或者查看 dat2[3,1] 的值 3 来自 dat1。取 dat2[3,1]（值为 20）并在 dat3 中除以 300，因为它在 dat1 中被赋值为 3。
有办法吗？
这是我的数据输出，分别使用 dput() 表示 dat1 和 dat3。
structure(list(DNB = c(1, 1, 1), `NORSK HYDRO` = c(2, 2, 2),
ORKLA = c(1, 1, 1), STOREBRAND = c(1, 1, 2), ATEA = c(1,
1, 1), `SCHIBSTED A` = c(2, 2, 2), BONHEUR = c(1, 1, 1),
EKORNES = c(1, 1, 1), `KONGSBERG GRUPPEN` = c(2, 2, 2), `TOMRA SYSTEMS` = c(1,
1, 1), VEIDEKKE = c(1, 1, 2), `ARENDALS FOSSEKOMPANI` = c(2,
2, 2), `OLAV THON EIEP.` = c(2, 2, 2), `PETROLEUM GEO SERVICES` = c(2,
2, 2), `SPAREBANK 1 SR BANK` = c(2, 2, 2), `STOLT-NIELSEN` = c(2,
2, 2), `ODFJELL 'A'` = c(1, 1, 1), `SPAREBANK 1 NORD-NORGE` = c(2,
2, 2), `SPAREBANK 1 SMN` = c(2, 2, 2), `WILHS.WILHELMSEN HDG.'A'` = c(2,
2, 2), `NORDEA BANK (~NK)` = c(2, 2, 2), `ATLAS COPCO 'A' (~NK)` = c(2,
2, 2), `VOLVO 'B' (~NK)` = c(2, 2, 2), `SANDVIK (~NK)` = c(2,
2, 2), `SWEDBANK 'A' (~NK)` = c(1, 1, 1), `ERICSSON 'B' (~NK)` = c(1,
1, 1), `SVENSKA HANDBKN.'A' (~NK)` = c(1, 1, 1), `HENNES & MAURITZ 'B' (~NK)` = c(2,
2, 2), `SEB 'A' (~NK)` = c(2, 2, 2), `INVESTOR 'B' (~NK)` = c(1,
1, 1), `SWEDISH MATCH (~NK)` = c(1, 1, 1), `ELECTROLUX 'B' (~NK)` = c(2,
2, 2), `SKANSKA 'B' (~NK)` = c(2, 2, 1), `SCA 'B' (~NK)` = c(2,
2, 2), `SECURITAS 'B' (~NK)` = c(2, 2, 2), `HOLMEN 'B' (~NK)` = c(2,
2, 2), `SSAB 'A' (~NK)` = c(1, 1, 1), `ERICSSON 'A' (~NK)` = c(2,
2, 2), `INVESTOR 'A' (~NK)` = c(2, 2, 2), `VOLVO 'A' (~NK)` = c(2,
2, 2), `NOVO NORDISK 'B' (~NK)` = c(2, 2, 2), `DANSKE BANK (~NK)` = c(1,
1, 1), `COLOPLAST 'B' (~NK)` = c(2, 2, 3), `CARLSBERG 'B' (~NK)` = c(2,
2, 2), `A P MOLLER - MAERSK 'B' (~NK)` = c(2, 2, 2), `TDC (~NK)` = c(2,
2, 2), `TOPDANMARK (~NK)` = c(2, 2, 2), `WILLIAM DEMANT HLDG. (~NK)` = c(3,
3, 2), `JYSKE BANK (~NK)` = c(1, 1, 1), `KOBENHAVNS LUFTHAVNE (~NK)` = c(2,
2, 1), `NKT (~NK)` = c(1, 1, 1), `ROCKWOOL 'B' (~NK)` = c(2,
2, 2), `SYDBANK (~NK)` = c(2, 2, 2), `FLSMIDTH & CO.'B' (~NK)` = c(2,
2, 1), `GN STORE NORD (~NK)` = c(2, 2, 2), `ALK-ABELLO (~NK)` = c(2,
2, 2), `BANG & OLUFSEN 'B' (~NK)` = c(3, 3, 2), `SANTA FE GROUP (~NK)` = c(2,
2, 2), `CARLSBERG 'A' (~NK)` = c(2, 2, 2), `ROCKWOOL 'A' (~NK)` = c(2,
2, 2), `NOKIA (~NK)` = c(1, 1, 1), `SAMPO 'A' (~NK)` = c(1,
1, 1), `KONE 'B' (~NK)` = c(2, 2, 2), `UPM-KYMMENE (~NK)` = c(1,
1, 1), `WARTSILA (~NK)` = c(1, 1, 1), `METSO (~NK)` = c(1,
