如何从逻辑上减去行
如何从逻辑上减去行
提问于 2019-06-19 20:59:17
我的df看起来有点像这样：
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)实际上，两者之间的逻辑非常简单。对于每个id，获取已付价值的金额，然后按优先顺序“尽可能多地支付”费用-基本费用、营销费用、效用费用。请注意，任何费用的支付金额都不能高于其实际价值。
我的代码下面工作，但它是非常丑陋的重复代码的一部分。现在我有了更复杂的包含100+列的数据帧。我不想创建更多更复杂的代码，如果有成千上万行的话。
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)
)回答 2
Stack Overflow用户
回答已采纳
发布于 2019-06-19 21:24:29
我不确定这是否比您现有的解决方案简单得多，但这里有一种方法可以获得额外的列
library(tidyverse)
fee_data <- select_at(data, vars(contains('fee')))
fee_data %>%
accumulate(`+`) %>%
map2_df(data$paid + fee_data, ~ .y - .x) %>%
map2_df(fee_data, ~ pmax(0, pmin(.x, .y))) %>%
rename_all(~ paste0('paid_', sub('_fee', '', .x))) %>%
bind_cols(data, .)
# id paid basic_fee marketing_fee utility_fee paid_basic paid_marketing paid_utility
# 1 2 80.00 500.00 151.51 65.00 80.00 0.00 0.00
# 2 4 293.64 140.59 10.12 99.29 140.59 10.12 99.29
# 3 5 157.00 21.49 562.50 102.35 21.49 135.51 0.00Stack Overflow用户
发布于 2019-06-19 21:34:18
我最初的答案不能推广到任意数量的行数，所以这里是另一个尝试：
r <- data$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, .)它返回：
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我不使用mutate，而是使用mutate_all将带有pmin的map2_dbl应用到子集的每一列。
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原文链接：
https://stackoverflow.com/questions/56668247
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