r - R - 使用 ifelse 语句在不同的列上分配一个数字的份额

Question

我有以下数据集：

observation <- c(1:10)
pop.d.rank  <- c(1:10)
cost.1  <- c(101:110)
cost.2  <- c(102:111)
cost.3  <- c(103:112)
all <- data.frame(observation,pop.d.rank,cost.1, cost.2, cost.3)

我想在三年内分配以下金额：

annual.investment <- 500

我可以使用以下脚本在第一年执行此操作：

library(dplyr)

all <- all %>%  
 mutate(capital_allocated.5G = diff(c(0, pmin(cumsum(cost), annual.investment)))) %>%
 mutate(capital_percentage.5G = capital_allocated.5G / cost * 100) %>%
 mutate(year = ifelse(capital_percentage.5G >= 50, "Year.1",0))

但是当我第二年尝试这样做时，考虑到上一年的投资，代码不起作用。这是我尝试在 mutate 循环中放置一条 ifelse 语句，这样它就不会覆盖上一年分配的资金：

all <- all %>%  
 mutate(capital_allocated.5G = ifelse(year == 0, diff(c(0, pmin(cumsum(cost), annual.investment))), 0) %>%
 mutate(capital_percentage.5G = capital_allocated.5G / cost * 100) %>%
 mutate(year = ifelse(capital_percentage.5G >= 50, "Year.2",0))

我希望数据如下所示，其中分配的金额首先分配到上一年尚未 100% 完成的任何行。

capital_allocated.5G <- c(101, 102, 103, 104, 105, 106, 107, 108, 109, 55)
capital_percentage.5G <- c(100, 100, 100, 100, 100, 100, 100, 100, 100, 50)
year <- c("Year.1", "Year.1","Year.1", "Year.1","Year.1", "Year.2", "Year.2","Year.2", "Year.2","Year.2")
example.output <- data.frame(observation,pop.d.rank,cost,   capital_allocated.5G, capital_percentage.5G, year)

编辑：cost.1 是第 1 年的成本变量，cost.2 是第 2 年的变量，cost.3 是第 3 年的成本变量

编辑：以前接受的答案有问题

我意识到这最终会为 capital_percentage.5G 变量分配超过 100 个。我创建了一个可重现的示例。我认为这与一些成本随着时间的推移而降低而一些成本随着时间的推移而增加的事实有关。

这背后的逻辑是，当在一年内进行投资时，5G 移动网络的部署会产生特定的成本，这就是与该时间点相关的成本列。一旦在一年内进行了投资，我希望该功能提供一个 capital_percentage.5G 100%，然后在未来几年不再向它分配任何资金。

如何获得它以使百分比值达到 100 的限制，并且以后不会分配更多的资本分配给它？

observation <- c(1:10)
pop.d.rank  <- c(1:10)
cost.1  <- c(101:110)
cost.2  <- c(110:101)
cost.3  <- c(100:91)
all <- data.frame(observation,pop.d.rank,cost.1, cost.2, cost.3) 

capital_allocated.5G <- rep(0,10)   ## initialize to zero
capital_percentage.5G <- rep(0,10)  ## initialize to zero
year <- rep(NA,10)                  ## initialize to NA
all <- data.frame(observation,pop.d.rank,cost.1, cost.2, cost.3,   capital_allocated.5G,capital_percentage.5G,year) 

alloc.invest <- function(df, ann.invest, y) {
  df %>% mutate_(cost=paste0("cost.",y)) %>%
    mutate(capital_percentage.5G = capital_allocated.5G / cost * 100,
           year = ifelse(capital_percentage.5G < 50, NA, year),
           not.yet.alloc = ifelse(capital_percentage.5G < 100,cost-capital_allocated.5G,0),
           capital_allocated.5G = capital_allocated.5G +     ifelse(capital_percentage.5G < 100,diff(c(0, pmin(cumsum(not.yet.alloc), ann.invest))), 0),
       capital_percentage.5G = capital_allocated.5G / cost * 100,
       year = ifelse(is.na(year) & capital_percentage.5G >= 50, paste0("Year.",y), year)) %>%
select(-cost,-not.yet.alloc)
}

annual.investment <- 500
all <- alloc.invest(all,annual.investment,1)
print(all)
all <- alloc.invest(all,annual.investment,2)
print(all)
all <- alloc.invest(all,annual.investment,3)
print(all)

第 3 年，在这里的最终投资分配中，capital_percentage.5G 突然飙升至 110%。

score 5 · Accepted Answer

更新了可能增加或减少的同比成本

对于每年可能减少和增加的不同成本，我们根本不需要capital_percentage.5G在更新时检查是否超过 100%not.yet.alloc并且capital_allocated.5G：

library(dplyr)
alloc.invest <- function(df, ann.invest, y) {
  df %>% mutate_(cost=paste0("cost.",y)) %>%
    mutate(capital_percentage.5G = capital_allocated.5G / cost * 100,
           year = ifelse(capital_percentage.5G < 50, NA, year),
           not.yet.alloc = cost-capital_allocated.5G,
           capital_allocated.5G = capital_allocated.5G + diff(c(0, pmin(cumsum(not.yet.alloc), ann.invest))),
           capital_percentage.5G = capital_allocated.5G / cost * 100,
           year = ifelse(is.na(year) & capital_percentage.5G >= 50, paste0("Year.",y), year)) %>%
    select(-cost,-not.yet.alloc)
}