1, 1), `STORA ENSO 'R' (~NK)` = c(2, 2, 2), `HUHTAMAKI (~NK)` = c(1,
1, 1), `FINNAIR (~NK)` = c(2, 2, 2), `KEMIRA (~NK)` = c(1,
1, 1), `UPONOR (~NK)` = c(1, 1, 1), `KESKO 'B' (~NK)` = c(1,
1, 2), `ORION 'B' (~NK)` = c(2, 2, 2), `OUTOKUMPU 'A' (~NK)` = c(2,
2, 2), `RAISIO (~NK)` = c(2, 2, 2), `TIETO OYJ (~NK)` = c(1,
1, 1), `METSA BOARD 'B' (~NK)` = c(2, 2, 2), `ORION 'A' (~NK)` = c(2,
2, 2), `STOCKMANN 'A' (~NK)` = c(2, 2, 2), `STORA ENSO 'A' (~NK)` = c(2,
2, 2)), .Names = c("DNB", "NORSK HYDRO", "ORKLA", "STOREBRAND",
"ATEA", "SCHIBSTED A", "BONHEUR", "EKORNES", "KONGSBERG GRUPPEN",
"TOMRA SYSTEMS", "VEIDEKKE", "ARENDALS FOSSEKOMPANI", "OLAV THON EIEP.",
"PETROLEUM GEO SERVICES", "SPAREBANK 1 SR BANK", "STOLT-NIELSEN",
"ODFJELL 'A'", "SPAREBANK 1 NORD-NORGE", "SPAREBANK 1 SMN", "WILHS.WILHELMSEN HDG.'A'",
"NORDEA BANK (~NK)", "ATLAS COPCO 'A' (~NK)", "VOLVO 'B' (~NK)",
"SANDVIK (~NK)", "SWEDBANK 'A' (~NK)", "ERICSSON 'B' (~NK)",
"SVENSKA HANDBKN.'A' (~NK)", "HENNES & MAURITZ 'B' (~NK)", "SEB 'A' (~NK)",
"INVESTOR 'B' (~NK)", "SWEDISH MATCH (~NK)", "ELECTROLUX 'B' (~NK)",
"SKANSKA 'B' (~NK)", "SCA 'B' (~NK)", "SECURITAS 'B' (~NK)",
"HOLMEN 'B' (~NK)", "SSAB 'A' (~NK)", "ERICSSON 'A' (~NK)", "INVESTOR 'A' (~NK)",
"VOLVO 'A' (~NK)", "NOVO NORDISK 'B' (~NK)", "DANSKE BANK (~NK)",
"COLOPLAST 'B' (~NK)", "CARLSBERG 'B' (~NK)", "A P MOLLER - MAERSK 'B' (~NK)",
"TDC (~NK)", "TOPDANMARK (~NK)", "WILLIAM DEMANT HLDG. (~NK)",
"JYSKE BANK (~NK)", "KOBENHAVNS LUFTHAVNE (~NK)", "NKT (~NK)",
"ROCKWOOL 'B' (~NK)", "SYDBANK (~NK)", "FLSMIDTH & CO.'B' (~NK)",
"GN STORE NORD (~NK)", "ALK-ABELLO (~NK)", "BANG & OLUFSEN 'B' (~NK)",
"SANTA FE GROUP (~NK)", "CARLSBERG 'A' (~NK)", "ROCKWOOL 'A' (~NK)",
"NOKIA (~NK)", "SAMPO 'A' (~NK)", "KONE 'B' (~NK)", "UPM-KYMMENE (~NK)",
"WARTSILA (~NK)", "METSO (~NK)", "STORA ENSO 'R' (~NK)", "HUHTAMAKI (~NK)",
"FINNAIR (~NK)", "KEMIRA (~NK)", "UPONOR (~NK)", "KESKO 'B' (~NK)",
"ORION 'B' (~NK)", "OUTOKUMPU 'A' (~NK)", "RAISIO (~NK)", "TIETO OYJ (~NK)",
"METSA BOARD 'B' (~NK)", "ORION 'A' (~NK)", "STOCKMANN 'A' (~NK)",
"STORA ENSO 'A' (~NK)"), row.names = c(NA, 3L), class = "data.frame")
structure(c(572008.53, 617720.24, 654277.81, 686839.49, 736058.9,
714108.91, 8344.65, 9753.26, 5407.72), .Dim = c(3L, 3L), .Dimnames = list(
c("1", "2", "3"), c("1", "2", "3")))
这里是实际数据的维度。
dat1 使用 dim() 得到
[1] 252 80
