我必须将excel VBA代码“翻译”为R代码。我是编程新手，可能需要一点帮助。 VBA代码如下所示：
lastrow = .UsedRange.Rows.Count
lastcolumn = .UsedRange.Columns.Count
Nominal = ThisWorkbook.Worksheets("Admin").Cells(15, 3)
Coupons = ThisWorkbook.Worksheets("Admin").Cells(16, 3)
For a = 1 To lastcolumn
If .Cells(1, a) = "X" Then
For b = 2 To lastrow
For c = a + 1 To lastcolumn
If .Cells(b, a).Value = Year(Date) - 1 + c - a Then
.Cells(b, c).Value = .Cells(b, Nominal) + .Cells(b, Nominal) * .Cells(b, Coupons) / 100
ElseIf .Cells(b, a).Value > Year(Date) - 1 + c - a Then
.Cells(b, c).Value = .Cells(b, Nominal) * .Cells(b, Coupons) / 100
End If
Next
If .Cells(b, a) = vbEmpty Or .Cells(b, a) - Year(Date) > lastcolumn - a - 1 Then
.Cells(b, c - 1).Value = .Cells(b, Nominal)
End If
Next
Exit For
End If
Next
MsgBox ("Hello World")
第一个If条件是指名为X的列，该列将左侧的“旧”列与右侧的“新”列分开。我可以为您提供一些与R中必须使用的数据类似的数据。
data <- data.frame("nominal" = c(500000,1000000,2000000,1470000,500000,1000000), "coupon" = c(0,0,0,0.01,0.03,0.04), "year of maturity" = c(2023,2020,2019,2021,2022,2023), "X" = c(rep("X",6)))
我想要的输出看起来像这样：
data_final <- data.frame("nominal" = c(500000,1000000,2000000,1470000,500000,1000000), "coupon" = c(0,0,0,0.01,0.03,0.04), "year of maturity" = c(2023,2020,2019,2021,2022,2023), "X" = c(rep("X",6)) , "2019" = c(0,0,2000000,147,150,400), "2020" = c(0,1000000,0,147,150,400), "2021" = c(0,0,0,1470147,150,400), "2022" = c(0,0,0,0,500150,400), "2023" = c(500000,0,0,0,0,1000400))
请注意，实际数据要大得多（> 2000 obs。和> 50个变量[最大的到期年份是2068]），因此我必须使用循环。我尝试转换VBA代码的过程看起来像这样（不需要R代码中的消息框）：
year <- as.numeric(format(Sys.Date(), "%Y"))
for (a in 1:ncol(data)) {
if (data[1,a] == "X") {
for (b in 1:nrow(data)) {
for (c in a+1:48) {
if (data[b,a] == year - 1 + c - a) {
data[b,c] <- data[b,"nominal"] + data[b,"nominal"]*data[b, "coupons"] / 100
} else {
data[b,c] <- data[b,"nominal"]*data[b, "coupons"] / 100
}
if (data[b,a] == is.na() | data[b,a] - jahr > nrow(data) - a - 1) {
data[b,c-1] <- data[b, "nominal"]
}
}
}
}
}
我要么收到以下错误，要么什么都没有（意味着没有警告/错误，也没有结果）：
Error in matrix(unlist(value, recursive = FALSE, use.names = FALSE), nrow = nr, : 'data' must be of a vector type, was 'NULL'
[请注意，我不会允许将软件包用作jrvFinance。如果您需要任何进一步的信息，请告诉我。提前致谢。
这里是使用dplyr和tidyr的方法：
library(dplyr); library(tidyr)
year <- as.numeric(format(Sys.Date(), "%Y"))
data %>%
rowid_to_column() %>% # Add a row to keep track of original order
uncount(year.of.maturity - year + 1) %>% # Make copy for each year until maturity
group_by(nominal, coupon) %>%
mutate(cur_year = year + row_number() - 1, # Assign year number
value = nominal * (coupon/100 + if_else(cur_year == year.of.maturity,
1, 0))) %>% # debt service
ungroup() %>%
spread(cur_year, value, fill = 0) # Convert year rows into year columns
## A tibble: 6 x 10
# rowid nominal coupon year.of.maturity X `2019` `2020` `2021` `2022` `2023`
# <int> <dbl> <dbl> <dbl> <fct> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 1 500000 0 2023 X 0 0 0 0 500000
#2 2 1000000 0 2020 X 0 1000000 0 0 0
#3 3 2000000 0 2019 X 2000000 0 0 0 0
#4 4 1470000 0.01 2021 X 147 147 1470147 0 0
#5 5 500000 0.03 2022 X 150 150 150 500150 0
#6 6 1000000 0.04 2023 X 400 400 400 400 1000400
使用dplyr和tidyr可以做
library(dplyr)
library(tidyr)
year_max <- max(my_df$year.of.maturity)
year_start <- as.numeric(format(Sys.Date(), "%Y"))
my_df %>%
# For each combination create the year column from start to maximum year
group_by(nominal, coupon, year.of.maturity, X) %>%
mutate(year = year.of.maturity) %>%
expand(year = year_start:year_max) %>%
# Set value for each year
mutate(value = case_when(year < year.of.maturity ~ nominal * coupon / 100,
year == year.of.maturity ~ nominal + nominal * coupon / 100,
TRUE ~ 0)) %>%
# Spread transforms the data from long to wide to have the desired representation
spread(year, value) %>%
ungroup()
# A tibble: 6 x 9
# nominal coupon year.of.maturity X `2019` `2020` `2021` `2022` `2023`
# <dbl> <dbl> <dbl> <fct> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 500000 0 2023 X 0 0 0 0 500000
# 2 500000 0.03 2022 X 150 150 150 500150 0
# 3 1000000 0 2020 X 0 1000000 0 0 0
# 4 1000000 0.04 2023 X 400 400 400 400 1000400
# 5 1470000 0.01 2021 X 147 147 1470147 0 0
# 6 2000000 0 2019 X 2000000 0 0 0 0
数据
my_df <-
data.frame("nominal" = c(500000, 1000000, 2000000, 1470000, 500000, 1000000),
"coupon" = c(0, 0, 0, 0.01, 0.03, 0.04),
"year of maturity" = c(2023, 2020, 2019, 2021, 2022, 2023),
"X" = c(rep("X", 6)))