如何用现有的xyz数据制作矩阵

问题描述

我想使用matplotlib.pyplot.pcolormesh绘制深度图.

I want to use matplotlib.pyplot.pcolormesh to plot a depth plot.

我有一个xyz文件三列，即x(lat)，y(lon)，z(dep).

What I have is a xyz fileThree columns i.e. x(lat), y(lon), z(dep).

所有列的长度相等

pcolormesh需要矩阵作为输入.因此，使用numpy.meshgrid，我可以将x和y转换为矩阵:

pcolormesh require matrices as input.So using numpy.meshgrid I can transform the x and y into matrices:

xx,yy = numpy.meshgrid(x_data,y_data)

这很好用...但是，我不知道如何创建我的深度(z)数据的矩阵...如何为z_data创建与x_data和y_data矩阵相对应的矩阵?

This works great...However, I don't know how to create Matrix of my depth (z) data...How do I create a matrix for my z_data that corresponds to my x_data and y_data matrices?

推荐答案

根据是否生成z，您至少有两个不同的选择.

Depending on whether you're generating z or not, you have at least two different options.

如果要生成z(例如，您知道它的公式)，则非常简单(请参见下面的method_1()).

If you're generating z (e.g. you know the formula for it) it's very easy (see method_1() below).

如果仅列出(x，y，z)个元组，则难度会更大(请参见下面的method_2()，也许还有method_3()).

If you just have just a list of (x,y,z) tuples, it's harder (see method_2() below, and maybe method_3()).

常量

# min_? is minimum bound, max_? is maximum bound,
#   dim_? is the granularity in that direction
min_x, max_x, dim_x = (-10, 10, 100)
min_y, max_y, dim_y = (-10, 10, 100)

方法1:生成z

Method 1: Generating z

# Method 1:
#   This works if you are generating z, given (x,y)
def method_1():
    x = np.linspace(min_x, max_x, dim_x)
    y = np.linspace(min_y, max_y, dim_y)

    X,Y = np.meshgrid(x,y)

    def z_function(x,y):
        return math.sqrt(x**2 + y**2)

    z = np.array([z_function(x,y) for (x,y) in zip(np.ravel(X), np.ravel(Y))])
    Z = z.reshape(X.shape)

    plt.pcolormesh(X,Y,Z)
    plt.show()

哪个会生成以下图形:

这相对容易，因为您可以在任意位置生成z.

This is relatively easy, since you can generate z at whatever points you want.

如果您没有该能力，则会获得固定的(x,y,z).您可以执行以下操作.首先，我定义一个生成假数据的函数:

If you don't have that ability, and are given a fixed (x,y,z). You could do the following. First, I define a function that generates fake data:

def gen_fake_data():
    # First we generate the (x,y,z) tuples to imitate "real" data
    # Half of this will be in the + direction, half will be in the - dir.
    xy_max_error = 0.2

    # Generate the "real" x,y vectors
    x = np.linspace(min_x, max_x, dim_x)
    y = np.linspace(min_y, max_y, dim_y)

    # Apply an error to x,y
    x_err = (np.random.rand(*x.shape) - 0.5) * xy_max_error
    y_err = (np.random.rand(*y.shape) - 0.5) * xy_max_error
    x *= (1 + x_err)
    y *= (1 + y_err)

    # Generate fake z
    rows = []
    for ix in x:
        for iy in y:
            z = math.sqrt(ix**2 + iy**2)
            rows.append([ix,iy,z])

    mat = np.array(rows)
    return mat

在这里，返回的矩阵如下:

Here, the returned matrix looks like:

mat = [[x_0, y_0, z_0],
       [x_1, y_1, z_1],
       [x_2, y_2, z_2],
       ...
       [x_n, y_n, z_n]]

方法2:在常规网格上插入给定的z点

Method 2: Interpolating given z points over a regular grid

# Method 2:
#   This works if you have (x,y,z) tuples that you're *not* generating, and (x,y) points
#   may not fall evenly on a grid.
def method_2():
    mat = gen_fake_data()

    x = np.linspace(min_x, max_x, dim_x)
    y = np.linspace(min_y, max_y, dim_y)

    X,Y = np.meshgrid(x, y)

    # Interpolate (x,y,z) points [mat] over a normal (x,y) grid [X,Y]
    #   Depending on your "error", you may be able to use other methods
    Z = interpolate.griddata((mat[:,0], mat[:,1]), mat[:,2], (X,Y), method='nearest')

    plt.pcolormesh(X,Y,Z)
    plt.show()

此方法产生以下图形:

错误= 0.2

错误= 0.8

方法3:不插值(对采样数据有约束)

还有第三个选项，具体取决于您的(x,y,z)的设置方式.此选项需要两件事:

There's a third option, depending on how your (x,y,z) is set up. This option requires two things:

1. 不同的x个采样位置的数量等于不同的y个采样位置的数量.
2. 对于每个可能的唯一(x，y)对，您的数据中都有一个对应的(x，y，z).

由此，(x,y,z)对的数量必须等于唯一x点数量的平方(其中唯一x位数量等于唯一y位数量).

From this, it follows that the number of (x,y,z) pairs must be equal to the square of the number of unique x points (where the number of unique x positions equals the number of unique y positions).

通常，对于采样数据，此不会为真.但如果是这样，则可以避免插值:

In general, with sampled data, this will not be true. But if it is, you can avoid having to interpolate:

def method_3():
    mat = gen_fake_data()

    x = np.unique(mat[:,0])
    y = np.unique(mat[:,1])

    X,Y = np.meshgrid(x, y)

    # I'm fairly sure there's a more efficient way of doing this...
    def get_z(mat, x, y):
        ind = (mat[:,(0,1)] == (x,y)).all(axis=1)
        row = mat[ind,:]
        return row[0,2]

    z = np.array([get_z(mat,x,y) for (x,y) in zip(np.ravel(X), np.ravel(Y))])
    Z = z.reshape(X.shape)

    plt.pcolormesh(X,Y,Z)
    plt.xlim(min(x), max(x))
    plt.ylim(min(y), max(y))
    plt.show()

错误= 0.2

错误= 0.8

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