dat2 使用 dim() 得到
[1] 252 80
dat3 使用 dim() 得到
[1] 252 3
我相信你正在寻找这样的东西：
您需要将 dat1 和 dat2 转换为矩阵而不是数据帧。
dat1 <- as.matrix(dat1)
dat2 <- as.matrix(dat2)
dfx <- do.call('rbind', lapply(1:nrow(dat1),function(x)dat3[x,dat1[x,]]))
要获得比率，您需要将 dat2 除以 dfx（因为没有提供 dat2，我将此留给 OP 解决）：
dat2/dfx #This should give you final answer
输出:
> dat2/dfx
a b c
[1,] 0.10000000 0.15 0.25
[2,] 0.20000000 0.15 0.25
[3,] 0.06666667 0.10 0.25
示例输出
> do.call('rbind', lapply(1:nrow(dat1),function(x)dat3[x,c(dat1[x,])]))
1 2 1 1
[1,] 572008.5 686839.5 572008.5 572008.5
[2,] 617720.2 736058.9 617720.2 617720.2
[3,] 654277.8 714108.9 654277.8 714108.9
1 2 1 1
[1,] 572008.5 686839.5 572008.5 572008.5
[2,] 617720.2 736058.9 617720.2 617720.2
[3,] 654277.8 714108.9 654277.8 654277.8
我有三个带值的数据框。它们彼此相关并且具有相同的列名但显示不同的内容。
dat1 = cbind(a = c(2,1,3),
b = c(2,2,3),
c = c(2,2,2))
a b c
[1,] 2 2 2
[2,] 1 2 2
[3,] 3 3 2
dat2 = cbind(a = c(20, 20, 20),
b = c(30, 30, 30),
c = c(50, 50, 50))
a b c
[1,] 20 30 50
[2,] 20 30 50
[3,] 20 30 50
dat3 = cbind("1" = c(100,100,100),
"2" = c(200, 200, 200),
"3" = c(300, 300, 300))
1 2 3
[1,] 100 200 300
[2,] 100 200 300
[3,] 100 200 300
忽略 dat2 和 dat3 中值的模式。这只是为了说明。我的数据集有更多的随机值。
获得result
a b c
[1,] 0.100 0.150 0.25
[2,] 0.200 0.150 0.25
[3,] 0.067 0.067 0.25
这基本上是
1 2 3
[1,] "20/200" "30/200" "50/200"
[2,] "20/100" "30/200" "50/200"
[3,] "20/300" "30/300" "50/200"
基本原理是，如果 dat1 中的值具有值 1，则从 dat2 中取相应的值并除以 [=18= 中的相应值和列].同样，如果dat1中的值在dat1中的值为2，则从dat2中取对应的值（同"position"）除以dat3 中相应的值和列。 3.
例如，让我们看一下 dat1 和 1 的值。我们只有一个数据点，值为 1，它是 dat2[2,1]，值为 20。取 20 除以 dat3[2,1]。或者查看 dat2[3,1] 的值 3 来自 dat1。取 dat2[3,1]（值为 20）并在 dat3 中除以 300，因为它在 dat1 中被赋值为 3。
有办法吗？
这是我的数据输出，分别使用 dput() 表示 dat1 和 dat3。
structure(list(DNB = c(1, 1, 1), `NORSK HYDRO` = c(2, 2, 2),
ORKLA = c(1, 1, 1), STOREBRAND = c(1, 1, 2), ATEA = c(1,
1, 1), `SCHIBSTED A` = c(2, 2, 2), BONHEUR = c(1, 1, 1),
EKORNES = c(1, 1, 1), `KONGSBERG GRUPPEN` = c(2, 2, 2), `TOMRA SYSTEMS` = c(1,
1, 1), VEIDEKKE = c(1, 1, 2), `ARENDALS FOSSEKOMPANI` = c(2,
2, 2), `OLAV THON EIEP.` = c(2, 2, 2), `PETROLEUM GEO SERVICES` = c(2,
2, 2), `SPAREBANK 1 SR BANK` = c(2, 2, 2), `STOLT-NIELSEN` = c(2,
2, 2), `ODFJELL 'A'` = c(1, 1, 1), `SPAREBANK 1 NORD-NORGE` = c(2,
2, 2), `SPAREBANK 1 SMN` = c(2, 2, 2), `WILHS.WILHELMSEN HDG.'A'` = c(2,
2, 2), `NORDEA BANK (~NK)` = c(2, 2, 2), `ATLAS COPCO 'A' (~NK)` = c(2,
