在R中，将浮点坐标离散到最近的坐标

我正在尝试将坐标转换为最接近的坐标。从某种意义上说，我正在做一次kmeans聚类迭代，有1222个质心。在下面，我有一个功能可以做到这一点，不完美和太慢。我寻求帮助改善这一功能：在R中，将浮点坐标离散到最近的坐标

discretizeCourt <- function(x_loc, y_loc) { 

    # create the dataframe of points that I want to round coordinates to 
    y <- seq(0, 50, by = 2) 
    x1 <- seq(1, 93, by = 2) 
    x2 <- seq(2, 94, by = 2) 
    x <- c(x1, x2) 

    coordinates <- data.frame(
    x = rep(x, 13), 
    y = rep(y, each = length(x1)), 
    count = 0 
) 

    # loop over each point in x_loc and y_loc 
    # increment the count column whenever a point is 'near' that column  
    for(i in 1:length(x_loc)) { 
    this_x = x_loc[i] 
    this_y = y_loc[i] 

    coordinates[coordinates$x > this_x-1 & 
       coordinates$x < this_x+1 & 
       coordinates$y > this_y-1 & 
       coordinates$y < this_y+1, ]$count = 
     coordinates[coordinates$x > this_x-1 & 
        coordinates$x < this_x+1 & 
        coordinates$y > this_y-1 & 
        coordinates$y < this_y+1, ]$count + 1 
    } 
}

这里是我工作的一些测试数据：

> dput(head(x_loc, n = 50)) 
c(13.57165, 13.61702, 13.66478, 13.70833, 13.75272, 13.7946, 
13.83851, 13.86792, 13.8973, 13.93906, 13.98099, 14.02396, 14.06338, 
14.10872, 14.15412, 14.2015, 14.26116, 14.30871, 14.35056, 14.39536, 
14.43964, 14.48442, 14.5324, 14.57675, 14.62267, 14.66972, 14.71443, 
14.75383, 14.79012, 14.82455, 14.85587, 14.87557, 14.90737, 14.9446, 
14.97763, 15.01079, 15.04086, 15.06752, 15.09516, 15.12394, 15.15191, 
15.18061, 15.20413, 15.22896, 15.25411, 15.28108, 15.3077, 15.33578, 
15.36507, 15.39272) 

> dput(head(y_loc, n = 50)) 
c(25.18298, 25.17431, 25.17784, 25.18865, 25.20188, 25.22865, 
25.26254, 25.22778, 25.20162, 25.25191, 25.3044, 25.35787, 25.40347, 
25.46049, 25.5199, 25.57132, 25.6773, 25.69842, 25.73877, 25.78383, 
25.82168, 25.86067, 25.89984, 25.93067, 25.96943, 26.01083, 26.05861, 
26.11965, 26.18428, 26.25347, 26.3352, 26.35756, 26.4682, 26.55412, 
26.63745, 26.72157, 26.80021, 26.8691, 26.93522, 26.98879, 27.03783, 
27.07818, 27.03786, 26.9909, 26.93697, 26.87916, 26.81606, 26.74908, 
26.67815, 26.60898)

我的实际x_loc和y_loc文件〜60000个坐标，我有成千上万的文件每个都有〜60000个坐标，所以这是很多工作。我很确定函数运行缓慢的原因是我索引/递增的方式。

计数不完美。一个技术上更好的方法是遍历所有60000个点（对于这个例子只有50个点以上），并且对于每个点，计算该点与坐标数据框中每个点之间的距离（1222点）。然而这就是60000 * 1222的计算，只是针对这一组点，这太高了。

将不胜感激任何帮助！谢谢，

编辑：我正在将我的数据框/向量转换为2矩阵，并向量化整个方法，会让你知道它是否工作。

来源

2017-03-17 Canovice

如果要以比解决方案更快的速度处理矩阵，请考虑使用data.table库。请看下面的例子：

df <- data.table(x_loc, y_loc) # Your data.frame is turned into a data.table 
df$row.idx <- 1:nrow(df) # This column is used as ID for each sample point.

现在，我们可以找到每个点的正确坐标。稍后我们可以计算出某个坐标点有多少个点。我们首先保持coordinates数据帧：

y <- seq(0, 50, by = 2) 
x1 <- seq(1, 93, by = 2) 
x2 <- seq(2, 94, by = 2) 
x <- c(x1, x2) 

coordinates <- data.frame(
    x = rep(x, 13), 
    y = rep(y, each = length(x1)), 
    count = 0 
) 
coordinates$row <- 1:nrow(coordinates) # Similar to yours. However, this time we are interested in seeing which points belong to this coordinate.

现在，我们定义它检查的坐标，并返回问题点的一个单位距离内的一个函数。

f <- function(this_x, this_y, coordinates) { 
    res <- coordinates[coordinates$x > this_x-1 & 
          coordinates$x < this_x+1 & 
          coordinates$y > this_y-1 & 
          coordinates$y < this_y+1, ]$row 
    res 
}

对于每一个点，我们发现其右侧坐标：

df[, coordinate.idx := f(x_loc, y_loc), by = row.idx] 
df[, row.idx := NULL]

df包含以下变量：(x_loc, y_loc, coordinate.idx)。您可以使用它填充coordinates$count。即使是60000分，也不会超过1秒。

for(i in 1:nrow(coordinates)) { 
    coordinates$count = length(which(df$coordinate.idx == i)) 
}

来源

2017-03-17 23:03:02

在R中，将浮点坐标离散到最近的坐标

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