水平因子时得到错误的值答案

【问题标题】：get wrong values when level factors水平因子时得到错误的值
【发布时间】：2020-02-25 03:34:14
【问题描述】：

我有一个数据，我用 10 个观察对 5 个变量进行了子集化，以解释我的问题。

originaldata <- read_table2("Desktop/originaldata.txt", col_names = FALSE)
str(originaldata)
Classes ‘spec_tbl_df’, ‘tbl_df’, ‘tbl’ and 'data.frame':    5822 obs. of  86 variables:
- attr(*, "spec")=
  .. cols(
  ..   X1 = col_double(),
  ..   X2 = col_double(),
  ..............   
  ..   X86 = col_double()
  .. )


dt <- subset(originaldata, select = c(6:10))
dt <- dt[1:10,]
str(dt)

Classes ‘tbl_df’, ‘tbl’ and 'data.frame':   10 obs. of  5 variables:
 $ X6 : num  0 1 0 2 1 0 2 0 0 3
 $ X7 : num  5 4 4 3 4 5 2 7 1 5
 $ X8 : num  1 1 2 2 1 0 0 0 3 0
 $ X9 : num  3 4 4 4 4 5 5 2 6 2
 $ X10: num  7 6 3 5 7 0 7 7 6 7

> dput(dt)

structure(list(X6 = c(0, 1, 0, 2, 1, 0, 2, 0, 0, 3), X7 = c(5, 
4, 4, 3, 4, 5, 2, 7, 1, 5), X8 = c(1, 1, 2, 2, 1, 0, 0, 0, 3, 
0), X9 = c(3, 4, 4, 4, 4, 5, 5, 2, 6, 2), X10 = c(7, 6, 3, 5, 
7, 0, 7, 7, 6, 7)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-10L))

我想做的是重构（重新调整）数据。我使用for loop 进行但得到缺失值。

for (i in which(colnames(dt)=="X6"):which(colnames(dt)=="X10")){
    dt[,i] <- factor(dt[,i],
                              levels=c(0:9), 
                              labels=c("0%",
                                       "1-10%",
                                       "11-23%",
                                       "24-36%",
                                       "37-49%",
                                       "50-62%",
                                       "63-75%",
                                       "76-88%",
                                       "89-99%",
                                       "100%"))
}

str(dt)

Classes ‘tbl_df’, ‘tbl’ and 'data.frame':   10 obs. of  5 variables:
 $ X6 : Factor w/ 10 levels "0%","1-10%","11-23%",..: NA NA NA NA NA NA NA NA NA NA
 $ X7 : Factor w/ 10 levels "0%","1-10%","11-23%",..: NA NA NA NA NA NA NA NA NA NA
 $ X8 : Factor w/ 10 levels "0%","1-10%","11-23%",..: NA NA NA NA NA NA NA NA NA NA
 $ X9 : Factor w/ 10 levels "0%","1-10%","11-23%",..: NA NA NA NA NA NA NA NA NA NA
 $ X10: Factor w/ 10 levels "0%","1-10%","11-23%",..: NA NA NA NA NA NA NA NA NA NA

> dput(dt)

structure(list(X6 = structure(c(NA_integer_, NA_integer_, NA_integer_, 
NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_, 
NA_integer_, NA_integer_), class = "factor", .Label = c("0%", 
"1-10%", "11-23%", "24-36%", "37-49%", "50-62%", "63-75%", "76-88%", 
"89-99%", "100%")), X7 = structure(c(NA_integer_, NA_integer_, 
NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_, 
NA_integer_, NA_integer_, NA_integer_), class = "factor", .Label = c("0%", 
"1-10%", "11-23%", "24-36%", "37-49%", "50-62%", "63-75%", "76-88%", 
"89-99%", "100%")), X8 = structure(c(NA_integer_, NA_integer_, 
NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_, 
NA_integer_, NA_integer_, NA_integer_), class = "factor", .Label = c("0%", 
"1-10%", "11-23%", "24-36%", "37-49%", "50-62%", "63-75%", "76-88%", 
"89-99%", "100%")), X9 = structure(c(NA_integer_, NA_integer_, 
NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_, 
NA_integer_, NA_integer_, NA_integer_), class = "factor", .Label = c("0%", 
"1-10%", "11-23%", "24-36%", "37-49%", "50-62%", "63-75%", "76-88%", 
"89-99%", "100%")), X10 = structure(c(NA_integer_, NA_integer_, 
NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_, 
NA_integer_, NA_integer_, NA_integer_), class = "factor", .Label = c("0%", 
"1-10%", "11-23%", "24-36%", "37-49%", "50-62%", "63-75%", "76-88%", 
"89-99%", "100%"))), row.names = c(NA, -10L), class = c("tbl_df", 
"tbl", "data.frame"))

