Question

我的df看起来像这样：

data <- data.frame(
  "id" = c(2, 4, 5), 
  "paid" = c(80, 293.64, 157),
  "basic_fee" = c(500, 140.59, 21.49),
  "marketing_fee" = c(151.51, 10.12, 562.50),
  "utility_fee" = c(65, 99.29, 102.35),
stringsAsFactors = F)

我想要实现的是

final <- data.frame(
    "id" = c(2, 4, 5), 
    "paid" = c(80, 293.64, 157),
    "basic_fee" = c(500, 140.59, 21.49),
    "marketing_fee" = c(151.51, 10.12, 562.50),
    "utility_fee" = c(65, 99.29, 102.35),
    "paid_basic" = c(80, 140.59, 21.49),
    "paid_marketing" = c(0, 10.12, 135.51),
    "paid_utlity" = c(0, 99.29, 0),
    stringsAsFactors = F)

两者之间的逻辑实际上很简单。对于每个ID，获得付款价值的数量，然后“尽可能多地”支付优先级的费用-基本，营销，公用事业。请注意，任何费用所支付的金额都不能超过其实际价值。

下面的代码可以工作，但是重复的代码部分很难看。现在，我有100多个列甚至更复杂的数据框。如果其他行有成千上万的行，我不会创建越来越复杂的代码。

final <- 
  data %>% 
  mutate(
    paid_basic = if_else(basic_fee - paid > 0, basic_fee - (basic_fee - paid), basic_fee),
    overpayment_basic = if_else(paid-paid_basic > 0, 1, 0),

    paid_marketing = if_else(overpayment_basic == 1, (paid-paid_basic), 0),
    paid_marketing = if_else(paid_marketing > marketing_fee, marketing_fee, paid_marketing),
    overpayment_marketing = if_else(paid-paid_basic-paid_marketing > 0, 1, 0),

    paid_utility = if_else(overpayment_marketing == 1, (paid-paid_basic-paid_marketing), 0),
    paid_utility = if_else(paid_utility > utility_fee, utility_fee, paid_utility)
)

Answer 1

我不确定这是否比您现有的解决方案复杂得多，但这是获取更多列的一种方法

library(tidyverse)

fee_data <- select_at(data, vars(contains('fee')))

fee_data %>% 
  accumulate(`+`) %>% 
  map2_df(data$paid + fee_data, ~ .y - .x) %>% 
  map2_df(fee_data, ~ pmax(0, pmin(.x, .y))) %>% 
  rename_all(~ paste0('paid_', sub('_fee', '', .x))) %>% 
  bind_cols(data, .)

#   id   paid basic_fee marketing_fee utility_fee paid_basic paid_marketing paid_utility
# 1  2  80.00    500.00        151.51       65.00      80.00           0.00         0.00
# 2  4 293.64    140.59         10.12       99.29     140.59          10.12        99.29
# 3  5 157.00     21.49        562.50      102.35      21.49         135.51         0.00

Answer 2

我的原始答案无法推广到任意数量的行，因此这是另一种尝试：

r <- data$paid # keep track of remaining money
select(data, ends_with("_fee")) %>%
    set_names(sub("(.*)_.*", "paid_\\1", names(.))) %>%
    mutate_all( ~ {`<-`(x, map2_dbl(., r, ~ pmin(.x, .y))); `<<-`(r, r-x); x}) %>%
    bind_cols(data, .)

哪个返回：

  id   paid basic_fee marketing_fee utility_fee paid_basic paid_marketing paid_utility
1  2  80.00    500.00        151.51       65.00      80.00           0.00         0.00
2  4 293.64    140.59         10.12       99.29     140.59          10.12        99.29
3  5 157.00     21.49        562.50      102.35      21.49         135.51         0.00

我使用mutate而不是mutate_all将map2_dbl和pmin应用于子集中的每一列。

如何逻辑上减去行

2 个答案: