在python matplotlib中绘制(x, y)坐标列表。

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.

我有一个对(a, b)的列表，我想用python中的matplotlib作为实际的x-y坐标。目前，它正在做两个图，其中列表的索引给出了x坐标，第一个图的y值是成对的，第二个图的y值是对的b。

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:

为了澄清，我的数据是这样的:li = [(a,b)， (c,d)，…我想做一个只叫pl .plot()不正确的一行程序。如果我不需要一艘客轮，我可以做的是:

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?
• 我如何得到matplotlib来把这些对作为x-y坐标?

Thanks for all the help!

谢谢你的帮助!

3 个解决方案

#1

As per this example:

根据这个例子:

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()

will produce:

会产生:

To unpack your data from pairs into lists use zip:

要将你的数据从成对打开到列表，请使用zip:

x, y = zip(*li)

So, the one-liner:

所以,一行程序:

plt.scatter(*zip(*li))

#2

If you have a numpy array you can do this:

如果你有一个numpy数组，你可以这样做:

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)

#3

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

如果你想画一条连接列表中所有点的单线。

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.

这将绘制一条线，将列表中的所有对点连接到从列表开始到结束的笛卡尔平面上的点。我希望这就是你想要的。