Question

我有两个向量，如：

one <- c("A", "B", "C", "D", "E", "F")
two <- c("1", "2", "3", "4", "5", "6")

然后我有一个包含原始值的数据框：

我想替换原始df中的所有条目，如：

from   to
one    two
two    one
two    one
one    one
one    two
one    two
two    two

我正在尝试关注this，但它只适用于我做的第一次检查，然后转换所有数字：

df[] <- as.data.frame(lapply(df, function(x) ifelse(x%in%one, "one", x)))

当然这只是一个假的例子，在我的实际案例中，这里报告的时间太长，原始df中的所有值都是字符串

Answer 1

我们将'one'，'two'向量放在list，stack中创建一个键，值data.frame（'d1'），循环遍历初始数据集（{ {1}}）和lapply(df1,..带有'values'列的向量来获取'index'并替换为与索引匹配的'ind'列。

match

数据

d1 <- stack(list(one=one, two=two))
df2 <- df1
df2[] <- lapply(df1, function(x) d1$ind[match(x, d1$values)])
df2
#  from  to
#1  one two
#2  two one
#3  two one
#4  one one
#5  one two
#6  one two
#7  two two

Answer 2

我认为这就是你要做的事情：

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或类似地，使用> df <- data.frame(from=c("A",2,3,"F","C","D",6),to=c(1,"A","B","B",4:6)) > one <- c("A", "B", "C", "D", "E", "F") > two <- c("1", "2", "3", "4", "5", "6") > as.data.frame(apply(df,2,function(x)ifelse(x%in%one,"one",ifelse(x%in%two,"two",x)))) from to 1 one two 2 two one 3 two one 4 one one 5 one two 6 one two 7 two two技巧：

df[]

或者再次使用> df[]<-apply(df,2,function(x)ifelse(x%in%one,"one",ifelse(x%in%two,"two",x))) > df from to 1 one two 2 two one 3 two one 4 one one 5 one two 6 one two 7 two two：

lapply

如果value是％vector中的％，如何替换列中的值

2 个答案:

数据