2, 2), `VOLVO 'B' (~NK)` = c(2, 2, 2), `SANDVIK (~NK)` = c(2,
2, 2), `SWEDBANK 'A' (~NK)` = c(1, 1, 1), `ERICSSON 'B' (~NK)` = c(1,
1, 1), `SVENSKA HANDBKN.'A' (~NK)` = c(1, 1, 1), `HENNES & MAURITZ 'B' (~NK)` = c(2,
2, 2), `SEB 'A' (~NK)` = c(2, 2, 2), `INVESTOR 'B' (~NK)` = c(1,
1, 1), `SWEDISH MATCH (~NK)` = c(1, 1, 1), `ELECTROLUX 'B' (~NK)` = c(2,
2, 2), `SKANSKA 'B' (~NK)` = c(2, 2, 1), `SCA 'B' (~NK)` = c(2,
2, 2), `SECURITAS 'B' (~NK)` = c(2, 2, 2), `HOLMEN 'B' (~NK)` = c(2,
2, 2), `SSAB 'A' (~NK)` = c(1, 1, 1), `ERICSSON 'A' (~NK)` = c(2,
2, 2), `INVESTOR 'A' (~NK)` = c(2, 2, 2), `VOLVO 'A' (~NK)` = c(2,
2, 2), `NOVO NORDISK 'B' (~NK)` = c(2, 2, 2), `DANSKE BANK (~NK)` = c(1,
1, 1), `COLOPLAST 'B' (~NK)` = c(2, 2, 3), `CARLSBERG 'B' (~NK)` = c(2,
2, 2), `A P MOLLER - MAERSK 'B' (~NK)` = c(2, 2, 2), `TDC (~NK)` = c(2,
2, 2), `TOPDANMARK (~NK)` = c(2, 2, 2), `WILLIAM DEMANT HLDG. (~NK)` = c(3,
3, 2), `JYSKE BANK (~NK)` = c(1, 1, 1), `KOBENHAVNS LUFTHAVNE (~NK)` = c(2,
2, 1), `NKT (~NK)` = c(1, 1, 1), `ROCKWOOL 'B' (~NK)` = c(2,
2, 2), `SYDBANK (~NK)` = c(2, 2, 2), `FLSMIDTH & CO.'B' (~NK)` = c(2,
2, 1), `GN STORE NORD (~NK)` = c(2, 2, 2), `ALK-ABELLO (~NK)` = c(2,
2, 2), `BANG & OLUFSEN 'B' (~NK)` = c(3, 3, 2), `SANTA FE GROUP (~NK)` = c(2,
2, 2), `CARLSBERG 'A' (~NK)` = c(2, 2, 2), `ROCKWOOL 'A' (~NK)` = c(2,
2, 2), `NOKIA (~NK)` = c(1, 1, 1), `SAMPO 'A' (~NK)` = c(1,
1, 1), `KONE 'B' (~NK)` = c(2, 2, 2), `UPM-KYMMENE (~NK)` = c(1,
1, 1), `WARTSILA (~NK)` = c(1, 1, 1), `METSO (~NK)` = c(1,
1, 1), `STORA ENSO 'R' (~NK)` = c(2, 2, 2), `HUHTAMAKI (~NK)` = c(1,
1, 1), `FINNAIR (~NK)` = c(2, 2, 2), `KEMIRA (~NK)` = c(1,
1, 1), `UPONOR (~NK)` = c(1, 1, 1), `KESKO 'B' (~NK)` = c(1,
1, 2), `ORION 'B' (~NK)` = c(2, 2, 2), `OUTOKUMPU 'A' (~NK)` = c(2,
2, 2), `RAISIO (~NK)` = c(2, 2, 2), `TIETO OYJ (~NK)` = c(1,
1, 1), `METSA BOARD 'B' (~NK)` = c(2, 2, 2), `ORION 'A' (~NK)` = c(2,
2, 2), `STOCKMANN 'A' (~NK)` = c(2, 2, 2), `STORA ENSO 'A' (~NK)` = c(2,
2, 2)), .Names = c("DNB", "NORSK HYDRO", "ORKLA", "STOREBRAND",
"ATEA", "SCHIBSTED A", "BONHEUR", "EKORNES", "KONGSBERG GRUPPEN",
"TOMRA SYSTEMS", "VEIDEKKE", "ARENDALS FOSSEKOMPANI", "OLAV THON EIEP.",
"PETROLEUM GEO SERVICES", "SPAREBANK 1 SR BANK", "STOLT-NIELSEN",
"ODFJELL 'A'", "SPAREBANK 1 NORD-NORGE", "SPAREBANK 1 SMN", "WILHS.WILHELMSEN HDG.'A'",
