的主要问题是,你是治疗x
当索引时是一个类似数组的对象。即您正在使用x[row, col]
索引,您应该使用x[element]
。
将结果插入到df1$diff
时,还需要编入索引。你有欧几里得距离方程是错误的;你需要加起来的平方差,而不是减去它们。
df1 <- data.frame(list(x = 1:5, y = (1:5)^2, z = 6:10))
df1$diff <- NA
for (i in 2:nrow(df1)) {
df1$diff[i] <- with(df1, sqrt((x[i] - x[i-1])^2 +
(y[i] - y[i-1])^2 +
(z[i] - z[i-1])^2))
}
> df1
x y z diff
1 1 1 6 NA
2 2 4 7 3.316625
3 3 9 8 5.196152
4 4 16 9 7.141428
5 5 25 10 9.110434
你并不需要为这个循环,虽然,你可以依靠R上做元素乘元素的操作,因此这样做在一个单一的步骤:
df1 <- data.frame(list(x = 1:5, y = (1:5)^2, z = 6:10))
df1$diff <- c(NA, sqrt(rowSums((df1[-1, 1:3] - df1[-5, 1:3])^2)))
df1
> df1
x y z diff
1 1 1 6 NA
2 2 4 7 3.316625
3 3 9 8 5.196152
4 4 16 9 7.141428
5 5 25 10 9.110434
你可能会如果你真正的问题很大,想要用df1
强制执行此操作,因为数据帧非常慢。
m1 <- as.matrix(df1[, 1:3])
m1 <- cbind(m1, diff = c(NA, sqrt(rowSums((m1[-1, 1:3] - m1[-5, 1:3])^2))))
> m1
x y z diff
[1,] 1 1 6 NA
[2,] 2 4 7 3.316625
[3,] 3 9 8 5.196152
[4,] 4 16 9 7.141428
[5,] 5 25 10 9.110434
你可以用这个进入使用head()
和tail()
所以你do't功能需要担心的原始数据有多少行有:
myEuc <- function(x) {
if (isdf <- is.data.frame(x)) {
x <- data.matrix(x)
}
dij <- c(NA, sqrt(rowSums((tail(x, -1) - head(x, -1))^2)))
x <- cbind(x, diff = dij)
if (isdf) {
x <- as.data.frame(x)
}
x
}
df1 <- data.frame(list(x = 1:5, y = (1:5)^2, z = 6:10))
myEuc(df1)
> myEuc(df1)
x y z diff
1 1 6 NA
[2,] 2 4 7 3.316625
[3,] 3 9 8 5.196152
[4,] 4 16 9 7.141428
[5,] 5 25 10 9.110434
也许'sqrt(abs(Reduce(' - ',lapply(df1,function(x)(x-lag(x,default = x [1]))^ 2))))'或'c(0,sqrt (abs(Reduce(' - ',as.data.frame((sapply(df1,diff))^ 2)))))' – akrun