我有以下问题，我想合并距离矩阵，考虑以下数据：
x1 <- c(2,2,2,3,1,2,4,6,1,2,4)
y1 <- c(5,4,3,3,4,2,1,6,4,2,3)
x2 <- c(8,2,7,3,1,2,2,2,1,2,6)
y2 <- c(1,3,3,3,1,2,4,3,1,2,8)
x3 <- c(4,4,1,2,4,6,3,2,4,6,9)
y3 <- c(1,2,3,3,1,2,4,6,1,2,1)
x4 <- c(4,4,1,2,4,6,3,2,4,6,9)
y4 <- c(1,2,3,3,1,2,4,6,1,2,1)
x5 <- c(4,1,2,4,6,2,3,3,6,2,9)
y5 <- c(1,3,3,3,1,2,4,6,1,2,1)
id1 <- c("a","a","a","a","b","b","b","b","b","b","b")
dat <- data.frame(x1,y1,x2,y2,x3,y3,id1)
我想创建每组中各点之间的距离：
distance <- by(dat, list(id1=dat$id1), function(x){
dist(x, upper=TRUE, diag = TRUE)
})
如何合并它们（rbind，与NAs，因为矩阵的列数较少），以便我可以将原始数据集的距离作为新变量添加？
最终输出应如下所示
id1 Dist1 Dist2 Dist3 Dist4 Dist5 Dist6
a 0 7 5.066228 7 NA and so on ---
a 7 0 6.480741 3.05505 NA
a 5.066228 6.480741 0 4.582576 NA
a 7 3.05505 4.582576 0 NA
b 0 3.741657 6.658328 8.573214 0
b 3.741657 0 5.066228 8.708234 3.741657
b 6.658328 5.066228 0 6.390097 6.658328
b 8.573214 8.708234 6.390097 0 8.573214
b 0 3.741657 6.658328 8.573214 0
b 3.741657 0 5.066228 8.708234 3.741657
b 11.27682 8.841191 9.721111 12.220202 11.27682
请注意，实际上大约有20组，总共有10000行数据。这只是一个简化版本。谢谢！
答案 0 :(得分：1)
以下结合了许多简单的步骤，但它们有效：
> aa = dist(dat[dat$id1=='a',], upper=T)
Warning message:
In dist(dat[dat$id1 == "a", ], upper = T) : NAs introduced by coercion
> bb = dist(dat[dat$id1=='b',], upper=T)
Warning message:
In dist(dat[dat$id1 == "b", ], upper = T) : NAs introduced by coercion
>
> aa
1 2 3 4
1 7.000000 5.066228 7.000000
2 7.000000 6.480741 3.055050
3 5.066228 6.480741 4.582576
4 7.000000 3.055050 4.582576
> bb
5 6 7 8 9 10 11
5 3.741657 6.658328 8.573214 0.000000 3.741657 11.276820
6 3.741657 5.066228 8.708234 3.741657 0.000000 8.841191
7 6.658328 5.066228 6.390097 6.658328 5.066228 9.721111
8 8.573214 8.708234 6.390097 8.573214 8.708234 12.220202
9 0.000000 3.741657 6.658328 8.573214 3.741657 11.276820
10 3.741657 0.000000 5.066228 8.708234 3.741657 8.841191
11 11.276820 8.841191 9.721111 12.220202 11.276820 8.841191
>
> aadf = data.frame(as.matrix(aa))
> aadf
X1 X2 X3 X4
1 0.000000 7.000000 5.066228 7.000000
2 7.000000 0.000000 6.480741 3.055050
3 5.066228 6.480741 0.000000 4.582576
4 7.000000 3.055050 4.582576 0.000000
>
> bbdf = data.frame(as.matrix(bb))
> bbdf
X5 X6 X7 X8 X9 X10 X11
5 0.000000 3.741657 6.658328 8.573214 0.000000 3.741657 11.276820
6 3.741657 0.000000 5.066228 8.708234 3.741657 0.000000 8.841191
7 6.658328 5.066228 0.000000 6.390097 6.658328 5.066228 9.721111
8 8.573214 8.708234 6.390097 0.000000 8.573214 8.708234 12.220202
9 0.000000 3.741657 6.658328 8.573214 0.000000 3.741657 11.276820
10 3.741657 0.000000 5.066228 8.708234 3.741657 0.000000 8.841191
11 11.276820 8.841191 9.721111 12.220202 11.276820 8.841191 0.000000
>
>
> aadf$newcol = 1:4
> aadf
X1 X2 X3 X4 newcol
1 0.000000 7.000000 5.066228 7.000000 1
2 7.000000 0.000000 6.480741 3.055050 2
3 5.066228 6.480741 0.000000 4.582576 3
4 7.000000 3.055050 4.582576 0.000000 4
>
>
> bbdf$newcol = 5:11
> bbdf
