根据数据子集创建新的分类变量

时间:2016-05-18 14:32:04

标签: r data-manipulation data-cleaning dummy-variable

我有一个如下所示的数据框:

         cnt    bnk qst ans
1  Country 1 Bank 1  q1   1
2  Country 2 Bank 2  q1   1
3  Country 3 Bank 3  q1   3
4  Country 4 Bank 4  q1   1
5  Country 1 Bank 1  q2   1
6  Country 2 Bank 2  q2   2
7  Country 3 Bank 3  q2   3
8  Country 4 Bank 4  q2   4
9  Country 1 Bank 1  q3   1
10 Country 2 Bank 2  q3   1
11 Country 3 Bank 3  q3   2
12 Country 4 Bank 4  q3   1

为了您的信息,q代表"问题"。所以,q2是"问题2"。同样,ans是回复。

现在,我想根据q2中的回复创建一个分类变量。特别是我想分配以下类别:

  1. 公开
  2. 私人
  3. 混合
  4. 其他
  5. 所以,如果ans=1qst=q2,这是"公开",如果ans=2qst=q2这是"私有&# 34;所以,我之后的数据框应该是这样的:

             cnt    bnk qst ans   dummy
    1  Country 1 Bank 1  q1   1  Public
    2  Country 2 Bank 2  q1   1 Private
    3  Country 3 Bank 3  q1   3   Mixed
    4  Country 4 Bank 4  q1   1  Other'
    5  Country 1 Bank 1  q2   1  Public
    6  Country 2 Bank 2  q2   2 Private
    7  Country 3 Bank 3  q2   3   Mixed
    8  Country 4 Bank 4  q2   4  Other'
    9  Country 1 Bank 1  q3   1  Public
    10 Country 2 Bank 2  q3   1 Private
    11 Country 3 Bank 3  q3   2   Mixed
    12 Country 4 Bank 4  q3   1  Other'
    

    我尝试使用ifelse,但是我没能按照自己的意愿行事。有人可以给我一些关于如何做到这一点的建议吗?

    数据

    dput(df)
    structure(list(cnt = c("Country 1", "Country 2", "Country 3", 
    "Country 4", "Country 1", "Country 2", "Country 3", "Country 4", 
    "Country 1", "Country 2", "Country 3", "Country 4"), bnk = c("Bank 1", 
    "Bank 2", "Bank 3", "Bank 4", "Bank 1", "Bank 2", "Bank 3", "Bank 4", 
    "Bank 1", "Bank 2", "Bank 3", "Bank 4"), qst = structure(c(1L, 
    1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c("q1", 
    "q2", "q3"), class = "factor"), ans = c(1L, 1L, 3L, 1L, 1L, 2L, 
    3L, 4L, 1L, 1L, 2L, 1L), dummy = c(NA, NA, NA, NA, "Public", 
    "Private", "Mixed", "Other", NA, NA, NA, NA)), .Names = c("cnt", 
    "bnk", "qst", "ans", "dummy"), row.names = c("1", "2", "3", "4", 
    "5", "6", "7", "8", "9", "10", "11", "12"), class = "data.frame")
    

2 个答案:

答案 0 :(得分:1)

以下内容将NA用于所有其他Q,

df$dummy <- ifelse(df$ans == 1 & df$qst == 'q2', 'Public', 
               ifelse(df$ans == 2 & df$qst == 'q2', 'Private', 
                   ifelse(df$ans == 3 & df$qst == 'q2', 'Mixed', 
                        ifelse(df$ans == 4 & df$qst == 'q2', 'Other', NA))))

#         cnt    bnk qst ans   dummy
#1  Country 1 Bank 1  q1   1    <NA>
#2  Country 2 Bank 2  q1   1    <NA>
#3  Country 3 Bank 3  q1   3    <NA>
#4  Country 4 Bank 4  q1   1    <NA>
#5  Country 1 Bank 1  q2   1  Public
#6  Country 2 Bank 2  q2   2 Private
#7  Country 3 Bank 3  q2   3   Mixed
#8  Country 4 Bank 4  q2   4   Other
#9  Country 1 Bank 1  q3   1    <NA>
#10 Country 2 Bank 2  q3   1    <NA>
#11 Country 3 Bank 3  q3   2    <NA>
#12 Country 4 Bank 4  q3   1    <NA>

答案 1 :(得分:1)

以下内容适用于名为df的data.frame。没有数据就很难测试:

# construct dummy variable in subset data.frame
dfCountryQ2 <- df[df$qst=="q2", c("cnt", "ans")]
dfCountryQ2$dummy <- factor(dfCountryQ2$ans, levels=1:4,
                            labels=c("Public", "Private", "Mixed", "Other"))

# merge on by country
df <- merge(df, dfCountryQ2[, c("cnt", "dummy")], by="cnt")