当 r 中的 diff <0 时将状态值更改为 1答案

【问题标题】：changing the status value to 1 when the diff is <0 in r当 r 中的 diff <0 时将状态值更改为 1
【发布时间】：2017-12-14 11:23:55
【问题描述】：

当 Diff 值

df <- as.data.frame(list(diff=c("4","0","0","0","0","0","0","0","0","0",
                       "0","-30","0","0","0","0","0","0","0","0",
                       "14","0","0","0","0","0","0","-55","0","0",
                       "0","0","0","0","0","0","0","0","0","0",
                       "0","0","0","-40","0","0","0","0","0","0",
                       "0","0","0","0","0","0","0","0","0","0",
                       "0","-30","0","0","0","0","0","0","0","0",
                       "0","0","0","0","0"),
             status=c("0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0")))

【问题讨论】：

标签： r for-loop dplyr

【解决方案1】：

通过为 data.frame 函数提供一组字符值，您创建了一个难题。默认情况下，除非您设置 stringsAsFactors=FALSE，否则将创建因子变量。有两件事是必然的：与 0 的比较将失败，并且分配给 df$status 列中不存在的值同样会失败。

df <- data.frame(diff=c("4","0","0","0","0","0","0","0","0","0",
                       "0","-30","0","0","0","0","0","0","0","0",
                       "14","0","0","0","0","0","0","-55","0","0",
                       "0","0","0","0","0","0","0","0","0","0",
                       "0","0","0","-40","0","0","0","0","0","0",
                       "0","0","0","0","0","0","0","0","0","0",
                       "0","-30","0","0","0","0","0","0","0","0",
                       "0","0","0","0","0"),
             status=c("0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0"), stringsAsFactors=FALSE)

一旦这些问题被搁置一旁，您就可以通过以下逻辑索引分配获得成功：

df$status[as.numeric(df$diff) < 0 ] <- 1

【讨论】：

【解决方案2】：

library(dplyr)    
df <- df %>%
      mutate(status = ifelse(diff < 0, 1, 0))

【讨论】：

你能解释一下这将如何解决问题的代码吗？
虽然此代码可以解决问题，including an explanation 说明如何以及为什么解决问题将真正有助于提高您的帖子质量，并可能导致更多的赞成票。请记住，您正在为将来的读者回答问题，而不仅仅是现在提出问题的人。请编辑您的答案以添加解释，并说明适用的限制和假设。
感谢 Joseph，这是一个梦想，我必须先删除 NA 的值，以免出现任何错误
我不认为代码真的有效。本来预计 diff 列的因子性质会导致数字比较出现问题......确实如此。