使用新的成本数据：

observation <- c(1:10)
pop.d.rank  <- c(1:10)
cost.1  <- c(101:110)
cost.2  <- c(110:101)
cost.3  <- c(100:91)

像以前一样增加初始值列：

capital_allocated.5G <- rep(0,10)   ## initialize to zero
capital_percentage.5G <- rep(0,10)  ## initialize to zero
year <- rep(NA,10)                  ## initialize to NA
all <- data.frame(observation,pop.d.rank,cost.1, cost.2, cost.3, capital_allocated.5G,capital_percentage.5G,year)

第一年：

annual.investment <- 500
all <- alloc.invest(all,annual.investment,1)
print(all)
##   observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G   year
##1            1          1    101    110    100                  101             100.00000 Year.1
##2            2          2    102    109     99                  102             100.00000 Year.1
##3            3          3    103    108     98                  103             100.00000 Year.1
##4            4          4    104    107     97                  104             100.00000 Year.1
##5            5          5    105    106     96                   90              85.71429 Year.1
##6            6          6    106    105     95                    0               0.00000   <NA>
##7            7          7    107    104     94                    0               0.00000   <NA>
##8            8          8    108    103     93                    0               0.00000   <NA>
##9            9          9    109    102     92                    0               0.00000   <NA>
##10          10         10    110    101     91                    0               0.00000   <NA>

第 2 年：

all <- alloc.invest(all,annual.investment,2)
print(all)
##   observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G   year
##1            1          1    101    110    100                  110             100.00000 Year.1
##2            2          2    102    109     99                  109             100.00000 Year.1
##3            3          3    103    108     98                  108             100.00000 Year.1
##4            4          4    104    107     97                  107             100.00000 Year.1
##5            5          5    105    106     96                  106             100.00000 Year.1
##6            6          6    106    105     95                  105             100.00000 Year.2
##7            7          7    107    104     94                  104             100.00000 Year.2
##8            8          8    108    103     93                  103             100.00000 Year.2
##9            9          9    109    102     92                  102             100.00000 Year.2
##10          10         10    110    101     91                   46              45.54455   <NA>

第 3 年：

all <- alloc.invest(all,annual.investment,3)
print(all)
##   observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G   year
##1            1          1    101    110    100                  100                   100 Year.1
##2            2          2    102    109     99                   99                   100 Year.1
##3            3          3    103    108     98                   98                   100 Year.1
##4            4          4    104    107     97                   97                   100 Year.1
##5            5          5    105    106     96                   96                   100 Year.1
##6            6          6    106    105     95                   95                   100 Year.2
##7            7          7    107    104     94                   94                   100 Year.2
##8            8          8    108    103     93                   93                   100 Year.2
##9            9          9    109    102     92                   92                   100 Year.2
##10          10         10    110    101     91                   91                   100 Year.3

您的代码的原始问题是仅根据条件提供输出ifelse开关，而不是在. 因此，计算整体，而不仅仅是计算. 为了解决这个问题，我们可以定义以下函数，然后每年依次执行。costTRUEifelsecumsum(cost)cumsumcostTRUEifelse

library(dplyr)
alloc.invest <- function(df, ann.invest, y) {
  df %>% mutate(not.yet.alloc = ifelse(capital_percentage.5G < 100,cost-capital_allocated.5G,0),
                capital_allocated.5G = capital_allocated.5G + ifelse(capital_percentage.5G < 100,diff(c(0, pmin(cumsum(not.yet.alloc), ann.invest))), 0),
                capital_percentage.5G = capital_allocated.5G / cost * 100,
                year = ifelse(is.na(year) & capital_percentage.5G >= 50, paste0("Year.",y), year)) %>%
         select(-not.yet.alloc)
}

笔记：

创建一个新的临时列not.yet.alloc，我们从中计算cumsum年度分配的结果。
不需要单独的mutate语句。
is.na(year)设置前还需要检查year。否则，先前year已标记的将被覆盖。

capital_allocated.5G要使用此函数，我们必须首先用、capital_percentage.5G和的一些初始值来扩充输入数据year：

capital_allocated.5G <- rep(0,10)   ## initialize to zero
capital_percentage.5G <- rep(0,10)  ## initialize to zero
year <- rep(NA,10)                  ## initialize to NA
all <- data.frame(observation,pop.d.rank,cost,capital_allocated.5G,capital_percentage.5G,year)

