R：使用不存在的变量进行子集设置不会产生错误

在创建lead变量时，我无意中遗漏了用于分组数据的lead变量。我使用括号插入NA，没有报告错误。为了检查我的理智，我对ifelse做了同样的操作，并创建了一条错误消息。我担心的是，如果没有仔细的回顾和一些运气，我可能永远不会知道我的错误

其他人如何以不同的编码方式在未来降低这种可能性（以最小的时间成本）？另外，还有其他类似的问题需要我注意吗？谢谢，下面是可复制的示例

dt <- data.frame(
group_name = c("D44", "D44","D44", "D45", "D45", "D47", "D47", "D47", "D47", "D48"),
order_number = sample(1:10))

dt$group_name <- as.character(dt$group_name) # so not a factor

dt <- dt[order(dt$group_name, dt$order_number),] # sort data

dt$lead1order_number <- c(dt$order_number[-1], NA)

# COMMENT OUT NEXT LINE AND RUN, no error with brackets, but one with ifelse
dt$lead1group_name <- c(dt$group_name[-1], NA) 

# done two different ways below
    # if group_name doesn't match lead1group_name, then lead1order_number NA
dt$lead1order_number[dt$group_name != dt$lead1group_name] <- NA  

dt$lead1order_number <- ifelse(dt$group_name != dt$lead1group_name, NA, dt$lead1order_number)

---

dt你的问题很深。括号或子集的问题是R的关键特征之一。很难全面回答您的问题。我只是提出一个可能的最简单的解决方案：

# `stringsAsFactors = FALSE` ensures that strings will not be transformed to factors
dt <- data.frame(group_name = c("D44", "D44","D44", "D45", 
    "D45", "D47", "D47", "D47", "D47", "D48"),
    order_number = sample(1:10), stringsAsFactors = FALSE)
dt <- dt[order(dt$group_name, dt$order_number),] # sort data
dt$lead1order_number <- c(dt$order_number[-1], NA)
# the example was slightly modified to demonstrate subsetting with NA
dt$lead1group_name <- c(dt$group_name[-c(1:2)], NA, "D")

使用
[
保留子集将导致错误：

print(dt[ ,"lead2group_name", drop = FALSE])

[.data.frame（dt，“lead2group_name”）中出错：未定义列
精选

我将使用此问题来确保data.frame中存在请求的列：

ind_of_non_match <- which(dt[ ,"group_name", drop = FALSE] != dt[ ,"lead1group_name", drop = FALSE])
ind_of_na <- which(is.na(dt[ , "lead1group_name", drop = FALSE]))
dt$lead1order_number[c(ind_of_non_match, ind_of_na)] <- NA

ind\u of\u non\u match你到底在问什么？不确定你指的是什么括号。有或没有分组变量，lead/lag都有意义。这一切都取决于你在做什么。使用
dplyr
会容易得多。然后你可以使用
group\u by（）
命令和
mutate（）
和
lead（）
函数。@MrFlick我使用lags来查看组名中下一个订单号的订单号。当该订单号与下一个组名重叠时，尽管我需要将其设置为NA。重叠由组名！=lead1group\u名称标识。当我在括号内设置该条件时，[dt$group\U name！=dt$lead1group\U name]，即使lead1group\U name不存在，也不会给我错误。我同意dplyr是一个选项，但重点是在使用括号与ifelse（或dplyr）比较时显示错误。
ind_of_non_match <- which(dt[ ,"group_name", drop = FALSE] != dt[ ,"lead1group_name", drop = FALSE])
ind_of_na <- which(is.na(dt[ , "lead1group_name", drop = FALSE]))
dt$lead1order_number[c(ind_of_non_match, ind_of_na)] <- NA

dt$lead1order_number[(dt[ ,"group_name", drop = FALSE] != dt[ ,"lead1group_name", drop = FALSE])] <- NA