2

我正在处理对种族有疑问的调查数据。每个种族类别都是它自己的变量。这是我想做的事情：

  1. 创建一个新变量，p.race。
  2. 为种族/民族（下）分配p.race八个变量之一的值。
  3. 确定个人是否标记了两个或更多种族，并p.race在这种情况下分配值“两个或更多种族”。
  4. 当他们表示该种族时，分配p.race值“西班牙裔或拉丁裔”。
  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，包括西班牙裔/拉丁裔）。这应为 0 或 1。

八个种族类别是白人*、黑人*、亚洲*、AIAN*、NHPI*、其他种族*、两个或更多种族*和西班牙裔；其中 * 表示不是西班牙裔或拉丁裔。

到目前为止，这是我尝试解析“两场或多场比赛”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，它返回此错误：

> 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race八个种族类别之一的新变量，而不使它成为一个很长的代码。

如果有帮助，以下是调查问题：

Q1。您认为自己是西班牙裔、拉丁裔还是西班牙裔？

  • 是的
  • 不

Q2。您认同哪个种族（勾选所有适用项）？

  • 白人、欧洲人、中东人或高加索人
  • 黑人、非裔美国人或非洲人
  • 亚洲人或亚裔美国人
  • 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))
4

2 回答 2

2

这不是很优雅，但我认为它有效。使用循环，尤其是嵌套循环，不是很“R”，因为它们很慢，但也有副作用，比如让你的工作空间变得混乱。

如果未指定种族，您可能希望更改其处理p.poc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))
于 2014-10-15T04:28:12.063 回答
1

我带来工作的方式，如果数据是长格式而不是宽格式，这种任务似乎总是更容易。但是，这意味着每个响应都需要一个唯一的 ID - 在这种情况下，您只需为每一行分配一个整数。

library(tidyr)
library(dplyr)

# Add individual ID to each row
p = mutate(p, id = 1:n())

完成后，我会做一些工作使该p.hispanic列看起来更像其他比赛列，将数据集放入长格式，删除所有NA/空白，然后创建两个新变量。一旦创建了新变量，就可以将它们连接到原始变量。我使用包tidyr进行整形，使用dplyr进行操作。

p %>%
    mutate(p.hispanic = ifelse(p.hispanic == "No", NA, "Hispanic or Latino")) %>% # change p.hispanic column
    gather(category, answer, p.hispanic:p.race_other, na.rm = TRUE) %>%
    filter(answer != "") %>% # get rid of blanks (if were NA would have removed in "gather")
    group_by(id) %>%
    # Create new variable p.race and p.pop based on rules
    mutate(p.race = ifelse(n_distinct(answer) > 1, "Two or more races", answer),
          p.poc = as.integer(p.race == "White, European, Middle Eastern, or Caucasian")) %>%
    slice(1) %>% # take only 1 record for the duplicate id's
    select(-category, - answer) %>% # remove columns that aren't needed
    left_join(p, ., by = "id") %>% # join new columns with original dataset
    select(-id) # remove ID column if not wanted

拥有此数据集后，如果您希望级别以某种方式显示，则可以重置p.racewith的级别。factor

于 2014-10-15T17:24:10.960 回答