将因子级别应用于缺少因子级别的多个列 [英] Apply factor levels to multiple columns with missing factor levels

问题描述

我有一个包含许多因素的数据框，并希望创建统计表以显示每个因素的分布，包括零观测值的因素水平.例如，这些数据:

I have a dataframe with many factors and want to create statistical tables that show the distribution for each factor, including factor levels with zero observations. For instance, these data:

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，但每列仅观察到一些因子水平.生成表格时，我不仅要显示观察到的水平的计数，还要显示未观察到的水平的计数(例如强烈不同意").像这样:

show 6 rows, but only some of the factor levels are observed for each column. When I produce a table, I'd like to display not only counts for the levels observed, but also levels NOT observed (such as "Strongly Disagree"). Like this:

# 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)

这将产生一个频率表，该表显示每个级别的计数，包括未报告/未观察到的级别的零.

Which results in a frequency table that shows counts for each level, including zeroes for levels not reported / observed.

但是我有几十个要转换的变量.在Stackoverflow上推荐的一个简单的 lapply 函数似乎不起作用，例如以下代码:

But I have dozens of variables to convert. A simple lapply function recommended on Stackoverflow doesn't seem to work, such as this one:

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

我也为此尝试了一个简单的函数(n =列列表)，但失败了:

I also tried a simple function (n=list of columns) for this, but failed:

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?

Can someone offer a solution?

推荐答案

lapply 应该可以正常工作，您只需要指定 factor()函数:

An lapply should work fine, you just need to specify the factor() function:

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

然后您可以像这样调用表:

Then you can call table like:

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 

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