"NORDEA BANK (~NK)", "ATLAS COPCO 'A' (~NK)", "VOLVO 'B' (~NK)",
"SANDVIK (~NK)", "SWEDBANK 'A' (~NK)", "ERICSSON 'B' (~NK)",
"SVENSKA HANDBKN.'A' (~NK)", "HENNES & MAURITZ 'B' (~NK)", "SEB 'A' (~NK)",
"INVESTOR 'B' (~NK)", "SWEDISH MATCH (~NK)", "ELECTROLUX 'B' (~NK)",
"SKANSKA 'B' (~NK)", "SCA 'B' (~NK)", "SECURITAS 'B' (~NK)",
"HOLMEN 'B' (~NK)", "SSAB 'A' (~NK)", "ERICSSON 'A' (~NK)", "INVESTOR 'A' (~NK)",
"VOLVO 'A' (~NK)", "NOVO NORDISK 'B' (~NK)", "DANSKE BANK (~NK)",
"COLOPLAST 'B' (~NK)", "CARLSBERG 'B' (~NK)", "A P MOLLER - MAERSK 'B' (~NK)",
"TDC (~NK)", "TOPDANMARK (~NK)", "WILLIAM DEMANT HLDG. (~NK)",
"JYSKE BANK (~NK)", "KOBENHAVNS LUFTHAVNE (~NK)", "NKT (~NK)",
"ROCKWOOL 'B' (~NK)", "SYDBANK (~NK)", "FLSMIDTH & CO.'B' (~NK)",
"GN STORE NORD (~NK)", "ALK-ABELLO (~NK)", "BANG & OLUFSEN 'B' (~NK)",
"SANTA FE GROUP (~NK)", "CARLSBERG 'A' (~NK)", "ROCKWOOL 'A' (~NK)",
"NOKIA (~NK)", "SAMPO 'A' (~NK)", "KONE 'B' (~NK)", "UPM-KYMMENE (~NK)",
"WARTSILA (~NK)", "METSO (~NK)", "STORA ENSO 'R' (~NK)", "HUHTAMAKI (~NK)",
"FINNAIR (~NK)", "KEMIRA (~NK)", "UPONOR (~NK)", "KESKO 'B' (~NK)",
"ORION 'B' (~NK)", "OUTOKUMPU 'A' (~NK)", "RAISIO (~NK)", "TIETO OYJ (~NK)",
"METSA BOARD 'B' (~NK)", "ORION 'A' (~NK)", "STOCKMANN 'A' (~NK)",
"STORA ENSO 'A' (~NK)"), row.names = c(NA, 3L), class = "data.frame")
structure(c(572008.53, 617720.24, 654277.81, 686839.49, 736058.9,
714108.91, 8344.65, 9753.26, 5407.72), .Dim = c(3L, 3L), .Dimnames = list(
c("1", "2", "3"), c("1", "2", "3")))
这里是实际数据的维度。
dat1 使用 dim() 得到
[1] 252 80
dat2 使用 dim() 得到
[1] 252 80
dat3 使用 dim() 得到
[1] 252 3
我相信你正在寻找这样的东西：
您需要将 dat1 和 dat2 转换为矩阵而不是数据帧。
dat1 <- as.matrix(dat1)
dat2 <- as.matrix(dat2)
dfx <- do.call('rbind', lapply(1:nrow(dat1),function(x)dat3[x,dat1[x,]]))
要获得比率，您需要将 dat2 除以 dfx（因为没有提供 dat2，我将此留给 OP 解决）：
dat2/dfx #This should give you final answer
输出:
> dat2/dfx
a b c
[1,] 0.10000000 0.15 0.25
[2,] 0.20000000 0.15 0.25
[3,] 0.06666667 0.10 0.25
示例输出
> do.call('rbind', lapply(1:nrow(dat1),function(x)dat3[x,c(dat1[x,])]))
1 2 1 1
[1,] 572008.5 686839.5 572008.5 572008.5
[2,] 617720.2 736058.9 617720.2 617720.2
[3,] 654277.8 714108.9 654277.8 714108.9
1 2 1 1
[1,] 572008.5 686839.5 572008.5 572008.5
[2,] 617720.2 736058.9 617720.2 617720.2
[3,] 654277.8 714108.9 654277.8 654277.8