我该如何解决这个问题？我不知道是什么原因导致这些 NA.
如果我单独修改，它是正确的。见专栏X6。

dt$X6 <- factor(dt$X6,
       levels=c(0:9), 
       labels=c("0%",
                "1-10%",
                "11-23%",
                "24-36%",
                "37-49%",
                "50-62%",
                "63-75%",
                "76-88%",
                "89-99%",
                "100%"))
str(dt)

Classes ‘tbl_df’, ‘tbl’ and 'data.frame':   10 obs. of  5 variables:
 $ X6 : Factor w/ 10 levels "0%","1-10%","11-23%",..: 1 2 1 3 2 1 3 1 1 4
 $ X7 : num  5 4 4 3 4 5 2 7 1 5
 $ X8 : num  1 1 2 2 1 0 0 0 3 0
 $ X9 : num  3 4 4 4 4 5 5 2 6 2
 $ X10: num  7 6 3 5 7 0 7 7 6 7

> dput(dt)

structure(list(X6 = structure(c(1L, 2L, 1L, 3L, 2L, 1L, 3L, 1L, 
1L, 4L), .Label = c("0%", "1-10%", "11-23%", "24-36%", "37-49%", 
"50-62%", "63-75%", "76-88%", "89-99%", "100%"), class = "factor"), 
    X7 = c(5, 4, 4, 3, 4, 5, 2, 7, 1, 5), X8 = c(1, 1, 2, 2, 
    1, 0, 0, 0, 3, 0), X9 = c(3, 4, 4, 4, 4, 5, 5, 2, 6, 2), 
    X10 = c(7, 6, 3, 5, 7, 0, 7, 7, 6, 7)), row.names = c(NA, 
-10L), class = c("tbl_df", "tbl", "data.frame"))

我也试过这个并得到NA。

dt[,2] <- factor(dt[,2],
       levels=c(0:9), 
       labels=c("0%",
                "1-10%",
                "11-23%",
                "24-36%",
                "37-49%",
                "50-62%",
                "63-75%",
                "76-88%",
                "89-99%",
                "100%"))
str(dt)

Classes ‘tbl_df’, ‘tbl’ and 'data.frame':   10 obs. of  5 variables:
 $ X6 : Factor w/ 10 levels "0%","1-10%","11-23%",..: 1 2 1 3 2 1 3 1 1 4
 $ X7 : Factor w/ 10 levels "0%","1-10%","11-23%",..: NA NA NA NA NA NA NA NA NA NA
 $ X8 : num  1 1 2 2 1 0 0 0 3 0
 $ X9 : num  3 4 4 4 4 5 5 2 6 2
 $ X10: num  7 6 3 5 7 0 7 7 6 7

【问题讨论】：

不要打印出您的数据。使用dput(dt) 并将输出复制到您的问题中。
@dcarlson 我编辑了。

标签： r for-loop data-manipulation

【解决方案1】：

你可以这样做：

library(dplyr)

labs <- c("0%", "1-10%", "11-23%", "24-36%", "37-49%", "50-62%", "63-75%", "76-88%", "89-99%", "100%")

df %>%
  mutate_at(vars(starts_with("X")), factor, levels = 0:9, labels = labs)

或者：

df[grep("^X", names(df))] <- lapply(df[grep("^X", names(df))], factor, levels = 0:9, labels = labs)

df
# A tibble: 10 x 5
   X6     X7     X8     X9     X10   
   <fct>  <fct>  <fct>  <fct>  <fct> 
 1 0%     50-62% 1-10%  24-36% 76-88%
 2 1-10%  37-49% 1-10%  37-49% 63-75%
 3 0%     37-49% 11-23% 37-49% 24-36%
 4 11-23% 24-36% 11-23% 37-49% 50-62%
 5 1-10%  37-49% 1-10%  37-49% 76-88%
 6 0%     50-62% 0%     50-62% 0%    
 7 11-23% 11-23% 0%     50-62% 76-88%
 8 0%     76-88% 0%     11-23% 76-88%
 9 0%     1-10%  24-36% 63-75% 63-75%
10 24-36% 50-62% 0%     11-23% 76-88%

循环不起作用，因为您使用 [,i] 进行索引，对于 tibble 返回单列 tibble - 如果它是常规数据框，它会起作用。您需要使用[[i]] 或[, i, drop = TRUE]。

for (i in which(colnames(df)=="X6"):which(colnames(df)=="X10")){
  df[i] <- factor(df[[i]],
                   levels=c(0:9), 
                   labels=c("0%",
                            "1-10%",
                            "11-23%",
                            "24-36%",
                            "37-49%",
                            "50-62%",
                            "63-75%",
                            "76-88%",
                            "89-99%",
                            "100%"))
}

【讨论】：