r - 我应该如何格式化 R mlogit 包的数据？

Question

我正在将 mlogit 包与 R 结合使用。

使用以下方式导入我的数据后:

t <-read.csv('junk.csv',header=TRUE, sep=",", dec=".")

并调用:

x <- mlogit.data(t,choice="D",shape="long",id.var="key",alt.var="altkey")

我收到以下错误:

Error in `row.names<-.data.frame`(`*tmp*`, value = c("1.1", "1.2", "1.3",  : 
  duplicate 'row.names' are not allowed
In addition: Warning message:
non-unique values when setting 'row.names': ‘1.1’, ‘1.2’, ‘1.3’, ‘1.4’, ‘1.5’, ‘1.6’

有什么解决办法吗？

我的数据以以下格式存在于 csv 文件中:

[垃圾.csv]

key,altkey,A,B,C,D
201005131,1,2.6,118.17,117,0
201005131,2,1.4,117.11,115,0
201005131,3,1.1,117.38,122,1
201005131,4,24.6,,122,0
201005131,5,48.6,91.90,122,0
201005131,6,59.8,,122,0
201005132,1,20.2,118.23,113,0
201005132,2,2.5,123.67,120,1
201005132,3,7.4,116.30,120,0
201005132,4,2.8,118.86,120,0
201005132,5,6.9,124.72,120,0
201005132,6,2.5,123.81,120,0
201005132,7,8.5,119.23,115,

Answer 1

我对 mlogit 的体验是，它对于不正确的数据不太宽容。

就您的情况而言，我注意到第一个受访者有 6 个备选方案，而第二个受访者有 7 个备选方案。如果您将数据格式化为为每个受访者提供相同数量的替代方案，则 mlogit.data 函数有效:

dat <- read.table(sep=",",text="
key,altkey,A,B,C,D
201005131,1, 2.6,118.17,117,0
201005131,2,1.4,117.11,115,0
201005131,3,1.1,117.38,122,1
201005131,4,24.6,,122,0
201005131,5,48.6,91.90,122,0
201005131,6,59.8,,122,0
201005132,1,20.2,118.23,113,0
201005132,2,2.5,123.67,120,1
201005132,3,7.4,116.30,120,0
201005132,4,2.8,118.86,120,0
201005132,5,6.9,124.72,120,0
201005132,6,2.5,123.81,120,0
201005132,7,8.5,119.23,115,0
", header=TRUE)

对所有数据运行 mlogit 会重现错误:

> mlogit.data(dat, choice="D", shape="long", id.var="key", alt.var="altkey")
Error in `row.names<-.data.frame`(`*tmp*`, value = c("1.1", "1.2", "1.3",  : 
  duplicate 'row.names' are not allowed
In addition: Warning message:
non-unique values when setting 'row.names': '1.1', '1.2', '1.3', '1.4', '1.5', '1.6' 

但是，删除第 13 行(即第七个替代方案)是可行的:

> mlogit.data(dat[-13, ], choice="D", shape="long", id.var="key", alt.var="altkey")
          key altkey    A      B   C     D
1.1 201005131      1  2.6 118.17 117 FALSE
1.2 201005131      2  1.4 117.11 115 FALSE
1.3 201005131      3  1.1 117.38 122  TRUE
1.4 201005131      4 24.6     NA 122 FALSE
1.5 201005131      5 48.6  91.90 122 FALSE
1.6 201005131      6 59.8     NA 122 FALSE
2.1 201005132      1 20.2 118.23 113 FALSE
2.2 201005132      2  2.5 123.67 120  TRUE
2.3 201005132      3  7.4 116.30 120 FALSE
2.4 201005132      4  2.8 118.86 120 FALSE
2.5 201005132      5  6.9 124.72 120 FALSE
2.6 201005132      6  2.5 123.81 120 FALSE

<小时/>

当然，这不是很令人满意，因为它破坏了一些数据。更好的解决方案是按照 mlogit() 期望的格式构造数据，然后直接调用 mlogit():

dat$key <- factor(as.numeric(as.factor(dat$key)))
dat$altkey <- as.factor(dat$altkey)
dat$D <- as.logical(dat$D)
row.names(dat) <- paste(dat$key, dat$altkey, sep = ".")

现在数据如下所示:

    key altkey    A      B   C     D
1.1   1      1  2.6 118.17 117 FALSE
1.2   1      2  1.4 117.11 115 FALSE
1.3   1      3  1.1 117.38 122  TRUE
1.4   1      4 24.6     NA 122 FALSE
1.5   1      5 48.6  91.90 122 FALSE
1.6   1      6 59.8     NA 122 FALSE
2.1   2      1 20.2 118.23 113 FALSE
2.2   2      2  2.5 123.67 120  TRUE
2.3   2      3  7.4 116.30 120 FALSE
2.4   2      4  2.8 118.86 120 FALSE
2.5   2      5  6.9 124.72 120 FALSE
2.6   2      6  2.5 123.81 120 FALSE
2.7   2      7  8.5 119.23 115 FALSE

并且您可以直接调用mlogit():

mlogit(D ~ A + B + C, dat, 
       chid.var = "key", 
       alt.var = "altkey", 
       choice = "D", 
       shape = "long")

结果:

Call:
mlogit(formula = D ~ A + B + C, data = dat, chid.var = "key",     alt.var = "altkey", choice = "D", shape = "long", method = "nr",     print.level = 0)

Coefficients:
2:(intercept)  3:(intercept)  4:(intercept)  5:(intercept)  6:(intercept)  
      10.7774         4.8129         5.2257       -17.2522        -7.7364  
7:(intercept)              A              B              C  
      10.0389         1.6010         2.7156         2.9888