在 R 中重新编码分散在多个变量中的单个变量

Question

我正在处理对种族有疑问的调查数据。每个种族类别都有其自己的变量。这是我想做的:

1. 创建一个新变量，p.race。
2. 为种族/民族的八个变量之一分配 p.race 值(如下)。
3. 确定个人是否标记了两个或多个种族，并在这种情况下为 p.race 分配值“两个或多个种族”。
4. 当 p.race 表示此种族时，为其分配值“西类牙裔或拉丁裔”。
5. 创建一个新变量 p.poc，以指示他们是否是有色人种(即不是白人，包括西类牙裔/拉丁裔)。这应该是 0 或 1。

八个种族类别是白人*、黑人*、亚洲人*、AIAN*、NHPI*、其他种族*、两个或多个种族*以及西类牙裔；其中 * 表示不是西类牙裔或拉丁裔种族。

---

这是我迄今为止尝试解析“两个或更多种族”的方法:

p['p.race'] <- NA # create new variable for race

# list of variable names that store a string indicating the race
## e.g., `race_white` would be either blank or contain "White, European, Middle Eastern, or Caucasian"
race.list <- c('p.race_white', 'p.race_black', 'p.race_asian', 'p.race_aian', 'p.race_nhpi', 'p.race_other')

# iterate through each record
for ( n in 1:length(p) ) {
  multiflag = 0

  # iterate through the race list
  for ( i in race.list ) {

    # if it is not blank, +1 to multiflag
    if ( p$i[n] != '' ) {
      multiflag <- multiflag + 1
    }
  }

  # if multiflag was flagged more than once, assign "Two or more races" to `race`
  if ( multiflag > 1 ) {
    p$p.race[n] <- 'Two or more races'
  }
}

执行时，返回此错误:

> Error in if (p$i[n] != "") { : argument is of length zero

这是我的 poc 变量编码，错误如下:

p['p.poc'] <- 0 # create a new variable for whether they are a person of color
for ( n in 1:length(p) ) {
  if ( p$p.race_black[n] == 'Black, African-American, or African'
       | p$p.race_asian[n] == 'Asian or Asian-American'
       | p$p.race_aian[n] == 'American Indian or Alaskan Native'
       | p$p.race_nhpi[n] == 'Native Hawaiian or other Pacific Islander'
       | p$p.race_other[n] == 'Other (please specify)'
       | p$p.hispanic[n] == 'Yes') {
    p$p.poc[n] <- 1
  }
}

> Error in if (p$p.race_black[n] == "Black, African-American, or African" |  : 
  missing value where TRUE/FALSE needed

我真的不知道从哪里开始分配八个种族类别之一的新 race 变量，而不使其成为非常长代码。

---

如果有帮助，请查看以下调查问题:

Q1。您认为自己有西类牙裔、拉丁裔或西类牙血统吗？

• 是
• 没有

第二季度。您认同哪个种族(勾选所有适用项)？

• 白人、欧洲人、中东人或白种人
• 黑人、非裔美国人或非洲人
• 亚裔或亚裔美国人
• 美洲印第安人或阿拉斯加原住民
• 夏威夷原住民或其他太平洋岛民
• 其他(请注明)

这是示例输出(文本被截断):

> p[264:271]
#    
#      p.hispanic  p.race_white p.race_black p.race_asian p.race_aian p.race_nhpi p.race_other
#   1  Yes         White
#   2  No          White
#   3  No                       Black
#   4  No          White                     Asian
#   5  Yes                                                                        Some other race

这是一个 dput 输出:

> dput(p[264:270])
structure(list(p.hispanic = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 2L, 
2L, 2L, 2L, 2L, 3L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("", "No", "Yes"
), class = "factor"), p.race_white = structure(c(2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 
1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 
2L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L), .Label = c("", 
"White, European, Middle Eastern, or Caucasian"), class = "factor"), 
    p.race_black = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 
    1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", 
    "Black, African-American, or African"), class = "factor"), 
    p.race_asian = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L), .Label = c("", 
    "Asian or Asian-American"), class = "factor"), p.race_aian = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 
    1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L), .Label = c("", "American Indian or Alaskan Native"
    ), class = "factor"), p.race_nhpi = c(NA, NA, NA, NA, NA, 
    NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
    NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
    NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
    NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
    NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), 
    p.race_other = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 
    1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", 
    "Other (please specify)"), class = "factor")), .Names = c("p.hispanic", 
"p.race_white", "p.race_black", "p.race_asian", "p.race_aian", 
"p.race_nhpi", "p.race_other"), class = "data.frame", row.names = c(NA, 
-79L))

Answer 1

这不是很优雅，但我认为它有效。使用循环，尤其是嵌套循环，并不是很“R”，因为它们很慢，而且还有副作用，例如使工作空间变得困惑。

如果未指定种族，您可能希望更改其处理 p.poc 的方式，因为它默认为 1，这可能不是您想要的。

所以这是一种方法:

tmp <- lapply(1:nrow(p), function(ii) {
  ## this checks for columns that aren't blank or NA, takes the colname
  ## and strips off the prefix
  tmp <- gsub('p.race_', '', names(p)[which(p[ii, -1] != '' & !is.na(p[ii, -1])) + 1])

  ## some special cases for > 1 race and blanks and p.poc
  tmp <- ifelse(length(tmp) > 1, 'Two or more', tmp)
  tmp[is.na(tmp)] <- 'Not specified'
  tmp <- ifelse(p[ii, 1] %in% 'Yes', 'Hispanic or Latino', tmp)
  p.poc <- (!grepl('white', tmp)) * 1

  return(list(p.race = tmp, p.poc = p.poc))
})

head(do.call(rbind, tmp), 20)

#   p.race               p.poc
# [1,] "white"               0    
# [2,] "white"               0    
# [3,] "white"               0    
# [4,] "white"               0    
# [5,] "white"               0    
# [6,] "white"               0    
# [7,] "white"               0    
# [8,] "white"               0    
# [9,] "asian"               1    
# [10,] "white"              0    
# [11,] "other"              1    
# [12,] "white"              0    
# [13,] "white"              0    
# [14,] "white"              0    
# [15,] "Hispanic or Latino" 1    
# [16,] "white"              0    
# [17,] "white"              0    
# [18,] "white"              0    
# [19,] "white"              0    
# [20,] "white"              0   

## and combine back to the data frame
p <- cbind(p, do.call(rbind, tmp))

数据:

p <- structure(list(p.hispanic = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 2L, 
2L, 2L, 2L, 2L, 3L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("", "No", "Yes"
), class = "factor"), p.race_white = structure(c(2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 
1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 
2L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L), .Label = c("", 
"White, European, Middle Eastern, or Caucasian"), class = "factor"), 
p.race_black = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 
1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", 
"Black, African-American, or African"), class = "factor"), 
p.race_asian = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L), .Label = c("", 
"Asian or Asian-American"), class = "factor"), p.race_aian = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 
1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L), .Label = c("", "American Indian or Alaskan Native"
), class = "factor"), p.race_nhpi = c(NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), 
p.race_other = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 
1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", 
"Other (please specify)"), class = "factor")), .Names = c("p.hispanic", 
"p.race_white", "p.race_black", "p.race_asian", "p.race_aian", 
"p.race_nhpi", "p.race_other"), class = "data.frame", row.names = c(NA, 
  -79L))