r - 将因子水平应用于缺少因子水平的多个列

Question

我有一个包含许多因素的数据框，想要创建统计表来显示每个因素的分布，包括观察值为零的因素水平。例如，这些数据:

structure(list(engag11 = structure(c(5L, 4L, 4L), .Label = c("Strongly Disagree", "Disagree", "Neither A or D", "Agree", "Strongly Agree"), class = "factor"), encor11 = structure(c(1L, 1L, 1L), .Label = c("Agree", "Neither Agree or Disagree", "Strongly Agree"), class = "factor"), know11 = structure(c(3L, 
1L, 1L), .Label = c("Agree", "Neither Agree or Disagree", "Strongly Agree"), class = "factor")), .Names = c("engag11", "encor11", "know11"), row.names = c(NA, 3L), class = "data.frame")

显示 6 行，但每列仅观察到部分因子水平。当我生成表格时，我不仅想显示观察到的级别的计数，还想显示未观察到的级别的计数(例如“强烈不同意”)。像这样:

# define the factor and levels
library(dplyr);library(pander);library(forcats)
eLevels<-factor(c(1,2,3,4,5), levels=1:5, labels=c("Strongly    Disagree","Disagree","Neither A or D","Agree","Strongly Agree"),ordered =TRUE )

# apply the factor to one variable
csc2$engag11<-factor(csc2$engag11,eLevels)

t1<-table(csc2$engag11)
pander(t1)

这会产生一个频率表，显示每个级别的计数，包括未报告/观察到的级别的零。

但是我有几十个变量需要转换。 Stackoverflow 上推荐的一个简单的 lapply 函数似乎不起作用，例如这个:

csc2[1:3]<-lapply(csc[1:3],eLevels)

我还为此尝试了一个简单的函数(n=列列表)，但失败了:

facConv<-function(df,n)
{   df$n<-factor(c(1,2,3,4,5), levels=1:5, labels=c("Strongly 
Disagree","Disagree","Neither A or D","Agree","Strongly Agree") )
return(result)   }

有人可以提供解决方案吗？

Answer 1

lapply 应该可以正常工作，您只需要指定 factor() 函数即可:

csc2[1:3] <- lapply(csc2[1:3], function(x) factor(x, eLevels))

然后你可以像这样调用表:

table(csc2[1])

#Strongly    Disagree             Disagree       Neither A or D                Agree       Strongly Agree 
#                   0                    0                    0                    2                    1 
table(csc2[2])

#Strongly    Disagree             Disagree       Neither A or D                Agree       Strongly Agree 
#                   0                    0                    0                    3                    0