使用 R 处理 CSV 以评估 if ((ColA != ColB) 并考虑 ColC

Question

我正在尝试跨两列进行简单的字符串比较。 （模拟）数据样本：

EMPLID,From_DeptCode,FromDept,To_DeptCode,To_Dept,TransactionTypeCode,TransactionType,EffectiveDate,ChangeType
0239583290,21,Sales,43,CustomerService,10,Promotion,12/12/2012
1230495829,21,Sales,21,Sales,10,Promotion,9/1/2013
4059503918,93,Operations,93,Operations,10,Demotion,11/18/2014
3040593021,19,Headquarters,23,International,11,Reorg,12/13/2011
7029406920,15,Marketing,84,Development,19,Reassignment,01/05/2010
2039052819,19,Headquarters,19,Headquarters,10,Promotion,4/15/2015

我要使用的逻辑是：

If From_DeptCode = To_DeptCode 
      then ChangeType="No Change" 
ElseIf From_DeptCode != To_DeptCode AND TransactionType = "Reorg" 
      then ChangeType="Reorg"
Else ChangeType="Transfer"

所以我的输出看起来像：

EMPLID,From_DeptCode,FromDept,To_DeptCode,To_Dept,TransactionTypeCode,TransactionType,EffectiveDate,ChangeType
0239583290,21,Sales,43,CustomerService,10,Promotion,12/12/2012,Transfer
1230495829,21,Sales,21,Sales,10,Promotion,9/1/2013,No Change
4059503918,93,Operations,93,Operations,10,Demotion,11/18/2014,No Change
3040593021,19,Headquarters,23,International,11,Reorg,12/13/2011,Reorg
7029406920,15,Marketing,84,Development,19,Reassignment,01/05/2010,Transfer
2039052819,19,Headquarters,19,Headquarters,10,Promotion,4/15/2015,No Change

这是我目前所知道的：

transfers <- read.csv(file="Transfers.csv", head=TRUE,
    sep=",",colClasses=c(NA,NA,NA,NA,NA,NA,NA,"Date",NA))

在这一点上，我假设，我会实现我的逻辑：

If From_DeptCode = To_DeptCode 
      then ChangeType="No Change" 
ElseIf From_DeptCode != To_DeptCode AND TransactionType = "Reorg" 
      then ChangeType="Reorg"
Else ChangeType="Transfer"

我假设我会在这里写出我的新 csv write.csv(transfers, file = "transfersprocessed.csv", row.names = FALSE)

有什么建议可以继续前进吗？

更新：

根据@josilber 的回答，我运行了以下代码：

transfers <- read.csv(file="Transfers.csv", head=TRUE, sep=",", colClasses=c(NA,NA,NA,NA,NA,NA,NA,"Date",NA))

dat$ChangeType <- ifelse(dat$From_DeptCode == dat$To_DeptCode, "No Change",ifelse(dat$TransactionType == "Reorg", "Reorg", "Transfer"))

View(transfers)

关于以下数据：

EMPLID,From_DeptCode,FromDept,To_DeptCode,To_Dept,TransactionTypeCode,TransactionType,EffectiveDate,ChangeType
0239583290,21,Sales,43,CustomerService,10,Promotion,12/12/2012
1230495829,21,Sales,21,Sales,10,Promotion,9/1/2013
4059503918,93,Operations,93,Operations,10,Demotion,11/18/2014
3040593021,19,Headquarters,23,International,11,Reorg,12/13/2011
7029406920,15,Marketing,84,Development,19,Reassignment,01/05/2010
2039052819,19,Headquarters,19,Headquarters,10,Promotion,4/15/2015

ChangeType 变量仍然是“NA”。

嵌套的 ifelse 语句语法是否正确？知道为什么 ChangeType 不起作用吗？

Answer 1

您可以使用嵌套的ifelse 语句来做到这一点：

dat$ChangeType <- ifelse(dat$From_DeptCode == dat$To_DeptCode, "No Change",
                         ifelse(dat$TransactionType == "Reorg", "Reorg", "Transfer"))
dat
#       EMPLID From_DeptCode     FromDept To_DeptCode         To_Dept TransactionTypeCode
# 1  239583290            21        Sales          43 CustomerService                  10
# 2 1230495829            21        Sales          21           Sales                  10
# 3 4059503918            93   Operations          93      Operations                  10
# 4 3040593021            19 Headquarters          23   International                  11
# 5 7029406920            15    Marketing          84     Development                  19
# 6 2039052819            19 Headquarters          19    Headquarters                  10
#   TransactionType EffectiveDate ChangeType
# 1       Promotion    12/12/2012   Transfer
# 2       Promotion      9/1/2013  No Change
# 3        Demotion    11/18/2014  No Change
# 4           Reorg    12/13/2011      Reorg
# 5    Reassignment    01/05/2010   Transfer
# 6       Promotion     4/15/2015  No Change

ifelse 被传递一个 TRUE/FALSE 值的向量作为其第一个参数，对于 TRUE 情况使用第二个参数，对于 FALSE 情况使用第三个参数。对于您的错误案例，您实际上想要运行另一个 ifelse，这就是逻辑嵌套在这里的原因。

请注意，对于大型数据帧，这比遍历数据并一次执行嵌套 if 语句要快得多。