R按标签差异聚合

Question

我有一个大型数据框，其中包含 28 个欧盟国家/地区的进出口数据以及其他详细信息。我想计算每个国家的贸易差额（出口减去进口）。

我使用以下代码成功汇总了每个国家/地区的进出口：

x = aggregate( value_in_euros ~ as_factor(flow) + as_factor(declarant), data, sum)

生成的数据集如下所示：

    as_factor(flow) as_factor(declarant) value_in_euros
1          Imports                   FR   1.019259e+12
2          Exports                   FR   8.539884e+11
3          Imports                   BE   6.936985e+11
4          Exports                   BE   7.350930e+11
5          Imports                   NL   1.043084e+12
6          Exports                   NL   1.180904e+12
7          Imports                   DE   2.048670e+12
8          Exports                   DE   2.415090e+12
9          Imports                   IT   7.399388e+11
10         Exports                   IT   8.671186e+11
11         Imports                   GB   9.705866e+10
12         Exports                   GB   6.669265e+10
13         Imports                   IE   1.725745e+11
14         Exports                   IE   3.150999e+11
15         Imports                   DK   1.707441e+11
16         Exports                   DK   1.897421e+11

现在我需要一种聚合方法来计算每个国家/地区的贸易差额（出口减去进口）。任何人都可以向我提供有关如何操作的提示吗？谢谢

Answer 1

我们可以使用split将数据帧分割成declarant，然后计算流之间的差异。

set.seed(1)
df <- data.frame(flow=rep(c('Imports','Exports'), 8),
                 declarant=rep(c('FR','BE','NL','DE','IT','GB','IE','DK'), each=2),
                 value_in_euros=rnorm(n=16L,mean=1e12,sd=1e4))
out <- sapply(split(df, df$declarant), function(x) diff(x$value_in_euros))
# ------------------------------------------
> out[1:4]
        BE         DE         DK         FR 
 24309.094   2508.957 -11698.645   8100.971

Answer 2

考虑聚合后重塑。那么，运行的区别。

agg_df <- aggregate(
    value_in_euros ~ flow + declarant, data=data, FUN=sum
) |> reshape(
  idvar = "declarant",
  v.names = "value_in_euros",
  timevar = "flow",
  direction = "wide",
  sep = "_"
) |> transform(
  balance = value_in_euros_Exports - value_in_euros_Imports
)

agg_df
   declarant value_in_euros_Imports value_in_euros_Exports       balance
1         FR           1.019259e+12           8.539884e+11 -165270600000
3         BE           6.936985e+11           7.350930e+11   41394500000
5         NL           1.043084e+12           1.180904e+12  137820000000
7         DE           2.048670e+12           2.415090e+12  366420000000
9         IT           7.399388e+11           8.671186e+11  127179800000
11        GB           9.705866e+10           6.669265e+10  -30366010000
13        IE           1.725745e+11           3.150999e+11  142525400000
15        DK           1.707441e+11           1.897421e+11   18998000000

数据

txt = '       flow              declarant value_in_euros
1          Imports                   FR   1.019259e+12
2          Exports                   FR   8.539884e+11
3          Imports                   BE   6.936985e+11
4          Exports                   BE   7.350930e+11
5          Imports                   NL   1.043084e+12
6          Exports                   NL   1.180904e+12
7          Imports                   DE   2.048670e+12
8          Exports                   DE   2.415090e+12
9          Imports                   IT   7.399388e+11
10         Exports                   IT   8.671186e+11
11         Imports                   GB   9.705866e+10
12         Exports                   GB   6.669265e+10
13         Imports                   IE   1.725745e+11
14         Exports                   IE   3.150999e+11
15         Imports                   DK   1.707441e+11
16         Exports                   DK   1.897421e+11'

agg_df <- read.table(text=txt)

Answer 3

您可以直接将此应用到您的 data 而不是进行汇总。

library(dplyr)
library(tidyr)

data %>%
  select(flow, declarant, value_in_euros) %>%
  pivot_wider(names_from = flow, 
              values_from = value_in_euros, values_fn = sum) %>%
  mutate(difference = Exports - Imports)

R按标签差异聚合

R aggregate by label difference

aggregate

r

dataframe

difference