# plot a circle with pyplot

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surprisingly I didn"t find a straight-forward description on how to draw a circle with matplotlib.pyplot (please no pylab) taking as input center (x,y) and radius r. I tried some variants of this:

``````import matplotlib.pyplot as plt
circle=plt.Circle((0,0),2)
# here must be something like circle.plot() or not?
plt.show()
``````

... but still didn"t get it working.

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## plot a circle with pyplot circle: Questions

How to do a scatter plot with empty circles in Python?

In Python, with Matplotlib, how can a scatter plot with empty circles be plotted? The goal is to draw empty circles around some of the colored disks already plotted by `scatter()`, so as to highlight them, ideally without having to redraw the colored circles.

I tried `facecolors=None`, to no avail.

204

From the documentation for scatter:

``````Optional kwargs control the Collection properties; in particular:

edgecolors:
The string ‚Äònone‚Äô to plot faces with no outlines
facecolors:
The string ‚Äònone‚Äô to plot unfilled outlines
``````

Try the following:

``````import matplotlib.pyplot as plt
import numpy as np

x = np.random.randn(60)
y = np.random.randn(60)

plt.scatter(x, y, s=80, facecolors="none", edgecolors="r")
plt.show()
``````

Note: For other types of plots see this post on the use of `markeredgecolor` and `markerfacecolor`.

## plot a circle with pyplot circle: Questions

plot a circle with pyplot

surprisingly I didn"t find a straight-forward description on how to draw a circle with matplotlib.pyplot (please no pylab) taking as input center (x,y) and radius r. I tried some variants of this:

``````import matplotlib.pyplot as plt
circle=plt.Circle((0,0),2)
# here must be something like circle.plot() or not?
plt.show()
``````

... but still didn"t get it working.

199

You need to add it to an axes. A `Circle` is a subclass of an `Patch`, and an `axes` has an `add_patch` method. (You can also use `add_artist` but it"s not recommended.)

Here"s an example of doing this:

``````import matplotlib.pyplot as plt

circle1 = plt.Circle((0, 0), 0.2, color="r")
circle2 = plt.Circle((0.5, 0.5), 0.2, color="blue")
circle3 = plt.Circle((1, 1), 0.2, color="g", clip_on=False)

fig, ax = plt.subplots() # note we must use plt.subplots, not plt.subplot
# (or if you have an existing figure)
# fig = plt.gcf()
# ax = fig.gca()

fig.savefig("plotcircles.png")
``````

This results in the following figure:

The first circle is at the origin, but by default `clip_on` is `True`, so the circle is clipped when ever it extends beyond the `axes`. The third (green) circle shows what happens when you don"t clip the `Artist`. It extends beyond the axes (but not beyond the figure, ie the figure size is not automatically adjusted to plot all of your artists).

The units for x, y and radius correspond to data units by default. In this case, I didn"t plot anything on my axes (`fig.gca()` returns the current axes), and since the limits have never been set, they defaults to an x and y range from 0 to 1.

Here"s a continuation of the example, showing how units matter:

``````circle1 = plt.Circle((0, 0), 2, color="r")
# now make a circle with no fill, which is good for hi-lighting key results
circle2 = plt.Circle((5, 5), 0.5, color="b", fill=False)
circle3 = plt.Circle((10, 10), 2, color="g", clip_on=False)

ax = plt.gca()
ax.cla() # clear things for fresh plot

# change default range so that new circles will work
ax.set_xlim((0, 10))
ax.set_ylim((0, 10))
# some data
ax.plot(range(11), "o", color="black")
# key data point that we are encircling
ax.plot((5), (5), "o", color="y")

fig.savefig("plotcircles2.png")
``````

which results in:

You can see how I set the fill of the 2nd circle to `False`, which is useful for encircling key results (like my yellow data point).

Finding the index of an item in a list

Given a list `["foo", "bar", "baz"]` and an item in the list `"bar"`, how do I get its index (`1`) in Python?

3740

``````>>> ["foo", "bar", "baz"].index("bar")
1
``````

Reference: Data Structures > More on Lists

# Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, `index` is a rather weak component of the `list` API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about `list.index` follow. It is probably worth initially taking a look at the documentation for it:

``````list.index(x[, start[, end]])
``````

Return zero-based index in the list of the first item whose value is equal to x. Raises a `ValueError` if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

## Linear time-complexity in list length

An `index` call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give `index` a hint. For instance, in this snippet, `l.index(999_999, 999_990, 1_000_000)` is roughly five orders of magnitude faster than straight `l.index(999_999)`, because the former only has to search 10 entries, while the latter searches a million:

``````>>> import timeit
>>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000)
9.356267921015387
>>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)
0.0004404920036904514

``````

## Only returns the index of the first match to its argument

A call to `index` searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

``````>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2
``````

Most places where I once would have used `index`, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for `index`, take a look at these excellent Python features.

## Throws if element not present in list

A call to `index` results in a `ValueError` if the item"s not present.

``````>>> [1, 1].index(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 2 is not in list
``````

If the item might not be present in the list, you should either

1. Check for it first with `item in my_list` (clean, readable approach), or
2. Wrap the `index` call in a `try/except` block which catches `ValueError` (probably faster, at least when the list to search is long, and the item is usually present.)

3740

One thing that is really helpful in learning Python is to use the interactive help function:

``````>>> help(["foo", "bar", "baz"])
Help on list object:

class list(object)
...

|
|  index(...)
|      L.index(value, [start, [stop]]) -> integer -- return first index of value
|
``````

which will often lead you to the method you are looking for.

3740

The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use `enumerate()`:

``````for i, j in enumerate(["foo", "bar", "baz"]):
if j == "bar":
print(i)
``````

The `index()` function only returns the first occurrence, while `enumerate()` returns all occurrences.

As a list comprehension:

``````[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]
``````

Here"s also another small solution with `itertools.count()` (which is pretty much the same approach as enumerate):

``````from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]
``````

This is more efficient for larger lists than using `enumerate()`:

``````\$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
\$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop
``````

We hope this article has helped you to resolve the problem. Apart from plot a circle with pyplot, check other circle-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

By the way, this material is also available in other languages:

Xu Emmerson

Moscow | 2022-11-30

Maybe there are another answers? What plot a circle with pyplot exactly means?. I just hope that will not emerge anymore

Ken Zelotti

California | 2022-11-30

I was preparing for my coding interview, thanks for clarifying this - plot a circle with pyplot in Python is not the simplest one. Checked yesterday, it works!

Julia Gonzalez

Moscow | 2022-11-30

Simply put and clear. Thank you for sharing. plot a circle with pyplot and other issues with sin was always my weak point 😁. I just hope that will not emerge anymore

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