r - 如果满足条件，则在行元素中创建空白

Question

我的数据框如下:

              df <- data.frame(Resource_Code = c("B-T234","B-T234","B-T234"),
             "Valid_To"= c("2021-10-17", "2021-11-28", "2021-10-31"),
             "V1"= c("2021-10-17", "2021-10-17", "2021-10-24"),
             "V2" = c("2021-10-24", "2021-10-31", "2021-10-31"),
             "V3" = c("2021-10-31", "2021-11-07", "2021-11-14"),
             "V4" = c("2021-11-14", "2021-11-21", "2021-11-28"),
             "V5" = c("2021-12-05", "2021-11-28", "2021-12-12"),
             "V6" = c("2021-12-12", "2021-12-19", "2021-12-26"),
             "V7" = c("2022-01-02", "2022-01-09", "2022-01-16"),
             "V8" = c("2022-01-23", "2022-01-30", "2022-02-06"),
             "V9" = c("2022-02-13", "2022-02-20", "2022-02-27"),
             "v10" = c("2022-02-27", "2022-03-06", "2022-03-13"))

在所需的输出中，如果任何行单元格(从 V1 列开始)等于相应的“Valid_To”行单元格，则后续单元格(从 V1 列开始)应变为空白

实际 DF 有数百行和列，以及日期格式的 V1 到 VN 列

期望的输出如下:

              df1 <- data.frame(Resource_Code = c("B-T234","B-T234","B- 
              T234"),
             "Valid_To"= c("2021-10-17", "2021-11-28", "2021-10-31"),
             "V1"= c("2021-10-17", "2021-10-17", "2021-10-24"),
             "V2" = c("", "2021-10-31", "2021-10-31"),
             "V3" = c("", "2021-11-07", ""),
             "V4" = c("", "2021-11-21", ""),
             "V5" = c("", "2021-11-28", ""),
             "V6" = c("", "", ""),
             "V7" = c("", "", ""),
             "V8" = c("", "", ""),
             "V9" = c("", "", ""),
             "v10" = c("", "", ""))

我的代码如下:我的代码没有给出所需的输出请帮忙

             for (i in 1:nrow(df1)){
                 for (j in 3:ncol(df1) ){
                     if (df[i,j] == df1[1,2]){
                         df[i,j + 1] <- "" 
                                             }

                                        }
                                    }

Answer 1

使用apply的基本R方式。

match 返回 Valid_To 值与 V1 至 V10 列中的日期匹配的位置。如果匹配存在，那么我们将从下一个位置到系列结束的值变为空白。

df[-1] <- t(apply(df[-1], 1, function(x) {
  inds <- match(x[1], x[-1])
  if(length(inds)) x[(inds + 2):length(x)] <- ''
  x
}))
df

#  Resource_Code   Valid_To         V1         V2         V3         V4         V5 V6 V7 V8 V9 v10
#1        B-T234 2021-10-17 2021-10-17                                                            
#2        B-T234 2021-11-28 2021-10-17 2021-10-31 2021-11-07 2021-11-21 2021-11-28                
#3        B-T234 2021-10-31 2021-10-24 2021-10-31