从一个字符串变量创建多个虚拟变量

Create several dummy variables from one string variable

我几乎尝试了 this similar question 中的所有方法,但我无法得到其他人似乎得到的结果。这是我的问题:

我有一个这样的数据框,列出了每个老师的成绩:

> profs <- data.frame(teaches = c("1st", "1st, 2nd",
                                  "2nd, 3rd",
                                  "1st, 2nd, 3rd"))
> profs
        teaches
1           1st
2      1st, 2nd
3      2nd, 3rd
4 1st, 2nd, 3rd

我一直在寻找将 teaches 变量分成列的解决方案,如下所示:

  teaches1st teaches2nd teaches3rd
1          1          0          0
2          1          1          0
3          0          1          1
4          1          1          1
考虑到回答者的解释,

I understand this solution 涉及 splitstackshape 库和显然已弃用的 concat.split.expanded 函数应该完全符合我的要求。但是,我似乎无法达到相同的结果:

> concat.split.expanded(profs, "teaches", fill = 0, drop = TRUE)
Fehler in seq.default(min(vec), max(vec)) : 
  'from' cannot be NA, NaN or infinite

使用 cSplit,我理解它取代了 "most of the earlier concat.split* functions",我得到这个:

> cSplit(profs, "teaches")
   teaches_1 teaches_2 teaches_3
1:       1st        NA        NA
2:       1st       2nd        NA
3:       2nd       3rd        NA
4:       1st       2nd       3rd

我试过使用 cSplit 的帮助并调整每一个参数,但我就是无法进行拆分。感谢您的帮助。

我找到了解决方法。如果您有一个只包含分隔符和数字的字符串变量,那么 concat.split.expanded 似乎可以工作,即:

> profs <- data.frame(teaches = c("1", "1, 2", "2, 3", "1, 2, 3"))
> profs
  teaches
1       1
2    1, 2
3    2, 3
4 1, 2, 3

现在 concat.split.expandedDummy variables from a string variable:

相同
> concat.split.expanded(profs, "teaches", fill = 0, drop = TRUE)
  teaches_1 teaches_2 teaches_3
1         1         0         0
2         1         1         0
3         0         1         1
4         1         1         1

但是,我仍在寻找一种解决方案,它不涉及从我的 teaches 变量中删除所有字母。

这是另一个选项:

Vectorize(grepl, 'pattern')(c('1st', '2nd', '3rd'), profs$teaches)
#        1st   2nd   3rd
# [1,]  TRUE FALSE FALSE
# [2,]  TRUE  TRUE FALSE
# [3,] FALSE  TRUE  TRUE
# [4,]  TRUE  TRUE  TRUE

您可以尝试 mtabulate 来自 qdapTools

library(qdapTools)
res <- mtabulate(strsplit(as.character(profs$teaches), ', '))
colnames(res) <- paste0('teaches', colnames(res))
res
#    teaches1st teaches2nd teaches3rd
#1          1          0          0
#2          1          1          0
#3          0          1          1
#4          1          1          1

或使用stringi

library(stringi)
(vapply(c('1st', '2nd', '3rd'), stri_detect_fixed, logical(4L), 
                          str=profs$teaches))+0L
#     1st 2nd 3rd
#[1,]   1   0   0
#[2,]   1   1   0
#[3,]   0   1   1
#[4,]   1   1   1

由于您的连接数据是连接的字符串(不是连接的数值),您需要添加 type = "character" 以使函数按预期工作。

该函数的默认设置是数值,因此出现有关 NaN 等的错误。

命名与同一系列中其他函数的缩写形式更加一致。因此,现在是 cSplit_e(尽管旧函数名称仍然有效)。

library(splitstackshape)
cSplit_e(profs, "teaches", ",", type = "character", fill = 0)
#         teaches teaches_1st teaches_2nd teaches_3rd
# 1           1st           1           0           0
# 2      1st, 2nd           1           1           0
# 3      2nd, 3rd           0           1           1
# 4 1st, 2nd, 3rd           1           1           1

?concat.split.expanded 的帮助页面与 cSplit_e 的帮助页面相同。如果您有任何使它更容易理解的提示,请在包的 GitHub 页面上提出问题。