使用 If else 检查多个列并根据字符串响应的响应创建新列

Question

我有以下数据集：

hairdf=data.frame(
id=c(1:4),
  typedad=c("straight*","curly"),
  colourdad=c("brown","black"),
  typemom=c("curly","wavy*"),
  colourmom=c("blonde","red"),
  typekid1=c("wavy","mixed*"),
  colourkid1=c("black","blonde"))

我想创建新的列来查看发型，如果头发类型出现在不带星号的“发型”列中，则给出值 1；如果带有星号，则给出值 2（如果不带星号，则为空白）出现在该行中）。它应该看起来像这样：

id	typedad	colordad	typemom	colormom	typekid1	colorkid1	Straight	curly	wavy	Mixed
1	striaght*	Brown	Curly	Blonde	wavy	Black	2	1	1
2	Curly	Black	wavy*	Red	Mixed*	Blonde		1	2	2

我的两个问题是所有其他示例都使用数字值和所有其他示例的感兴趣的列彼此相邻。我需要看起来匹配可以位于数据帧中任何位置的列中的字符串的代码。我已经尝试过以下操作：

straight<- hairdf %>% mutate(across(c("hairtypedad", "hairtypemom", "hairtypekid1"),
                                    ifelse(.=="straight", 1
                                             ifelse(.=="straight*",2, ""
                                             ))))
curly<- hairdf %>% mutate(across(c("hairtypedad", "hairtypemom", "hairtypekid1"),
                                        ifelse(.=="curly", 1
                                                 ifelse(.=="curly*",2, ""
 wavy<- hairdf %>% mutate(across(c("hairtypedad", "hairtypemom", "hairtypekid1"),
                                        ifelse(.=="wavy", 1
                                                 ifelse(.=="wavy*",2, ""
                                                 ))))      
mixed<- hairdf %>% mutate(across(c("hairtypedad", "hairtypemom", "hairtypekid1"),
                                        ifelse(.=="mixed", 1
                                                 ifelse(.=="mixed*",2, ""
                                                 )))) 

但我不确定这段代码是否有意义。另外，这会很乏味，因为我有更多的发型，所以任何使它更容易的建议也将不胜感激！谢谢你！！！

Answer 1

这不是更有效的答案，也不是更通用的解决方案，但可能满足一个解决方案：

#create columns
st <- rep(NA,nrow(hairdf));
cur <- rep(NA,nrow(hairdf));
wav <- rep(NA,nrow(hairdf));
mix <- rep(NA,nrow(hairdf));

#join and define words
hairdf <- cbind(hairdf,st,cur,wav,mix);
words <- c("straight","curly","wavy","mixed");
words_ast <- paste(words,"*",sep=""); #just get the "*" words

#make a loop according to positions of columns st,cur,wav,mix
for (j in 1:length(words_ast)){ #let's see if we can evaluate 2 in words_ast
  for (i in c(2,3,4)){ #but only in columns we selected
    a <- subset(hairdf,hairdf[,i]==words_ast[j]) #subset columns which satisfay condition. [Note that this can be written as hairdf %>% subset(.[,i]==words_ast[j]) ]
    hairdf[row.names(a),7+j] <- 2 #replace value from column 8
  }
}
#repeat process for "words"

for (j in 1:length(words)){
  for (i in c(2,3,4)){
    a <- subset(hairdf,hairdf[,i]==words[j])
    hairdf[row.names(a),7+j] <- 1
  }
}

这应该允许您获得预期的结果。或者，您可以使用 assign() 函数，即

assign(x,value=1)

其中 x 是单词中的每个元素。

所以在一个循环中：

assign(words[n],value=1) ; assign(words_ast[n],value=2)