在python matplotlib中绘制（x，y）坐标列表

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问题：在python matplotlib中绘制（x，y）坐标列表

我有一个列表，(a, b)希望matplotlib在python中绘制为实际的xy坐标。当前，它正在绘制两个图，其中列表的索引给出x坐标，第一个图的y值是对中的as，第二个图的y值是对中的bs。

澄清一下，我的数据看起来像这样：li = [(a,b), (c,d), ... , (t, u)] 我想做一个单行代码，只是调用plt.plot()不正确。如果我不需要单线，我可以轻松地做：

xs = [x[0] for x in li]
ys = [x[1] for x in li]
plt.plot(xs, ys)

如何获取matplotlib将这些对绘制为xy坐标？

I have a list of pairs (a, b) that I would like to plot with matplotlib in python as actual x-y coordinates. Currently, it is making two plots, where the index of the list gives the x-coordinate, and the first plot’s y values are the as in the pairs and the second plot’s y values are the bs in the pairs.

To clarify, my data looks like this: li = [(a,b), (c,d), ... , (t, u)] I want to do a one-liner that just calls plt.plot() incorrect. If I didn’t require a one-liner I could trivially do:

xs = [x[0] for x in li]
ys = [x[1] for x in li]
plt.plot(xs, ys)

How can I get matplotlib to plot these pairs as x-y coordinates?

回答 0

按照这个例子：

import numpy as np
import matplotlib.pyplot as plt

N = 50
x = np.random.rand(N)
y = np.random.rand(N)

plt.scatter(x, y)
plt.show()

将生成：

要将数据从成对的数据包解压缩为列表，请使用zip：

x, y = zip(*li)

因此，单线：

plt.scatter(*zip(*li))

As per this example:

import numpy as np
import matplotlib.pyplot as plt

N = 50
x = np.random.rand(N)
y = np.random.rand(N)

plt.scatter(x, y)
plt.show()

will produce:

To unpack your data from pairs into lists use zip:

x, y = zip(*li)

So, the one-liner:

plt.scatter(*zip(*li))

回答 1

如果您有一个numpy数组，则可以执行以下操作：

import numpy as np
from matplotlib import pyplot as plt

data = np.array([
    [1, 2],
    [2, 3],
    [3, 6],
])
x, y = data.T
plt.scatter(x,y)
plt.show()

If you have a numpy array you can do this:

import numpy as np
from matplotlib import pyplot as plt

data = np.array([
    [1, 2],
    [2, 3],
    [3, 6],
])
x, y = data.T
plt.scatter(x,y)
plt.show()

回答 2

如果要绘制一条连接列表中所有点的线

plt . plot ( li [ : ] )

plt . show ( )

这将绘制一条连接列表中所有对的直线，作为从列表起点到终点的笛卡尔平面上的点。我希望这就是您想要的。

If you want to plot a single line connecting all the points in the list

plt.plot(li[:])

plt.show()

This will plot a line connecting all the pairs in the list as points on a Cartesian plane from the starting of the list to the end. I hope that this is what you wanted.

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