Question

给定两个相同维度的数据框：

Categories.df1 <- data.frame(Categories)
Returns.df2 <- data.frame(Returns)

Categories.df1
#Fund             X31.05.2015     X30.04.2015     X31.03.2015
#1 3 Banken EU                                     Concentrated
#2 3 Banken GL                                     Stock Pickers
#3 3 Banken NAC                                    Stock Pickers
#4 3 Banken O                                      Factor Betters
#5 3V Invest      Stock Pickers   Stock Pickers    Moderately Active
#6 4Q EU          Stock Pickers   Stock Pickers    Stock Pickers

Returns.df2
#Fund             X31.05.2015    X30.04.2015      X31.03.2015
#1 3 Banken EU     0.01           0.02             
#2 3 Banken GL    -0.04          -0.01             0.03
#3 3 Banken NAC   -0.02          -0.01             0.04
#4 3 Banken O     -0.01                           -0.02
#5 3V Invest       0.01           0.02             0.00
#6 4Q EU          -0.01          -0.01             0.00

我可以使用Returns.df2中的值替换Categories.df1中的值，条件是“Stock Pickers”（否则为空白）？

因此，我想创建一个新的数据框或子集，如下所示：

Returns.StockPickers.df3
#Fund             X31.05.2015    X30.04.2015      X31.03.2015
#1 3 Banken EU                           
#2 3 Banken GL                                     0.03
#3 3 Banken NAC                                    0.04
#4 3 Banken O                                
#5 3V Invest       0.01           0.02             
#6 4Q EU          -0.01          -0.01             0.01

任何输入？

Answer 1

这应该这样做（我更新了片段并包含了一个函数，它接受两个不同的data.frames并返回带有新值的return_data.frame。但是，新值是一个字符串，而不是现在的数字）：

df1 <- data.frame(fund = c("Fund 1", "Fund 2", "Fund 3", "Fund 4"),
                  X31.05.2015 = c("Stock Picker", "", "Stock Picker", ""))

df2 <- data.frame(fund = c("Fund 1", "Fund 2", "Fund 3", "Fund 4"),
                  X31.05.2015 = c(1, 2, 3, 4))


fun <- function(cat_df, ret_df, col_name) {
  ret_df[, col_name] <- ifelse(cat_df[, col_name] == "Stock Picker", ret_df[, col_name],
                               "") # change '""' to 0 or NA if you want to get Numerics 
  return(ret_df)
}

df3 <- fun(df1, df2, col_name = "X31.05.2015")

df3
# fund X31.05.2015
# 1 Fund 1           1
# 2 Fund 2            
# 3 Fund 3           3
# 4 Fund 4

将df1中的值替换为df2中以名称

1 个答案: