条件乘法向量（货币换算）

Question

我有一个相当大的数据框，其中变量以当地年度货币（在下面的示例中为澳大利亚和奥地利货币）计价：

 Country   Var  _1995       _1996         _1997      _1998
     AUS    GO  1 014 828   1 059 326     1 119 101  1 194 995
     AUS    L   36 873      38 895        39 502     40 425
     AUS    K   41 498      45 008        48 683     47 252
     AUT    GO  289 923     299 487       309 734    323 273
     AUT    GO  8 032       7 849         8 049      7 815
     AUT    L   1 094       1 151         1 163      1 152
     AUT    K   12 032      11 760        11 743     11 611

我想使用以下乘数将此数据框中的值转换为 1995 美元：

Country   _1995     _1996     _1997  _1998 
AUS      0,7415   0,78295   0,74406  0,6294
AUT     1,36646   1,30031   1,12904 1,11319

因此，对于表 1 中包含变量 AUS 的每一行，每年的值乘以表 2 中包含 AUS 的行的适当的 $1995 乘数。同样的情况也适用于包含 AUT 的每一行作为我数据框中的其他 38 个国家/地区代码。

所以，在第一行我希望 R 执行这个计算：

 Country Var    _1995            _1996            _1997           _1998        
    AUS  GO  1014828*0,7415 1059326*0,78295   1119101*0,74406 1194995*0,6294 

等等。这是可行的吗？非常感谢您的帮助！

Answer 1

我建议将宽格式改成长格式，这将大大简化此操作。重塑是最复杂的部分。我在这里使用示例数据和reshape 命令展示它，但您也可以使用dplyr 或reshape2 或其他。

基本上，将两个数据集重新整形为长，然后合并它们，执行乘法（在长格式中，只是简单的向量乘法），然后重新整形为宽。

这是示例数据（与您的类似）：

set.seed(1)
dat <- data.frame(Country = rep(c("AUS", "AUT"), each = 3), 
                  Var = rep(c("GO", "L", "K"), times = 2), 
                  v_1996 = rnorm(6), v_1997 = rnorm(6), v_1998 = rnorm(6),  
                  stringsAsFactors = FALSE)

multipliers <- data.frame(Country = c("AUS", "AUT"), 
                          v_1995 = c(0.7415, 1.36646),
                          v_1996 = c(0.78295, 1.30031),
                          v_1997 = c(0.74406, 1.12904),
                          v_1998 = c(0.6294, 1.11319), stringsAsFactors = FALSE)

这是进行转换的代码：

long <- reshape(dat, times = 1996:1998, v.names = "Value", 
                varying = c("v_1996", "v_1997", "v_1998"), 
                direction = "long")
head(long, 3)
#        Country Var time      Value id
# 1.1996     AUS  GO 1996 -0.6264538  1
# 2.1996     AUS   L 1996  0.1836433  2
# 3.1996     AUS   K 1996 -0.8356286  3
# 4.1996     AUT  GO 1996  1.5952808  4

mlong <- reshape(multipliers, times = 1995:1998, v.names = "mult", 
                 varying = c("v_1995","v_1996", "v_1997", "v_1998"), 
                 direction = "long")
head(mlong, 3)
#        Country time    mult id
# 1.1995     AUS 1995 0.74150  1
# 2.1995     AUT 1995 1.36646  2
# 1.1996     AUS 1996 0.78295  1

merged <- merge(long, mlong, by = c("Country", "time"))
merged$converted <- merged$Value * merged$mult    
head(merged, 3)
#   Country time Var      Value id.x    mult id.y  converted
# 1     AUS 1996  GO -0.6264538    1 0.78295    1 -0.4904820
# 2     AUS 1996   L  0.1836433    2 0.78295    1  0.1437835
# 3     AUS 1996   K -0.8356286    3 0.78295    1 -0.6542554

reshape(merged, idvar = c("Country", "Var"), direction = "wide", 
        drop = c("id.x", "id.y","mult"))
#    Country Var Value.1996 converted.1996 Value.1997 converted.1997  Value.1998 converted.1998
# 1      AUS  GO -0.6264538     -0.4904820  0.4874291      0.3626765 -0.62124058    -0.39100882
# 2      AUS   L  0.1836433      0.1437835  0.7383247      0.5493579 -2.21469989    -1.39393211
# 3      AUS   K -0.8356286     -0.6542554  0.5757814      0.4284159  1.12493092     0.70803152
# 10     AUT  GO  1.5952808      2.0743596 -0.3053884     -0.3447957 -0.04493361    -0.05001964
# 11     AUT   L  0.3295078      0.4284623  1.5117812      1.7068614 -0.01619026    -0.01802284
# 12     AUT   K -0.8204684     -1.0668632  0.3898432      0.4401486  0.94383621     1.05066903

Answer 2

类似这样的：

（假设您的本地货币数据框称为“local”，而带有乘数的数据框称为“conv”。）

#unfactorise Country or you'll get very strange results
local$Country <- as.character(local$Country); conv$Country <- as.character(conv$Country)
countries <- unique(local$Country)
for(i in 1:length(countries)) {
        cy <- countries[i]
        rates <- matrix(conv[conv$Country==cy, -1])
        local[local$Country==cy, -c(1,2)] <- local[local$Country==cy, -c(1,2)] * rates
}

