我想将不同货币的某些价格转换为特定货币.假设我有这个:
library(data.table)
set.seed(100)
DT <- data.table(day=1:10, price=runif(10), currency=c("aud","eur"),
aud=runif(10) + 1, eur=runif(10) + 1.5)
DT
day price currency aud eur
1: 1 0.30776611 aud 1.624996 2.035811
2: 2 0.25767250 eur 1.882166 2.210804
3: 3 0.55232243 aud 1.280354 2.038349
4: 4 0.05638315 eur 1.398488 2.248972
5: 5 0.46854928 aud 1.762551 1.920101
6: 6 0.48377074 eur 1.669022 1.671420
7: 7 0.81240262 aud 1.204612 2.270302
8: 8 0.37032054 eur 1.357525 2.381954
9: 9 0.54655860 aud 1.359475 2.049097
10: 10 0.17026205 eur 1.690291 1.777724
每天的价格以货币栏中显示的相应货币表示.所以第一天的0.30776611是澳元(澳元),欧元(欧元)是0.25767250.列aud并eur以美元显示各货币的汇率.如何以某种data.table方式创建以美元表示的新价格列？
我需要price根据相应的列名称进行多次currency以获取此信息:
DT
day price currency aud eur price.in.usd
1: 1 0.30776611 aud 1.624996 2.035811 0.5001187
2: 2 0.25767250 eur 1.882166 2.210804 0.5696634
3: 3 0.55232243 aud 1.280354 2.038349 0.7071682
4: 4 0.05638315 eur 1.398488 2.248972 0.1268041
5: 5 0.46854928 aud 1.762551 1.920101 0.825842
6: 6 0.48377074 eur 1.669022 1.671420 0.8085841
7: 7 0.81240262 aud 1.204612 2.270302 0.9786299
8: 8 0.37032054 eur 1.357525 2.381954 0.8820865
9: 9 0.54655860 aud 1.359475 2.049097 0.7430328
10: 10 0.17026205 eur 1.690291 1.777724 0.3026789
因此,第一天我乘以price * aud = 0.30776611 * 1.624996,因为价格aud在currency列中,而在第二天price * eur = 0.25767250 * 2.210804出于同样的原因.
真实数据包括大约40种货币,因此多次ifelse()创建箭头反模式不是很方便.
目前,通过我的数据的子样本,我有这样的:
DT.all[, price := ifelse(curcdd=="AUD", adj.price * AUD,
ifelse(curcdd=="BEF", adj.price * BEF,
ifelse(curcdd=="BGN", adj.price * BGN,
ifelse(curcdd=="CHF", adj.price * CHF,
ifelse(curcdd=="CZK", adj.price * CZK,
ifelse(curcdd=="DEM", adj.price * DEM,
ifelse(curcdd=="EUR", adj.price * EUR,
ifelse(curcdd=="FRF", adj.price * FRF,
ifelse(curcdd=="GBP", adj.price * GBP,
ifelse(curcdd=="ILS", adj.price * ILS,
ifelse(curcdd=="JPY", adj.price * JPY,
ifelse(curcdd=="NLG", adj.price * NLG,
ifelse(curcdd=="NOK", adj.price * NOK,
ifelse(curcdd=="PLN", adj.price * PLN,
ifelse(curcdd=="SEK", adj.price * SEK,
ifelse(curcdd=="SGD", adj.price * SGD,
ifelse(curcdd=="USD", adj.price, NA)))))))))))))))))]
哪个有效,但它只有大约20种货币,做所有这些(约40种)肯定不优雅......
非常感谢你!
[编辑] 使用get我在 Matthew Dowle 的回答中看到的列名引用的值的想法，这似乎是有效的：
setkey(DT, currency)
DT[ , cvt := .SD[, get(currency)]*price, by=currency]
DT
day price currency aud eur cvt
1: 1 0.30776611 aud 1.624996 2.035811 0.5001188
2: 3 0.55232243 aud 1.280354 2.038349 0.7071681
3: 5 0.46854928 aud 1.762551 1.920101 0.8258420
4: 7 0.81240262 aud 1.204612 2.270302 0.9786301
5: 9 0.54655860 aud 1.359475 2.049097 0.7430328
6: 2 0.25767250 eur 1.882166 2.210804 0.5696634
7: 4 0.05638315 eur 1.398488 2.248972 0.1268041
8: 6 0.48377074 eur 1.669022 1.671420 0.8085842
9: 8 0.37032054 eur 1.357525 2.381954 0.8820863
10: 10 0.17026205 eur 1.690291 1.777724 0.3026789
这是一种方法，尽管它不能很好地推广到大量货币：
DT[ , cvt := ifelse (currency == 'aud', price*aud, price*eur) ]
> DT
day price currency aud eur cvt
1: 1 0.30776611 aud 1.624996 2.035811 0.5001188
2: 2 0.25767250 eur 1.882166 2.210804 0.5696634
3: 3 0.55232243 aud 1.280354 2.038349 0.7071681
4: 4 0.05638315 eur 1.398488 2.248972 0.1268041
5: 5 0.46854928 aud 1.762551 1.920101 0.8258420
6: 6 0.48377074 eur 1.669022 1.671420 0.8085842
7: 7 0.81240262 aud 1.204612 2.270302 0.9786301
8: 8 0.37032054 eur 1.357525 2.381954 0.8820863
9: 9 0.54655860 aud 1.359475 2.049097 0.7430328
10: 10 0.17026205 eur 1.690291 1.777724 0.3026789
您会收到警告（如果您尝试机智if(.){.}else{.}，则会得到不同的结果：
DT[ , cvt := if (currency == 'aud'){price*aud}else{price*eur}]
这完全类似于 data.frames 发生的情况。但是......ifelse在 data.table 中使用是众所周知的慢。