X5 X6 X7 X8 X9 X10 X11 newcol
5 0.000000 3.741657 6.658328 8.573214 0.000000 3.741657 11.276820 5
6 3.741657 0.000000 5.066228 8.708234 3.741657 0.000000 8.841191 6
7 6.658328 5.066228 0.000000 6.390097 6.658328 5.066228 9.721111 7
8 8.573214 8.708234 6.390097 0.000000 8.573214 8.708234 12.220202 8
9 0.000000 3.741657 6.658328 8.573214 0.000000 3.741657 11.276820 9
10 3.741657 0.000000 5.066228 8.708234 3.741657 0.000000 8.841191 10
11 11.276820 8.841191 9.721111 12.220202 11.276820 8.841191 0.000000 11
>
> dat$newcol = rownames(dat)
> dat
x1 y1 x2 y2 x3 y3 id1 newcol
1 2 5 8 1 4 1 a 1
2 2 4 2 3 4 2 a 2
3 2 3 7 3 1 3 a 3
4 3 3 3 3 2 3 a 4
5 1 4 1 1 4 1 b 5
6 2 2 2 2 6 2 b 6
7 4 1 2 4 3 4 b 7
8 6 6 2 3 2 6 b 8
9 1 4 1 1 4 1 b 9
10 2 2 2 2 6 2 b 10
11 4 3 6 8 9 1 b 11
>
> aa1 = merge(dat, aadf, id='newcol')
> bb1 = merge(dat, bbdf, id='newcol')
>
> aa1
newcol x1 y1 x2 y2 x3 y3 id1 X1 X2 X3 X4
1 1 2 5 8 1 4 1 a 0.000000 7.000000 5.066228 7.000000
2 2 2 4 2 3 4 2 a 7.000000 0.000000 6.480741 3.055050
3 3 2 3 7 3 1 3 a 5.066228 6.480741 0.000000 4.582576
4 4 3 3 3 3 2 3 a 7.000000 3.055050 4.582576 0.000000
> bb1
newcol x1 y1 x2 y2 x3 y3 id1 X5 X6 X7 X8 X9 X10 X11
1 10 2 2 2 2 6 2 b 3.741657 0.000000 5.066228 8.708234 3.741657 0.000000 8.841191
2 11 4 3 6 8 9 1 b 11.276820 8.841191 9.721111 12.220202 11.276820 8.841191 0.000000
3 5 1 4 1 1 4 1 b 0.000000 3.741657 6.658328 8.573214 0.000000 3.741657 11.276820
4 6 2 2 2 2 6 2 b 3.741657 0.000000 5.066228 8.708234 3.741657 0.000000 8.841191
5 7 4 1 2 4 3 4 b 6.658328 5.066228 0.000000 6.390097 6.658328 5.066228 9.721111
6 8 6 6 2 3 2 6 b 8.573214 8.708234 6.390097 0.000000 8.573214 8.708234 12.220202
7 9 1 4 1 1 4 1 b 0.000000 3.741657 6.658328 8.573214 0.000000 3.741657 11.276820
> datfinal = merge(aa1, bb1, id='newcol', all=T)
> datfinal
newcol x1 y1 x2 y2 x3 y3 id1 X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 1 2 5 8 1 4 1 a 0.000000 7.000000 5.066228 7.000000 NA NA NA NA NA NA
2 10 2 2 2 2 6 2 b NA NA NA NA 3.741657 0.000000 5.066228 8.708234 3.741657 0.000000
3 11 4 3 6 8 9 1 b NA NA NA NA 11.276820 8.841191 9.721111 12.220202 11.276820 8.841191
4 2 2 4 2 3 4 2 a 7.000000 0.000000 6.480741 3.055050 NA NA NA NA NA NA
5 3 2 3 7 3 1 3 a 5.066228 6.480741 0.000000 4.582576 NA NA NA NA NA NA
6 4 3 3 3 3 2 3 a 7.000000 3.055050 4.582576 0.000000 NA NA NA NA NA NA
7 5 1 4 1 1 4 1 b NA NA NA NA 0.000000 3.741657 6.658328 8.573214 0.000000 3.741657
8 6 2 2 2 2 6 2 b NA NA NA NA 3.741657 0.000000 5.066228 8.708234 3.741657 0.000000
9 7 4 1 2 4 3 4 b NA NA NA NA 6.658328 5.066228 0.000000 6.390097 6.658328 5.066228
10 8 6 6 2 3 2 6 b NA NA NA NA 8.573214 8.708234 6.390097 0.000000 8.573214 8.708234
11 9 1 4 1 1 4 1 b NA NA NA NA 0.000000 3.741657 6.658328 8.573214 0.000000 3.741657
X11
1 NA
2 8.841191
3 0.000000
4 NA
5 NA
6 NA
7 11.276820
8 8.841191
9 9.721111
10 12.220202
11 11.276820
>
编辑：
以下自动照顾所有群组：
automatic = function(dat){
dat$newcol = rownames(dat)
dat$newcol = as.numeric(dat$newcol)
gps = unique(dat$id1)
for(gp in gps){
aa = dist(dat[dat$id1==gp,], upper=T)
aadf = data.frame(as.matrix(aa))
aadf$newcol = rownames(aadf)
aadf$newcol = as.numeric(aadf$newcol)
dat = merge(dat, aadf, id='newcol', all=T)
}
dat
}
automatic(dat)
newcol x1 y1 x2 y2 x3 y3 id1 X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 1 2 5 8 1 4 1 a 0.000000 7.010197 5.451081 7.634508 NA NA NA NA NA NA
2 2 2 4 2 3 4 2 a 7.010197 0.000000 6.502747 3.703280 NA NA NA NA NA NA
3 3 2 3 7 3 1 3 a 5.451081 6.502747 0.000000 4.659859 NA NA NA NA NA NA
4 4 3 3 3 3 2 3 a 7.634508 3.703280 4.659859 0.000000 NA NA NA NA NA NA
5 5 1 4 1 1 4 1 b NA NA NA NA 0.000000 4.720775 8.485281 11.109841 5.237229 7.964206
6 6 2 2 2 2 6 2 b NA NA NA NA 4.720775 0.000000 6.279217 10.875924 6.000000 5.237229
7 7 4 1 2 4 3 4 b NA NA NA NA 8.485281 6.279217 0.000000 7.855844 8.485281 7.289915
8 8 6 6 2 3 2 6 b NA NA NA NA 11.109841 10.875924 7.855844 0.000000 10.474459 10.875924
9 9 1 4 1 1 4 1 b NA NA NA NA 5.237229 6.000000 8.485281 10.474459 0.000000 4.720775
10 10 2 2 2 2 6 2 b NA NA NA NA 7.964206 5.237229 7.289915 10.875924 4.720775 0.000000
11 11 4 3 6 8 9 1 b NA NA NA NA 15.766148 12.558435 12.895182 15.325050 13.918128 10.796825
X11
1 NA
2 NA
3 NA
4 NA
5 15.76615
6 12.55843
7 12.89518
8 15.32505
9 13.91813
10 10.79682
11 0.00000
Warning messages:
1: In dist(dat[dat$id1 == gp, ], upper = T) : NAs introduced by coercion
2: In dist(dat[dat$id1 == gp, ], upper = T) : NAs introduced by coercion
答案 1 :(得分：0)
这是另一种与你拥有多少x1，y1对无关的方式。它确实假设只有a和b组，但不是例如c。
这显然与rnso提供的内容不同，因为他为NAs组中的某些人包含b，而我的不包括a <- as.matrix(distance[["a"]])
b <- as.matrix(distance[["b"]])
a_na <- matrix(NA, nrow = nrow(a), ncol = ncol(b) - ncol(a))
a_merged <- cbind(a, a_na)
ab_merged <- rbind(a_merged, b)
dat2 <- cbind(dat, ab_merged)
names(dat2)[(ncol(dat) + 1):ncol(dat2)] <- paste0("Dist", 1:ncol(ab_merged))
。如果这是您正在寻找的内容，那么请不要使用此功能。
x1 y1 x2 y2 x3 y3 id1 Dist1 Dist2 Dist3 Dist4 Dist5 Dist6 Dist7
1 2 5 8 1 4 1 a 0.000000 7.000000 5.066228 7.000000 NA NA NA
2 2 4 2 3 4 2 a 7.000000 0.000000 6.480741 3.055050 NA NA NA
3 2 3 7 3 1 3 a 5.066228 6.480741 0.000000 4.582576 NA NA NA
4 3 3 3 3 2 3 a 7.000000 3.055050 4.582576 0.000000 NA NA NA
5 1 4 1 1 4 1 b 0.000000 3.741657 6.658328 8.573214 0.000000 3.741657 11.276820
6 2 2 2 2 6 2 b 3.741657 0.000000 5.066228 8.708234 3.741657 0.000000 8.841191
7 4 1 2 4 3 4 b 6.658328 5.066228 0.000000 6.390097 6.658328 5.066228 9.721111
8 6 6 2 3 2 6 b 8.573214 8.708234 6.390097 0.000000 8.573214 8.708234 12.220202
9 1 4 1 1 4 1 b 0.000000 3.741657 6.658328 8.573214 0.000000 3.741657 11.276820
10 2 2 2 2 6 2 b 3.741657 0.000000 5.066228 8.708234 3.741657 0.000000 8.841191
11 4 3 6 8 9 1 b 11.276820 8.841191 9.721111 12.220202 11.276820 8.841191 0.000000
结果如下：
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