[matplotlib] Adding an arbitrary line to a matplotlib plot in ipython notebook

I'm rather new to both python/matplotlib and using it through the ipython notebook. I'm trying to add some annotation lines to an existing graph and I can't figure out how to render the lines on a graph. So, for example, if I plot the following:

import numpy as np
np.random.seed(5)
x = arange(1, 101)
y = 20 + 3 * x + np.random.normal(0, 60, 100)
p =  plot(x, y, "o")

I get the following graph:

So how would I add a vertical line from (70,100) up to (70,250)? What about a diagonal line from (70,100) to (90,200)?

I've tried a few things with Line2D() resulting in nothing but confusion on my part. In R I would simply use the segments() function which would add line segments. Is there an equivalent in matplotlib?

Using vlines:

import numpy as np
np.random.seed(5)
x = arange(1, 101)
y = 20 + 3 * x + np.random.normal(0, 60, 100)
p =  plot(x, y, "o")
vlines(70,100,250)

The basic call signatures are:

vlines(x, ymin, ymax)
hlines(y, xmin, xmax)

It's not too late for the newcomers.

plt.axvline(x, color='r')

It takes the range of y as well, using ymin and ymax.

Rather than abusing plot or annotate, which will be inefficient for many lines, you can use matplotlib.collections.LineCollection:

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection

np.random.seed(5)
x = np.arange(1, 101)
y = 20 + 3 * x + np.random.normal(0, 60, 100)
plt.plot(x, y, "o")

# Takes list of lines, where each line is a sequence of coordinates
l1 = [(70, 100), (70, 250)]
l2 = [(70, 90), (90, 200)]
lc = LineCollection([l1, l2], color=["k","blue"], lw=2)

plt.gca().add_collection(lc)

plt.show()

It takes a list of lines [l1, l2, ...], where each line is a sequence of N coordinates (N can be more than two).

The standard formatting keywords are available, accepting either a single value, in which case the value applies to every line, or a sequence of M values, in which case the value for the ith line is values[i % M].

Matplolib now allows for 'annotation lines' as the OP was seeking. The annotate() function allows several forms of connecting paths and a headless and tailess arrow, i.e., a simple line, is one of them.

ax.annotate("",
            xy=(0.2, 0.2), xycoords='data',
            xytext=(0.8, 0.8), textcoords='data',
            arrowprops=dict(arrowstyle="-",
                      connectionstyle="arc3, rad=0"),
            )

In the documentation it says you can draw only an arrow with an empty string as the first argument.

From the OP's example:

%matplotlib notebook
import numpy as np
import matplotlib.pyplot as plt

np.random.seed(5)
x = np.arange(1, 101)
y = 20 + 3 * x + np.random.normal(0, 60, 100)
plt.plot(x, y, "o")


# draw vertical line from (70,100) to (70, 250)
plt.annotate("",
              xy=(70, 100), xycoords='data',
              xytext=(70, 250), textcoords='data',
              arrowprops=dict(arrowstyle="-",
                              connectionstyle="arc3,rad=0."), 
              )

# draw diagonal line from (70, 90) to (90, 200)
plt.annotate("",
              xy=(70, 90), xycoords='data',
              xytext=(90, 200), textcoords='data',
              arrowprops=dict(arrowstyle="-",
                              connectionstyle="arc3,rad=0."), 
              )

plt.show()

Just as in the approach in gcalmettes's answer, you can choose the color, line width, line style, etc..

Here is an alteration to a portion of the code that would make one of the two example lines red, wider, and not 100% opaque.

# draw vertical line from (70,100) to (70, 250)
plt.annotate("",
              xy=(70, 100), xycoords='data',
              xytext=(70, 250), textcoords='data',
              arrowprops=dict(arrowstyle="-",
                              edgecolor = "red",
                              linewidth=5,
                              alpha=0.65,
                              connectionstyle="arc3,rad=0."), 
              )

You can also add curve to the connecting line by adjusting the connectionstyle.