Question

dataf <- data.frame(Alert=logical(),IQR=integer(),Kurtosis=integer(),Entropy=integer(),Skewness=integer(),Sex=character(),Complex=character(),Picos=integer(),PicosFil=integer(),Umbral=integer(),Gama=character(),stringsAsFactors=FALSE)
dataf <- rbind(dataf,list(Alert=FALSE,IQR=2.6938,Kurtosis=1.73447,Entropy=1.76160,Skewness=0.140613,Sex="Mujer",Complex="Slim",Picos=0,PicosFill=0,Umbral=15.708,Gama="Alta"))
dataf <- rbind(dataf,list(Alert=FALSE,IQR=0.179574,Kurtosis=19.0538,Entropy=0.74779,Skewness=1.1355,Sex="Mujer",Complex="Slim",Picos=1,PicosFill=1,Umbral=18.975,Gama="Media"))

我遇到Gama的问题，当我输入新值（字符串）时，我收到以下错误：

警告讯息：在[<-.factor（*tmp*，ri，value =“Media”）中：无效因子水平，NA生成

Answer 1

您需要将stringsAsFactors=FALSE传递给您的列表。您可以看到添加第一行的时间，df上的所有character列都转为factor

dataf <- data.frame(Alert=logical(),IQR=integer(),Kurtosis=integer(),Entropy=integer(),Skewness=integer(),Sex=character(),Complex=character(),Picos=integer(),PicosFil=integer(),Umbral=integer(),Gama=character(),stringsAsFactors=FALSE)
dataf <- rbind(dataf,list(Alert=FALSE,IQR=2.6938,Kurtosis=1.73447,Entropy=1.76160,Skewness=0.140613,Sex="Mujer",Complex="Slim",Picos=0,PicosFill=0,Umbral=15.708,Gama="Alta"), stringsAsFactors=FALSE)
dataf <- rbind(dataf,list(Alert=FALSE,IQR=0.179574,Kurtosis=19.0538,Entropy=0.74779,Skewness=1.1355,Sex="Mujer",Complex="Slim",Picos=1,PicosFill=1,Umbral=18.975,Gama="Media"), stringsAsFactors=FALSE)

Answer 2

您可以使用do.call执行此操作。

lst <- list(
    list(Alert=FALSE,IQR=2.6938,Kurtosis=1.73447,Entropy=1.76160,Skewness=0.140613,Sex="Mujer",Complex="Slim",Picos=0,PicosFill=0,Umbral=15.708,Gama="Alta"),
    list(Alert=FALSE,IQR=0.179574,Kurtosis=19.0538,Entropy=0.74779,Skewness=1.1355,Sex="Mujer",Complex="Slim",Picos=1,PicosFill=1,Umbral=18.975,Gama="Media")
)
do.call(rbind.data.frame, lst)

请注意，do.call(rbind, lst)将返回matrix，您需要明确调用data.frame方法。

将数据帧列类型设置为字符而不是因子（默认）R

2 个答案: