blocks|key|418984|text|当然，这样就可以了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|418985|#rainbow

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

x+=+np.linspace(0,2*np.pi)
ax=+plt.subplot(1,1,1)
ax.plot(np.sin(x))
ax.plot(np.cos(x))

rainbow+=+ax._get_lines.color_cycle
print+rainbow
for+i,+color+in+enumerate(rainbow):
++++if+i<10:
++++++++print+color,|code-block|syntax|javascript|418986|提供：|418987|<itertools.cycle+object+at+0x034CB288>
r+c+m+y+k+b+g+r+c+m|418988|下面是matplotlib使用itertools.cycle的itertools函数|offset|length|418989|编辑:谢谢你的评论，似乎不可能复制迭代器。一个想法是转储一个完整的周期，并跟踪您正在使用的值，让我回到这一点上来。|418990|Edit2:好吧，这将给你下一个颜色，并生成一个新的迭代器，它的行为就像没有调用next一样。这不会保持颜色的顺序，只保留下一个颜色值，我把它留给你。|418991|这给出了下面的输出，注意图中的陡度对应于索引，例如第一个g是最底层的图形，依此类推。|418992|#rainbow

import+matplotlib.pyplot+as+plt
import+numpy+as+np
import+collections
import+itertools

x+=+np.linspace(0,2*np.pi)
ax=+plt.subplot(1,1,1)


def+create_rainbow():
++++rainbow+=+[ax._get_lines.color_cycle.next()]
++++while+True:
++++++++nextval+=+ax._get_lines.color_cycle.next()
++++++++if+nextval+not+in+rainbow:
++++++++++++rainbow.append(nextval)
++++++++else:
++++++++++++return+rainbow

def+next_color(axis_handle=ax):
++++rainbow+=+create_rainbow()
++++double_rainbow+=+collections.deque(rainbow)
++++nextval+=+ax._get_lines.color_cycle.next()
++++double_rainbow.rotate(-1)
++++return+nextval,+itertools.cycle(double_rainbow)


for+i+in+range(1,10):
++++nextval,+ax._get_lines.color_cycle+=+next_color(ax)
++++print+"Next+color+is:+",+nextval
++++ax.plot(i*(x))


plt.savefig("SO_rotate_color.png")
plt.show()|418993|控制台|418994|Next+color+is:++g
Next+color+is:++c
Next+color+is:++y
Next+color+is:++b
Next+color+is:++r
Next+color+is:++m
Next+color+is:++k
Next+color+is:++g
Next+color+is:++c|418995|​|418996|📷|atomic|418997|418998|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/itertools.html#itertools.cycle|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/14b972f7e19fc89db071043e99d53c7d.png|imageAlt^0|0|0|0|0|F|F|0|0|0|0|0|0|0|0|0|0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|1K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1M|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1N|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1O|8|@]|9|@$M|1P|N|1Q|1|1R]]|A|$]]|$1|O|3|P|5|6|7|1S|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1T|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1U|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1V|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1W|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1X|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|12|3|13|5|14|7|1Z|8|@]|9|@$M|20|N|21|1|22]]|A|$]]|$1|15|3|11|5|6|7|23|8|@]|9|@]|A|$]]|$1|16|3|-4|5|6|7|24|8|@]|9|@]|A|$]]]|17|$18|$5|19|1A|1B|A|$1C|1D]]|1E|$5|1F|1A|1G|A|$1H|1I|1J|-4]]]]

Sure, this will do it.

<pre><code>#rainbow

import matplotlib.pyplot as plt
import numpy as np

x = np.linspace(0,2*np.pi)
ax= plt.subplot(1,1,1)
ax.plot(np.sin(x))
ax.plot(np.cos(x))

rainbow = ax._get_lines.color_cycle
print rainbow
for i, color in enumerate(rainbow):
 if i&lt;10:
 print color,
</code></pre>

Gives:

<pre><code>&lt;itertools.cycle object at 0x034CB288&gt;
r c m y k b g r c m
</code></pre>

Here is the itertools function that matplotlib uses <a href="http://docs.python.org/2/library/itertools.html#itertools.cycle">itertools.cycle</a>

Edit: Thanks for the comment, it seems that it is not possible to copy an iterator. An idea would be to dump a full cycle and keep track of which value you are using, let me get back on that.

Edit2: Allright, this will give you the next color and make a new iterator that behaves as if next was not called. This does not preserve the order of coloring, just the next color value, I leave that to you.

This gives the following output, notice that steepness in the plot corresponds to index, eg first g is the bottomest graph and so on.

<pre><code>#rainbow

import matplotlib.pyplot as plt
import numpy as np
import collections
import itertools

x = np.linspace(0,2*np.pi)
ax= plt.subplot(1,1,1)


def create_rainbow():
 rainbow = [ax._get_lines.color_cycle.next()]
 while True:
 nextval = ax._get_lines.color_cycle.next()
 if nextval not in rainbow:
 rainbow.append(nextval)
 else:
 return rainbow

def next_color(axis_handle=ax):
 rainbow = create_rainbow()
 double_rainbow = collections.deque(rainbow)
 nextval = ax._get_lines.color_cycle.next()
 double_rainbow.rotate(-1)
 return nextval, itertools.cycle(double_rainbow)


for i in range(1,10):
 nextval, ax._get_lines.color_cycle = next_color(ax)
 print "Next color is: ", nextval
 ax.plot(i*(x))


plt.savefig("SO_rotate_color.png")
plt.show()
</code></pre>

Console

<pre><code>Next color is: g
Next color is: c
Next color is: y
Next color is: b
Next color is: r
Next color is: m
Next color is: k
Next color is: g
Next color is: c
</code></pre>

<img src="https://i.stack.imgur.com/cWsGY.png" alt="Rotate color">

blocks|key|419039|text|注意:在matplotlib+(>=+1.5)的最新版本中，_get_lines发生了变化。现在需要使用Python2或3中的next(ax._get_lines.prop_cycler)['color']+(或Python+2中的ax._get_lines.prop_cycler.next()['color']+)从颜色循环中获取下一种颜色。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|419040|只要有可能，请使用@joe-kington答案下半部分所示的更直接的方法。由于_get_lines不是面向应用程序接口的，因此将来可能会以不向后兼容的方式再次更改。|419041|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/5237611/itertools-cycle-next^0|U|A|1R|14|38|15|2Y|8|0|0|13|A|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@$9|W|A|X|1|Y]]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Note: In the latest versions of matplotlib (>= 1.5) <code>_get_lines</code> has changed. You now need to use <code>next(ax._get_lines.prop_cycler)['color']</code> in Python 2 or 3 (or <code>ax._get_lines.prop_cycler.next()['color']</code> in <a href="https://stackoverflow.com/questions/5237611/itertools-cycle-next">Python 2</a>) to get the next color from the color cycle.

Wherever possible use the more direct approach shown in the lower part of @joe-kington's answer. As <code>_get_lines</code> is not API-facing it might change again in a not backward compatible manner in the future.

blocks|key|200760|text|这里有一种在1.5中工作的方法，希望它能适应未来，因为它不依赖于带有下划线的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|200761|colors+=+plt.rcParams["axes.prop_cycle"].by_key()["color"]|code-block|syntax|javascript|200762|这将为您提供一个为当前样式定义的颜色列表。|200763|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Here's a way that works in 1.5 which will hopefully be future-proof as it doesn't rely on methods prepended with underscores:

<pre><code>colors = plt.rcParams["axes.prop_cycle"].by_key()["color"]
</code></pre>

This will give you a list of the colors defined in order for the present style.

blocks|key|419138|text|由于matplotlib使用itertools.cycle，我们实际上可以查看整个颜色循环，然后将迭代器恢复到以前的状态：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|419139|def+list_from_cycle(cycle):
++++first+=+next(cycle)
++++result+=+[first]
++++for+current+in+cycle:
++++++++if+current+==+first:
++++++++++++break
++++++++result.append(current)

++++#+Reset+iterator+state:
++++for+current+in+cycle:
++++++++if+current+==+result[-1]:
++++++++++++break
++++return+result|code-block|syntax|javascript|419140|这应该会在不改变迭代器状态的情况下返回列表。|419141|在matplotlib+>=+1.5中使用它：|419142|>>>+list_from_cycle(ax._get_lines.prop_cycler)
[{'color':+'r'},+{'color':+'g'},+{'color':+'b'}]|419143|或者matplotlib+<+1.5：|419144|>>>+list_from_cycle(ax._get_lines.color_cycle)
['r',+'g',+'b']|419145|entityMap^0|E|F|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|13|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|V|$]]

Since matplotlib uses <code>itertools.cycle</code> we can actually look through the entire color cycle and then restore the iterator to its previous state:

<pre><code>def list_from_cycle(cycle):
 first = next(cycle)
 result = [first]
 for current in cycle:
 if current == first:
 break
 result.append(current)

 # Reset iterator state:
 for current in cycle:
 if current == result[-1]:
 break
 return result
</code></pre>

This should return the list without changing the state of the iterator.

Use it with matplotlib >= 1.5:

<pre><code>&gt;&gt;&gt; list_from_cycle(ax._get_lines.prop_cycler)
[{'color': 'r'}, {'color': 'g'}, {'color': 'b'}]
</code></pre>

or with matplotlib &lt; 1.5:

<pre><code>&gt;&gt;&gt; list_from_cycle(ax._get_lines.color_cycle)
['r', 'g', 'b']
</code></pre>

blocks|key|200899|text|在matplotlib版本2.2.3中，在_get_lines属性上有一个get_next_color()方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|200900|import+from+matplotlib+import+pyplot+as+plt
fig,+ax+=+plt.subplots()
next_color+=+ax._get_lines.get_next_color()|code-block|syntax|javascript|200901|get_next_color()返回一个html颜色字符串，并推进颜色循环迭代器。|200902|entityMap^0|L|A|11|G|0|0|0|G|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

In matplotlib version 2.2.3 there is a <code>get_next_color()</code> method on the <code>_get_lines</code> property:

<pre><code>import from matplotlib import pyplot as plt
fig, ax = plt.subplots()
next_color = ax._get_lines.get_next_color()
</code></pre>

<code>get_next_color()</code> returns an html color string, and advances the color cycle iterator.

blocks|key|200937|text|我能找到的最简单的方法就是ax1.lines[-1].get_color()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|200938|entityMap^0|D|P|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

The simplest way possible I could find without doing the whole loop through the cycler is <code>ax1.lines[-1].get_color()</code>.

blocks|key|200991|text|最小工作示例|type|unstyled|depth|inlineStyleRanges|entityRanges|data|200992|我已经和它斗争过好几次了。这是Andis+answer的一个最小工作示例。|offset|length|200993|📷|atomic|200994|​|200995|代码|200996|import+numpy+as+np
import+matplotlib.pyplot+as+plt


xs+=+np.arange(10)


fig,+ax+=+plt.subplots()

for+ii+in+range(3):
++++color+=+next(ax._get_lines.prop_cycler)['color']
++++lbl+=+'line+{:d},+color+{:}'.format(ii,+color)
++++ys+=+np.random.rand(len(xs))
++++ax.plot(xs,+ys,+color=color,+label=lbl)
ax.legend()|code-block|syntax|javascript|200997|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/33608574/7128154|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/f9a656b7b6aaea25360864a17f5c1a8d.png|imageAlt^0|0|F|C|0|0|0|1|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|16|8|@]|9|@$D|17|E|18|1|19]]|A|$]]|$1|F|3|G|5|H|7|1A|8|@]|9|@$D|1B|E|1C|1|1D]]|A|$]]|$1|I|3|J|5|6|7|1E|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|1F|8|@]|9|@]|A|$]]|$1|M|3|N|5|O|7|1G|8|@]|9|@]|A|$P|Q]]|$1|R|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y]]|Z|$5|10|V|11|A|$12|13|14|-4]]]]

<h1>minimal working example</h1>
I struggelt with this quite a few times already.
This is a minimal working example for <a href="https://stackoverflow.com/a/33608574/7128154">Andis answer</a>.
<a href="https://i.stack.imgur.com/MG4L5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MG4L5.png" alt="enter image description here" /></a>
<h2>code</h2>
<pre><code>import numpy as np
import matplotlib.pyplot as plt


xs = np.arange(10)


fig, ax = plt.subplots()

for ii in range(3):
 color = next(ax._get_lines.prop_cycler)['color']
 lbl = 'line {:d}, color {:}'.format(ii, color)
 ys = np.random.rand(len(xs))
 ax.plot(xs, ys, color=color, label=lbl)
ax.legend()
</code></pre>

Is it possible to query the current state of the matplotlib color cycle? In other words is there a function <code>get_cycle_state</code> that will behave in the following way?

<pre><code>&gt;&gt;&gt; plot(x1, y1)
&gt;&gt;&gt; plot(x2, y2)
&gt;&gt;&gt; state = get_cycle_state()
&gt;&gt;&gt; print state
2
</code></pre>

Where I expect the state to be the index of the next color that will be used in a plot. Alternatively, if it returned the next color ("r" for the default cycle in the example above), that would be fine too.

Get matplotlib color cycle state

可以查询matplotlib颜色周期的当前状态吗？换句话说，有没有一个函数get_cycle_state的行为方式如下？>>> plot(x1, y1)>>> plot(x2, y2)>>> state = get_cycle_state()>>> print state2其中我期望状态是将在绘图中使用的下一种颜色的索...

问获取matplotlib颜色周期状态
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