Question

我有一个数据框如下所示，我在lse列中比较连续值。如果lse 0的值后跟1，我想将END_DATE lse的{{1}}值设为0 NA 输入：

END_DATE      lse   
2016-01-20     0  
2015-08-16     1

预期产出：

  END_DATE     lse   
        NA     0  
2015-08-16     1

我已经尝试过以下代码，但它无法正常工作而且没有任何错误。

for (i in length(loan_test1$lse)){
  if (length(loan_test1$lse) == 0 & length(loan_test1$lse)+1 == 1) {
    loan_test1$END_DATE[i] <- "NA"
  }
}

任何人都可以帮助我吗？

Answer 1

library(lubridate)
library(tidyverse)
df <- data.frame(END_DATE = seq.Date(from = (today()-99),to = today(),by=1),lse = sample(c(0,1), 100, replace = TRUE))

df <- df %>% mutate(leadlse = lead(lse))

condition <- df$lse == 0 & df$leadlse ==1

df$END_DATE[condition] = NA

df <- df %>% select(-leadlse)
df

或更简单：

df <- data.frame(END_DATE = seq.Date(from = (today()-99),to = today(),by=1),lse = sample(c(0,1), 100, replace = TRUE))

df$END_DATE[(df$lse == 0 & lead(df$lse) == 1)] <- NA

比较连续的列单元格值并替换其他列值

1 个答案: