Question

我有以下关于投资回报的数据集：

df <- structure(list(Date = structure(c(1620604800, 1620604800, 1620604800, 
1620604800, 1620604800, 1627257600, 1627257600), class = c("POSIXct", 
"POSIXt"), tzone = "UTC"), Client = c(56203, 56203, 56203, 56203, 
56203, 56203, 56203), FundName = c("SAFRA KEPLER EQUITY HEDGE FI MULTIMERCADO", 
"SAFRA KEPLER FI MULTIMERCADO", "SAFRA S&P SPECIAL FIC MULTIMERCADO", 
"SAFRA AÇÕES LIVRE FIC AÇÕES", "SAFRA CONSUMO AMERICANO FIC AÇÕES BDR NÍVEL I", 
"SAFRA S&P SPECIAL FIC MULTIMERCADO", "SAFRA CONSUMO AMERICANO FIC AÇÕES BDR NÍVEL I"
), Nature = c("Aplicação", "Aplicação", "Aplicação", "Aplicação", 
"Aplicação", "Resgate", "Resgate"), Quantity = c(145.927569, 
62.684383, 16.964545, 57.852278, 14.900635, 0, 0), Value = c(21240, 
12120, 8760, 9120, 8760, -9157.45, -1125), Saldo = c(21240, 12120, 
8760, 9120, 8760, 55.60576570794, 9198.89221970512), FundCode = c(25079578000106, 
21144577000147, 19107923000175, 32666326000149, 19436818000180, 
19107923000175, 19436818000180), NAVInitialDate = c(145.551660000114, 
193.349593000021, 521.980031999759, 157.489282999886, 639.441026999615, 
543.077092000283, 692.849144999869), NAVYesterday = c(147.526190999895, 
196.017907999922, 531.332205999643, 166.330892999889, 675.641246000305, 
531.332205999643, 675.641246000305)), row.names = c(NA, -7L), class = c("tbl_df", 
"tbl", "data.frame"))

df$NAVInitialDate = df$Value / df$Quantity

“NAVInitialDate”列是每行“价值/数量”的结果。我想对 R 说的是：“如果在一行中我们有 Quantity = 0，那么 'NAVInitialDate' = ABS(Value) 除以与行中 Fundname 具有相同 FundName 的所有行的数量总和数量等于0"

结果应该是，对于第 6 行 > NavInitialDate = 539,793 ；对于第 7 行 > NavInitialDate = 75,50

我该怎么做？我想它可能需要 3 个条件

Answer 1

我们可以在组内的所有 group_by Quantity mutate 0 中使用 dplyr、case_when FundName 和 == NAVInitialDate。

library(dplyr)

df%>%group_by(FundName)%>%
        mutate(NAVInitialDate=case_when(Quantity==0 ~ abs(Value)/sum(Quantity, na.rm=TRUE)))

# A tibble: 7 x 10
# Groups:   FundName [5]
  Date                Client FundName       Nature Quantity  Value  Saldo FundCode NAVInitialDate
  <dttm>               <dbl> <chr>          <chr>     <dbl>  <dbl>  <dbl>    <dbl>          <dbl>
1 2021-05-10 00:00:00  56203 SAFRA KEPLER … Aplic…    146.  21240  2.12e4  2.51e13           NA  
2 2021-05-10 00:00:00  56203 SAFRA KEPLER … Aplic…     62.7 12120  1.21e4  2.11e13           NA  
3 2021-05-10 00:00:00  56203 SAFRA S&P SPE… Aplic…     17.0  8760  8.76e3  1.91e13           NA  
4 2021-05-10 00:00:00  56203 SAFRA AÇÕES L… Aplic…     57.9  9120  9.12e3  3.27e13           NA  
5 2021-05-10 00:00:00  56203 SAFRA CONSUMO… Aplic…     14.9  8760  8.76e3  1.94e13           NA  
6 2021-07-26 00:00:00  56203 SAFRA S&P SPE… Resga…      0   -9157. 5.56e1  1.91e13          540. 
7 2021-07-26 00:00:00  56203 SAFRA CONSUMO… Resga…      0   -1125  9.20e3  1.94e13           75.5
# … with 1 more variable: NAVYesterday <dbl>

Answer 2

试试：

sum_0 <- sum(df[df$FundName%in%(df[df$Quantity==0,]$FundName),]$Quantity)
df$NAVInitialDate <- ifelse(df$Quantity == 0, abs(df$Value)/sum_0, df$Value / df$Quantity)

Answer 3

试试

library(tidyverse)


df <- df %>% group_by(FundName) %>% 
  mutate(sum_Quantity = sum(Quantity), 
         mycol = ifelse(Quantity==0, abs(Value)/sum_Quantity, Value/Quantity))

在这里您按 FundName 分组并应用一个简单的 mutate 命令。

为了检查，让我们看看感兴趣的行/列：

df %>% filter(Quantity==0) %>% select(FundName, Quantity, NAVInitialDate, mycol)
# A tibble: 2 x 4
# Groups:   FundName [2]
  FundName                                      Quantity NAVInitialDate mycol
  <chr>                                            <dbl>          <dbl> <dbl>
1 SAFRA S&P SPECIAL FIC MULTIMERCADO                   0           -Inf 540. 
2 SAFRA CONSUMO AMERICANO FIC AÇÕES BDR NÍVEL I        0           -Inf  75.5

在 R DataFrame 中使用多个条件进行计算

3 个答案: