r - 求解 "Error in if (obs <= ei) 2 * pv else 2 * (1 - pv) : missing value where TRUE/FALSE needed"for ape package Moran's I function in R

Question

开发商!
我遇到了错误消息

Error in if (obs <= ei) 2 * pv else 2 * (1 - pv) : missing value whereTRUE/FALSE needed

阻止我从 ape 包中获取 Moran 的 I 函数的值。这是我所做的:

library(ape)

nrstp <- data.frame(
     X = c(300226.9, 300224.6, 300226.4, 300226.1, 300224.0, 300226.4, 300225.7, 300226.4, 300226.1, 300226.4, 300226.3, 300226.3, 300227.1),
     Y = c(5057949, 5057952, 5057950, 5057950, 5057956, 5057950, 5057950, 5057950, 5057950, 5057950, 5057950, 5057950, 5057949),
     V3 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0))

nrstp = data.frame(nrstp)
dist = as.matrix(dist(cbind(nrstp$X, nrstp$Y)))
invdist = 1/dist
invdist[is.infinite(invdist)] <- 0
moranI = Moran.I(nrstp$V3, invdist)

此代码的目的是从一系列点计算 Moran's I 以检查空间自相关。到目前为止，这似乎是 R 中 Moran's I 的唯一功能。经过几次测试(我有数千组点)，这个错误似乎只发生在只有一个值的输入向量上(我尝试了其他数字而不是 0 ，它仍然会引发此错误)。
有人可以帮我改进这段代码吗？或者他们是否有更好的建议来计算 Moran's I 或从线串测试空间自相关(这些点组是一个线串的原点以及距该原点 10 米缓冲区内其他线串最近的点)？
感谢您提供任何帮助!

Answer 1

控制流选择if(condition) do something要求 condition 的值不是 NA .
在您的情况下，obs <= ei结果 NA .这就是错误消息 missing value where TRUE/FALSE needed 的原因生成。
了解如何obs <= ei结果 NA ，您可以在Moran.I中查看详细信息功能:

Moran.I
function (x, weight, scaled = FALSE, na.rm = FALSE, alternative = "two.sided") 
{
    if (dim(weight)[1] != dim(weight)[2]) 
        stop("'weight' must be a square matrix")
    n <- length(x)
    if (dim(weight)[1] != n) 
        stop("'weight' must have as many rows as observations in 'x'")
    ei <- -1/(n - 1)
    nas <- is.na(x)
    if (any(nas)) {
        if (na.rm) {
            x <- x[!nas]
            n <- length(x)
            weight <- weight[!nas, !nas]
        }
        else {
            warning("'x' has missing values: maybe you wanted to set na.rm = TRUE?")
            return(list(observed = NA, expected = ei, sd = NA, 
                p.value = NA))
        }
    }
    ROWSUM <- rowSums(weight)
    ROWSUM[ROWSUM == 0] <- 1
    weight <- weight/ROWSUM
    s <- sum(weight)
    m <- mean(x)
    y <- x - m
    cv <- sum(weight * y %o% y)
    v <- sum(y^2)
    obs <- (n/s) * (cv/v)
    if (scaled) {
        i.max <- (n/s) * (sd(rowSums(weight) * y)/sqrt(v/(n - 
            1)))
        obs <- obs/i.max
    }
    S1 <- 0.5 * sum((weight + t(weight))^2)
    S2 <- sum((apply(weight, 1, sum) + apply(weight, 2, sum))^2)
    s.sq <- s^2
    k <- (sum(y^4)/n)/(v/n)^2
    sdi <- sqrt((n * ((n^2 - 3 * n + 3) * S1 - n * S2 + 3 * s.sq) - 
        k * (n * (n - 1) * S1 - 2 * n * S2 + 6 * s.sq))/((n - 
        1) * (n - 2) * (n - 3) * s.sq) - 1/((n - 1)^2))
    alternative <- match.arg(alternative, c("two.sided", 
        "less", "greater"))
    pv <- pnorm(obs, mean = ei, sd = sdi)
    if (alternative == "two.sided") 
        pv <- if (obs <= ei) 
            2 * pv
        else 2 * (1 - pv)
    if (alternative == "greater") 
        pv <- 1 - pv
    list(observed = obs, expected = ei, sd = sdi, p.value = pv)
}
<bytecode: 0x000001cd5e0715d0>
<environment: namespace:ape>

通过分配 x = nrstp$V3和 weight = invdist , 你会得到 mean(x) = 0 .这导致 y=0 , cv = 0 , v=0 ，最后 obs = NaN .最后，

obs <= ei
[1] NA

要克服这个问题，您需要确保 obs 中的每一个和 ei不是 NA .在您的情况下，如果 mean(x)不为零， obs <= ei不会 NA .但是，因为我对这个特定主题一无所知，所以我不确定是否非零 mean(x)始终是正确的解决方案。