Question

我有多个data.frame（即“事故，车辆和伤亡”），这些数据将作为一个事故合并在一个data.frame中。我如何找到合并data.frame的因素，即如何找到事故的因素？

$ accident_severity           : char  "Serious" "Slight" "Slight" "Slight" ...
$ number_of_vehicles          : int  1 1 2 2 1 1 2 2 2 2 ...
$ number_of_casualties        : int  1 1 1 1 1 1 1 1 1 1 ...
$ date                        : char  "04/01/2005" "05/01/2005" "06/01/2005" "06/01/2005" ...
$ day_of_week                 : char  "Tuesday" "Wednesday" "Thursday" "Thursday" ...
$ time                        : char  "17:42" "17:36" "00:15" "00:15" ...

Answer 1

您可以使用character函数将选择的列从factor转换为lapply。有关列accident_severity和day_of_week的转换，请参见下面的代码：

df <- data.frame(accident_severity= c("Serious", "Slight", "Slight", "Slight"),
                 number_of_vehicles =  c(1, 1, 2, 2),
                 number_of_casualties =  c(1,  1,  1,  1),
                 date =  c("04/01/2005", "05/01/2005", "06/01/2005", "06/01/2005"),
                 day_of_week =  c("Tuesday", "Wednesday", "Thursday", "Thursday"),
                 time = c("17:42", "17:36", "00:15", "00:15"),
                 stringsAsFactors = FALSE)
str(df)
# 'data.frame': 4 obs. of  6 variables:
#   $ accident_severity   : Factor w/ 2 levels "Serious","Slight": 1 2 2 2
# $ number_of_vehicles  : num  1 1 2 2
# $ number_of_casualties: num  1 1 1 1
# $ date                : chr  "04/01/2005" "05/01/2005" "06/01/2005" "06/01/2005"
# $ day_of_week         : Factor w/ 3 levels "Thursday","Tuesday",..: 2 3 1 1
# $ time                : chr  "17:42" "17:36" "00:15" "00:15"

df[c("accident_severity", "day_of_week")] <- lapply(df[c("accident_severity", "day_of_week")], factor)
str(df)
# 'data.frame': 4 obs. of  6 variables:
#   $ accident_severity   : Factor w/ 2 levels "Serious","Slight": 1 2 2 2
# $ number_of_vehicles  : num  1 1 2 2
# $ number_of_casualties: num  1 1 1 1
# $ date                : chr  "04/01/2005" "05/01/2005" "06/01/2005" "06/01/2005"
# $ day_of_week         : Factor w/ 3 levels "Thursday","Tuesday",..: 2 3 1 1
# $ time                : chr  "17:42" "17:36" "00:15" "00:15"

要确定列名是否是因素，可以使用is.factor函数：

names(df)[unlist(lapply(df, is.factor))]
# [1] "accident_severity" "day_of_week"

如何执行系数函数

1 个答案: