2

我想计算两个数据帧的行之间的距离（差异），以便为每个观察找到最近的聚类。因为我有因子和数值变量，所以我使用的是高尔距离。因为我想比较两个数据帧（而不是一个矩阵的行之间的差异），所以 gower.dist 将是我需要的函数。然而，当我实现它时，我意识到结果与我使用 daisy 的 gower 时得到的结果不同，将行绑定在一起并查看感兴趣的相异矩阵的一部分。

我在这里只提供了我的数据样本，但是当我计算所有数据的差异时，gower.dist 经常导致差异为零，尽管相应的行彼此不相等。为什么？不同结果的原因可能是什么？在我看来，daisys 的 gower 工作正常，而 gower.dist 不是（在这个例子中）。

library(cluster)
library(StatMatch)

# Calculate distance using daisy's gower 
daisyDist <- daisy(rbind(df,cent),metric="gower")
daisyDist <- as.matrix(daisyDist)
daisyDist <- daisyDist[(nrow(df)+1):nrow(daisyDist),1:nrow(df)] #only look at part where rows from df are compared to (rows of) cent

# Calculate distance using dist.gower
gowerDist <- gower.dist(cent,df)

有以下数据

df <- structure(list(searchType = structure(c(NA, 1L, 1L, 1L, 1L), .Label = c("1", "2"), class = "factor"), roomMin = structure(c(4L, 1L, 1L, 6L, 6L), .Label = c("10", "100", "150", "20", "255", "30", "40", "50", "60", "70", "Missing[NoInput]"), class = "factor"), roomMax = structure(c(8L, 8L, NA, 10L, 9L), .Label = c("10", "100", "120", "150", "160", "20", "255", "30", "40", "50", "60", "70", "80", "90", "Missing[NoInput]"), class = "factor"), priceMin = c(NA, 73, 60, 29, 11), priceMax = c(35, 11, 1, 62, 23), sizeMin = structure(c(5L, 5L, 5L, 6L, 6L), .Label = c("100", "125", "150", "250", "50", "75", "Missing[NoInput]"), class = "factor"), sizeMax = structure(c(1L, 6L, 5L, 3L, 1L), .Label = c("100", "125", "150", "250", "50", "75", "Missing[NoInput]"), class = "factor"), longitude = c(6.6306, 7.47195, 8.5562, NA, 8.569), latitude = c(46.52425, 46.9512, 47.37515, NA, 47.3929), specificSearch = structure(c(1L, 1L, 1L, 1L, 1L), .Label = c("0", "1"), class = "factor"), objectType = structure(c(NA, 2L, 2L, 2L, 2L), .Label = c("1", "2", "3", "Missing[]"), class = "factor")), .Names = c("searchType", "roomMin", "roomMax", "priceMin", "priceMax", "sizeMin", "sizeMax", "longitude", "latitude", "specificSearch", "objectType"), row.names = c(112457L,  94601L, 78273L, 59172L, 117425L), class = "data.frame")                                                                                                                                                                
cent <- structure(list(searchType = structure(c(1L, 1L, 1L), .Label = c("1", "2"), class = "factor"), roomMin = structure(c(1L, 4L, 4L), .Label = c("10", "100", "150", "20", "255", "30", "40", "50", "60", "70", "Missing[NoInput]"), class = "factor"), roomMax = structure(c(6L, 9L, 8L), .Label = c("10", "100", "120", "150", "160", "20", "255", "30", "40", "50", "60", "70", "80", "90", "Missing[NoInput]"), class = "factor"), priceMin = c(60, 33, 73), priceMax = c(103, 46, 23), sizeMin = structure(c(1L, 5L, 5L), .Label = c("100", "125", "150", "250", "50", "75", "Missing[NoInput]"), class = "factor"), sizeMax = structure(c(1L, 2L, 1L), .Label = c("100", "125", "150", "250", "50", "75", "Missing[NoInput]"), class = "factor"), longitude = c(8.3015, 7.42765, 7.6104), latitude = c(47.05485, 46.9469, 46.75125), specificSearch = structure(c(1L, 1L, 1L), .Label = c("0", "1"), class = "factor"), objectType = structure(c(2L, 2L, 2L), .Label = c("1", "2", "3", "Missing[]"), class = "factor")), .Names = c("searchType", "roomMin", "roomMax", "priceMin", "priceMax", "sizeMin", "sizeMax", "longitude", "latitude", "specificSearch", "objectType"), row.names = c(60656L, 66897L, 130650L), class = "data.frame")

谢谢！

编辑：似乎出现错误/差异是因为数字列中有 NA 并且它们似乎被不同地对待。如何使 daisy 对 NA 的处理适应 gower.dist？

4

1 回答 1

2

这是由于数据框的数字列中的 NA 值。考虑下面的代码，看看这两个函数对于具有 NA 值的数字列的行为是如何完全不同的（daisy 比 gower.dist 更健壮）：

df1 <- rbind(df,cent)
head(df1)
       searchType roomMin roomMax priceMin priceMax sizeMin sizeMax longitude latitude specificSearch objectType
112457       <NA>      20      30       NA       35      50     100   6.63060 46.52425              0       <NA>
94601           1      10      30       73       11      50      75   7.47195 46.95120              0          2
78273           1      10    <NA>       60        1      50      50   8.55620 47.37515              0          2
59172           1      30      50       29       62      75     150        NA       NA              0          2
117425          1      30      40       11       23      75     100   8.56900 47.39290              0          2
60656           1      10      20       60      103     100     100   8.30150 47.05485              0          2

# only use the numeric column priceMin (4th column) to compute the distance
class(df1[,4])
# [1] "numeric"
df2 <- df1[4]

# daisy output
as.matrix(daisy(df2,metric="gower")) 
        112457     94601     78273      59172    117425     60656      66897    130650
112457      0        NA        NA         NA        NA        NA         NA        NA
94601      NA 0.0000000 0.2096774 0.70967742 1.0000000 0.2096774 0.64516129 0.0000000
78273      NA 0.2096774 0.0000000 0.50000000 0.7903226 0.0000000 0.43548387 0.2096774
59172      NA 0.7096774 0.5000000 0.00000000 0.2903226 0.5000000 0.06451613 0.7096774
117425     NA 1.0000000 0.7903226 0.29032258 0.0000000 0.7903226 0.35483871 1.0000000
60656      NA 0.2096774 0.0000000 0.50000000 0.7903226 0.0000000 0.43548387 0.2096774
66897      NA 0.6451613 0.4354839 0.06451613 0.3548387 0.4354839 0.00000000 0.6451613
130650     NA 0.0000000 0.2096774 0.70967742 1.0000000 0.2096774 0.64516129 0.0000000

# gower.dist output
gower.dist(df2)
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN
[2,]  NaN    0    0    0    0    0    0    0
[3,]  NaN    0    0    0    0    0    0    0
[4,]  NaN    0    0    0    0    0    0    0
[5,]  NaN    0    0    0    0    0    0    0
[6,]  NaN    0    0    0    0    0    0    0
[7,]  NaN    0    0    0    0    0    0    0
[8,]  NaN    0    0    0    0    0    0    0

使用 gower.dist 函数中的参数 rngs 修复此问题：

gower.dist(df2, rngs=max(df2, na.rm=TRUE) - min(df2, na.rm=TRUE))
     [,1]      [,2]      [,3]       [,4]      [,5]      [,6]       [,7]      [,8]
[1,]  NaN       NaN       NaN        NaN       NaN       NaN        NaN       NaN
[2,]  NaN 0.0000000 0.2096774 0.70967742 1.0000000 0.2096774 0.64516129 0.0000000
[3,]  NaN 0.2096774 0.0000000 0.50000000 0.7903226 0.0000000 0.43548387 0.2096774
[4,]  NaN 0.7096774 0.5000000 0.00000000 0.2903226 0.5000000 0.06451613 0.7096774
[5,]  NaN 1.0000000 0.7903226 0.29032258 0.0000000 0.7903226 0.35483871 1.0000000
[6,]  NaN 0.2096774 0.0000000 0.50000000 0.7903226 0.0000000 0.43548387 0.2096774
[7,]  NaN 0.6451613 0.4354839 0.06451613 0.3548387 0.4354839 0.00000000 0.6451613
[8,]  NaN 0.0000000 0.2096774 0.70967742 1.0000000 0.2096774 0.64516129 0.0000000

因此，当数值变量中存在 NA 时，使函数 gower.dist 像菊花一样工作的方法可以如下所示：

df1 <- rbind(df,cent)

# compute the ranges of the numeric variables correctly
cols <- which(sapply(df1, is.numeric))
rngs <- rep(1, ncol(df1))
rngs[cols] <- sapply(df1[cols], function(x) max(x, na.rm=TRUE) - min(x, na.rm=TRUE)) 

daisyDist <- as.matrix(daisy(df1,metric="gower"))
gowerDist <- gower.dist(df1)

daisyDist
          112457     94601     78273     59172    117425     60656     66897    130650
112457 0.0000000 0.3951059 0.6151851 0.7107843 0.6397059 0.6424374 0.3756990 0.1105551
94601  0.3951059 0.0000000 0.2355126 0.5788530 0.5629176 0.4235379 0.3651002 0.2199324
78273  0.6151851 0.2355126 0.0000000 0.5122549 0.4033046 0.3500130 0.3951874 0.3631533
59172  0.7107843 0.5788530 0.5122549 0.0000000 0.2969639 0.5446623 0.4690421 0.5657812
117425 0.6397059 0.5629176 0.4033046 0.2969639 0.0000000 0.4638003 0.4256891 0.4757460
60656  0.6424374 0.4235379 0.3500130 0.5446623 0.4638003 0.0000000 0.5063082 0.4272755
66897  0.3756990 0.3651002 0.3951874 0.4690421 0.4256891 0.5063082 0.0000000 0.2900150
130650 0.1105551 0.2199324 0.3631533 0.5657812 0.4757460 0.4272755 0.2900150 0.0000000

gowerDist
          [,1]      [,2]      [,3]      [,4]      [,5]      [,6]      [,7]      [,8]
[1,] 0.0000000 0.3951059 0.6151851 0.7107843 0.6397059 0.6424374 0.3756990 0.1105551
[2,] 0.3951059 0.0000000 0.2355126 0.5788530 0.5629176 0.4235379 0.3651002 0.2199324
[3,] 0.6151851 0.2355126 0.0000000 0.5122549 0.4033046 0.3500130 0.3951874 0.3631533
[4,] 0.7107843 0.5788530 0.5122549 0.0000000 0.2969639 0.5446623 0.4690421 0.5657812
[5,] 0.6397059 0.5629176 0.4033046 0.2969639 0.0000000 0.4638003 0.4256891 0.4757460
[6,] 0.6424374 0.4235379 0.3500130 0.5446623 0.4638003 0.0000000 0.5063082 0.4272755
[7,] 0.3756990 0.3651002 0.3951874 0.4690421 0.4256891 0.5063082 0.0000000 0.2900150
[8,] 0.1105551 0.2199324 0.3631533 0.5657812 0.4757460 0.4272755 0.2900150 0.0000000
于 2016-10-27T11:10:34.613 回答