然后是第 1 年：

annual.investment <- 500
all <- alloc.invest(all,annual.investment,1)
print(all)
##   observation pop.d.rank cost capital_allocated.5G capital_percentage.5G   year
##1            1          1  101                  101             100.00000 Year.1
##2            2          2  102                  102             100.00000 Year.1
##3            3          3  103                  103             100.00000 Year.1
##4            4          4  104                  104             100.00000 Year.1
##5            5          5  105                   90              85.71429 Year.1
##6            6          6  106                    0               0.00000   <NA>
##7            7          7  107                    0               0.00000   <NA>
##8            8          8  108                    0               0.00000   <NA>
##9            9          9  109                    0               0.00000   <NA>
##10          10         10  110                    0               0.00000   <NA>

第 2 年：

all <- alloc.invest(all,annual.investment,2)
print(all)
##   observation pop.d.rank cost capital_allocated.5G capital_percentage.5G   year
##1            1          1  101                  101                   100 Year.1
##2            2          2  102                  102                   100 Year.1
##3            3          3  103                  103                   100 Year.1
##4            4          4  104                  104                   100 Year.1
##5            5          5  105                  105                   100 Year.1
##6            6          6  106                  106                   100 Year.2
##7            7          7  107                  107                   100 Year.2
##8            8          8  108                  108                   100 Year.2
##9            9          9  109                  109                   100 Year.2
##10          10         10  110                   55                    50 Year.2

更新到每年变化成本的新要求

如果每年的成本不同，则该函数需要先重新调整列，capital_percentage.5G并且可能year首先调整列：

library(dplyr)
alloc.invest <- function(df, ann.invest, y) {
  df %>% mutate_(cost=paste0("cost.",y)) %>%
         mutate(capital_percentage.5G = capital_allocated.5G / cost * 100,
                year = ifelse(capital_percentage.5G < 50, NA, year),
                not.yet.alloc = ifelse(capital_percentage.5G < 100,cost-capital_allocated.5G,0),
                capital_allocated.5G = capital_allocated.5G + ifelse(capital_percentage.5G < 100,diff(c(0, pmin(cumsum(not.yet.alloc), ann.invest))), 0),
                capital_percentage.5G = capital_allocated.5G / cost * 100,
                year = ifelse(is.na(year) & capital_percentage.5G >= 50, paste0("Year.",y), year)) %>%
         select(-cost,-not.yet.alloc)
}

请注意，使用创建另一个临时列只是为了方便，因为需要根据输入动态选择成本列（否则，我们需要用于所有计算，这会有点混乱）。costmutate_ymutate_

更新后的数据同样增加了、和的初始值capital_allocated.5G，第 1 年：capital_percentage.5Gyear

annual.investment <- 500
all <- alloc.invest(all,annual.investment,1)
print(all)
##   observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G   year
##1            1          1    101    102    103                  101             100.00000 Year.1
##2            2          2    102    103    104                  102             100.00000 Year.1
##3            3          3    103    104    105                  103             100.00000 Year.1
##4            4          4    104    105    106                  104             100.00000 Year.1
##5            5          5    105    106    107                   90              85.71429 Year.1
##6            6          6    106    107    108                    0               0.00000   <NA>
##7            7          7    107    108    109                    0               0.00000   <NA>
##8            8          8    108    109    110                    0               0.00000   <NA>
##9            9          9    109    110    111                    0               0.00000   <NA>
##10          10         10    110    111    112                    0               0.00000   <NA>

第 2 年：请注意，最后一项资产的分配少于50%分配，因此它year仍然是NA。

all <- alloc.invest(all,annual.investment,2)
print(all)
##   observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G   year
##1            1          1    101    102    103                  102             100.00000 Year.1
##2            2          2    102    103    104                  103             100.00000 Year.1
##3            3          3    103    104    105                  104             100.00000 Year.1
##4            4          4    104    105    106                  105             100.00000 Year.1
##5            5          5    105    106    107                  106             100.00000 Year.1
##6            6          6    106    107    108                  107             100.00000 Year.2
##7            7          7    107    108    109                  108             100.00000 Year.2
##8            8          8    108    109    110                  109             100.00000 Year.2
##9            9          9    109    110    111                  110             100.00000 Year.2
##10          10         10    110    111    112                   46              41.44144   <NA>

第 3 年：

all <- alloc.invest(all,annual.investment,3)
print(all)
##   observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G   year
##1            1          1    101    102    103                  103                   100 Year.1
##2            2          2    102    103    104                  104                   100 Year.1
##3            3          3    103    104    105                  105                   100 Year.1
##4            4          4    104    105    106                  106                   100 Year.1
##5            5          5    105    106    107                  107                   100 Year.1
##6            6          6    106    107    108                  108                   100 Year.2
##7            7          7    107    108    109                  109                   100 Year.2
##8            8          8    108    109    110                  110                   100 Year.2
##9            9          9    109    110    111                  111                   100 Year.2
##10          10         10    110    111    112                  112                   100 Year.3

r - R - 使用 ifelse 语句在不同的列上分配一个数字的份额

1 回答 1

更新了可能增加或减少的同比成本

更新到每年变化成本的新要求

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