将自定义功能应用于DFS列表，将另一个列表作为输入-R
我有DFS清单和年度预算清单。 每个DF代表一个营业年，每个预算代表当年的总支出。
# the business year starts from Feb and ends in Jan.
# the budget column is first populated with the % of annual budget allocation
df <- data.frame(monthly_budget=c(0.06, 0.13, 0.07, 0.06, 0.1, 0.06, 0.06, 0.09, 0.06, 0.06, 0.1, 0.15),
month=month.abb[c(2:12, 1)])
# dfs for 3 years
df2019_20 <- df
df2020_21 <- df
df2021_22 <- df
# budgets for 3 years
budget2019_20 <- 6000000
budget2020_21 <- 7000000
budget2021_22 <- 8000000
# into lists
df_list <- list(df2019_20, df2020_21, df2021_22)
budget_list <- list(budget2019_20, budget2020_21, budget2021_22)
我已经写下了以下功能，以将正确的年份应用于Jan，并通过将相应的DFS名称删除来填写其余功能。 如果我提供单个DF和单个预算，它可以很好地工作。
budget_func <- function(df, budget){
df_name <- deparse(substitute(df))
df <- df %>%
mutate(year=ifelse(month=="Jan",
as.numeric(str_sub(df_name, -2)) + 2000,
as.numeric(str_extract(df_name, "\\d{4}(?=_)")))
)
for (i in 1:12){
df[i,1] <- df[i,1] * budget
i <- i+1
}
return(df)
}
为了加快事物的速度，我想将两个列表作为参数将其作为mapply 。但是我没有得到想要的结果 - 我在做什么错？
final_budgets <- mapply(budget_func, df_list, budget_list)
I have a list of dfs and a list of annual budgets.
Each df represents one business year, and each budget represents a total spend for that year.
# the business year starts from Feb and ends in Jan.
# the budget column is first populated with the % of annual budget allocation
df <- data.frame(monthly_budget=c(0.06, 0.13, 0.07, 0.06, 0.1, 0.06, 0.06, 0.09, 0.06, 0.06, 0.1, 0.15),
month=month.abb[c(2:12, 1)])
# dfs for 3 years
df2019_20 <- df
df2020_21 <- df
df2021_22 <- df
# budgets for 3 years
budget2019_20 <- 6000000
budget2020_21 <- 7000000
budget2021_22 <- 8000000
# into lists
df_list <- list(df2019_20, df2020_21, df2021_22)
budget_list <- list(budget2019_20, budget2020_21, budget2021_22)
I've written the following function to both apply the right year to Jan and fill in the rest by deparsing the respective dfs name.
It works perfectly if I supply a single df and a single budget.
budget_func <- function(df, budget){
df_name <- deparse(substitute(df))
df <- df %>%
mutate(year=ifelse(month=="Jan",
as.numeric(str_sub(df_name, -2)) + 2000,
as.numeric(str_extract(df_name, "\\d{4}(?=_)")))
)
for (i in 1:12){
df[i,1] <- df[i,1] * budget
i <- i+1
}
return(df)
}
To speed things up I want to pass both lists as arguments to mapply. However I don't get the results I want - what am I doing wrong?
final_budgets <- mapply(budget_func, df_list, budget_list)
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而不是使用
depars/替代（当我们传递单个数据集时它有效，并且在循环中是不同的，因为对象传递的对象不是对象名称），我们可以添加一个新参数以传递名称。此外，当我们创建list时，它也应该具有名称。我们可以使用list（DF2019_20 = DF2019_20，...）或使用setNames，或者更轻松的选项是dplyr :: lst确实返回用对象的名称传递-TESTING
-OUTPUT
Instead of using
deparse/substitute(which works when we are passing a single dataset, and is different in the loop because the object passed is not the object name), we may add a new argument to pass the names. In addition, when we create thelist, it should have the names as well. We can either uselist(df2019_20 = df2019_20, ...)or usesetNamesor an easier option isdplyr::lstwhich does return with the name of the object passed-testing
-output