1

我有以下代码

fig4 <- data.frame(chads=NA,age=NA,treatment=NA,mean=NA,lower=NA,upper=NA)
fig4$chads <- as.factor(fig4$chads)
levels(fig4$chads) <- c(0,1,2,3,4,5,6)
fig4$age <- as.factor(fig4$age)
levels(fig4$age ) <- c("u80","o80")
fig4$treatment <- as.factor(fig4$treatment)
levels(fig4$treatment) <- c("OAC","OAP")
fig4$mean <- as.numeric(fig4$mean)
fig4$lower <- as.numeric(fig4$lower)
fig4$upper <- as.numeric(fig4$upper)

> str(fig4)
'data.frame':   1 obs. of  6 variables:
 $ chads    : Factor w/ 7 levels "0","1","2","3",..: NA
 $ age      : Factor w/ 2 levels "u80","o80": NA
 $ treatment: Factor w/ 2 levels "OAC","OAP": NA
 $ mean     : num NA
 $ lower    : num NA
 $ upper    : num NA

到现在为止还挺好。但后来我这样做：

vc <- as.vector(c(6,"o80","OAC",0.1,0.02,0.25), mode = "any")
fig4 <- rbind(fig4,vc)

结果是：

> str(fig4)
'data.frame':   2 obs. of  6 variables:
 $ chads    : Factor w/ 7 levels "0","1","2","3",..: NA 7
 $ age      : Factor w/ 2 levels "u80","o80": NA 2
 $ treatment: Factor w/ 2 levels "OAC","OAP": NA 1
 $ mean     : chr  NA "0.1"
 $ lower    : chr  NA "0.02"
 $ upper    : chr  NA "0.25"

为什么数字向量变成字符向量？

4

2 回答 2

3

列表可以保存多种类型的对象，因此为避免将新数据转换为字符，您可以执行以下操作：

fig4[nrow(fig4) + 1, ] <- list(6,"o80","OAC",0.1,0.02,0.25)
于 2012-09-27T00:46:22.903 回答
2

出于同样的原因，矩阵会 --- 向量和矩阵都只能保存一种类型。当你强迫角色融入其中时，你就会得到角色。

使用 data.frame 来保存不同类型的“列”，然后对各个列进行子集化。

于 2012-09-27T00:22:55.913 回答