Answer 3

创建一个小辅助函数然后通过管道传输数据可能是最简单的。为了使其更简洁，请将转化的 row.names 设置为 Country 并删除该列。

df <- read.table(header = TRUE, text = '
                 Country Var  _1995       _1996         _1997      _1998
     AUS   GO  1014828   1059326     1119101  1194995
     AUS   L   36873      38895        39502     40425
     AUS   K    41498     45008        48683     47252
     AUT   GO  289923     299487       309734    323273
     AUT   GO  8032       7849         8049      7815
     AUT   L   1094       1151         1163      1152
     AUT   K   12032      11760        11743     11611
                 ')

conversions <- read.table(header = TRUE, text='
                          Country _1995     _1996   _1997   _1998 
 AUS     0.7415    0.78295 0.74406 0.6294
 AUT     1.36646   1.30031 1.12904 1.11319
                          ')

# the primary code to use
# set row.names, makes indexing cleaner below
row.names(conversions) <- conversions$Country
conversions <- conversions[,-1]

# helper function for conversions
myfun <- function(df1, df2) {
    df1[,3:6] <- df1[,3:6] * df2[df1$Country,]
    df1
}

library(dplyr)
df %>% 
   group_by(Country) %>% 
   do(myfun(., conversions))

Source: local data frame [7 x 6]
Groups: Country

  Country Var     X_1995     X_1996     X_1997     X_1998
1     AUS  GO 752494.962 829399.292 832678.290 752129.853
2     AUS   L  27341.330  30452.840  29391.858  25443.495
3     AUS   K  30770.767  35239.014  36223.073  29740.409
4     AUT  GO 396168.183 389425.941 349702.075 359864.271
5     AUT  GO  10975.407  10206.133   9087.643   8699.580
6     AUT   L   1494.907   1496.657   1313.074   1282.395
7     AUT   K  16441.247  15291.646  13258.317  12925.249

Answer 4

这是我使用dplyr 的尝试。我正在以各种方式进行实验并想出了这个。我首先将数据（即mydf）拆分为Country。对于列表中的每个数据框，我想应用适当的汇率。因此，我使用Country 对汇率数据（即rate）进行了子集化并创建了新数据。 （当代码运行时，R 正在获取每个国家/地区的汇率。）我在this question 中应用了我的答案，以便使用mutate_each() 计算多个列。最后我用bind_rows()把所有的数据框组合起来。

lapply(split(mydf, mydf$Country), function(i) {

        foo <- rate[rate$Country == unique(i$Country),]

        mutate_each(i, funs(. * foo$.), y_1995:y_1998)

    }) %>%
bind_rows

#  Country Var     y_1995     y_1996     y_1997     y_1998
#1     AUS  GO 752494.962 829399.292 832678.290 752129.853
#2     AUS   L  27341.330  30452.840  29391.858  25443.495
#3     AUS   K  30770.767  35239.014  36223.073  29740.409
#4     AUT  GO 396168.183 389425.941 349702.075 359864.271
#5     AUT  GO  10975.407  10206.133   9087.643   8699.580
#6     AUT   L   1494.907   1496.657   1313.074   1282.395
#7     AUT   K  16441.247  15291.646  13258.317  12925.249

数据

mydf <- structure(list(Country = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 
2L), .Label = c("AUS", "AUT"), class = "factor"), Var = structure(c(1L, 
3L, 2L, 1L, 1L, 3L, 2L), .Label = c("GO", "K", "L"), class = "factor"), 
y_1995 = c(1014828, 36873, 41498, 289923, 8032, 1094, 12032
), y_1996 = c(1059326, 38895, 45008, 299487, 7849, 1151, 
11760), y_1997 = c(1119101, 39502, 48683, 309734, 8049, 1163, 
11743), y_1998 = c(1194995, 40425, 47252, 323273, 7815, 1152, 
11611)), .Names = c("Country", "Var", "y_1995", "y_1996", 
"y_1997", "y_1998"), row.names = c(NA, -7L), class = "data.frame")

#  Country Var  y_1995  y_1996  y_1997  y_1998
#1     AUS  GO 1014828 1059326 1119101 1194995
#2     AUS   L   36873   38895   39502   40425
#3     AUS   K   41498   45008   48683   47252
#4     AUT  GO  289923  299487  309734  323273
#5     AUT  GO    8032    7849    8049    7815
#6     AUT   L    1094    1151    1163    1152
#7     AUT   K   12032   11760   11743   11611

rate <- structure(list(Country = structure(1:2, .Label = c("AUS", "AUT"
), class = "factor"), y_1995 = c(0.7415, 1.36646), y_1996 = c(0.78295, 
1.30031), y_1997 = c(0.74406, 1.12904), y_1998 = c(0.6294, 1.11319
)), .Names = c("Country", "y_1995", "y_1996", "y_1997", "y_1998"
), row.names = c(NA, -2L), class = "data.frame")

#  Country  y_1995  y_1996  y_1997  y_1998
#1     AUS 0.74150 0.78295 0.74406 0.62940
#2     AUT 1.36646 1.30031 1.12904 1.11319