blocks|key|1604112|text|我将使用numpy功能将您的线切成线段，然后分别绘制所有实线和虚线。在下面的示例中，我向您的数据中添加了两个额外的-1，以查看这是通用的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1604113|import+numpy+as+np
import+matplotlib.pyplot+as+plt

Y+=+np.array([5,6,-1,-1,+8,3,5,7,3,6,-1,3,8,5])
X+=+np.arange(len(Y))

idxs+=++np.where(Y==-1)[0]

sub_y+=+np.split(Y,idxs)
sub_x+=+np.split(X,idxs)

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

##replacing+-1+values+and+plotting+dotted+lines
for+i+in+range(1,len(sub_y)):
++++val+=+sub_y[i-1][-1]
++++sub_y[i][0]+=+val
++++ax.plot([sub_x[i-1][-1],+sub_x[i][0]],+[val,+val],+'r--')

##plotting+rest
for+x,y+in+zip(sub_x,+sub_y):
++++ax.plot(x,+y,+'r-o')

plt.show()|code-block|syntax|javascript|1604114|结果如下所示：|1604115|​|1604116|?|atomic|1604117|1604118|但是，请注意，如果第一个值为-1，则此操作将失败，因为您的问题没有明确定义(没有以前的值可供复制)。希望这能有所帮助。|1604119|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/4c3f55376f381864c2239eb8f4ac1301.png|imageAlt^0|4|5|1L|2|0|0|0|0|0|1|0|0|0|E|2|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|19|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|1A|8|@]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1B|8|@]|D|@$9|1C|A|1D|1|1E]]|E|$]]|$1|R|3|N|5|6|7|1F|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1G|8|@$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11|12|-4]]]]

I would use <code>numpy</code> functionality to cut your line into segments and then plot all solid and dashed lines separately. In the example below I added two additional <code>-1</code>s to your data to see that this works universally. 

<pre><code>import numpy as np
import matplotlib.pyplot as plt

Y = np.array([5,6,-1,-1, 8,3,5,7,3,6,-1,3,8,5])
X = np.arange(len(Y))

idxs = np.where(Y==-1)[0]

sub_y = np.split(Y,idxs)
sub_x = np.split(X,idxs)

fig, ax = plt.subplots()

##replacing -1 values and plotting dotted lines
for i in range(1,len(sub_y)):
 val = sub_y[i-1][-1]
 sub_y[i][0] = val
 ax.plot([sub_x[i-1][-1], sub_x[i][0]], [val, val], 'r--')

##plotting rest
for x,y in zip(sub_x, sub_y):
 ax.plot(x, y, 'r-o')

plt.show()
</code></pre>

The result looks like this:

<a href="https://i.stack.imgur.com/gD3LZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gD3LZ.png" alt="enter image description here"></a>

Note, however, that this will fail if the first value is <code>-1</code>, as then your problem is not well defined (no previous value to copy from). Hope this helps.

blocks|key|2056126|text|您可以在不使用循环的情况下实现类似的结果：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2056127|import+pandas+as+pd
import+matplotlib.pyplot+as+plt

#+Create+a+data+frame+from+the+list
a+=+pd.DataFrame([5,6,-1,-1,+8,3,5,7,3,6,-1,3,8,5])

#+Prepare+a+boolean+mask
mask+=+a+>+0

#+New+data+frame+with+missing+values+filled+with+the+last+element+of+++
#+the+previous+segment.+Choose+'bfill'+to+use+the+first+element+of+
#+the+next+segment.
a_masked+=+a[mask].fillna(method+=+'ffill')

#+Prepare+the+plot
fig,+ax+=+plt.subplots()
line,+=+ax.plot(a_masked,+ls+=+'--',+lw+=+1)
ax.plot(a[mask],+color=line.get_color(),+lw=1.5,+marker+=+'o')
plt.show()|code-block|syntax|javascript|2056128|​|2056129|?|atomic|offset|length|2056130|2056131|您还可以通过为线条选择不同的颜色来突出显示负片区域：|2056132|2056133|2056134|我的答案是基于2017年7月的一篇很棒的帖子。后者还处理第一个元素为NaN或在本例中为负数时的情况：Dotted+lines+instead+of+a+missing+value+in+matplotlib|2056135|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/ccbe69352268404326595cb911daa353.png|imageAlt|1|https://ask.qcloudimg.com/http-save/yehe-900000/072e71df68211a1e89e66cfab8dafa78.png|2|LINK|MUTABLE|url|https://stackoverflow.com/questions/45176584/dotted-lines-instead-of-a-missing-value-in-matplotlib^0|0|0|0|0|1|0|0|0|0|0|1|1|0|0|1E|1H|2|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1C|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|1D|8|@]|9|@$L|1E|M|1F|1|1G]]|A|$]]|$1|N|3|H|5|6|7|1H|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1I|8|@]|9|@]|A|$]]|$1|Q|3|J|5|K|7|1J|8|@]|9|@$L|1K|M|1L|1|1M]]|A|$]]|$1|R|3|H|5|6|7|1N|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1O|8|@]|9|@$L|1P|M|1Q|1|1R]]|A|$]]|$1|U|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11|12|-4]]|13|$5|X|Y|Z|A|$10|14|12|-4]]|15|$5|16|Y|17|A|$18|19]]]]

You can achieve something similar without the loops:

<pre><code>import pandas as pd
import matplotlib.pyplot as plt

# Create a data frame from the list
a = pd.DataFrame([5,6,-1,-1, 8,3,5,7,3,6,-1,3,8,5])

# Prepare a boolean mask
mask = a &gt; 0

# New data frame with missing values filled with the last element of 
# the previous segment. Choose 'bfill' to use the first element of 
# the next segment.
a_masked = a[mask].fillna(method = 'ffill')

# Prepare the plot
fig, ax = plt.subplots()
line, = ax.plot(a_masked, ls = '--', lw = 1)
ax.plot(a[mask], color=line.get_color(), lw=1.5, marker = 'o')
plt.show()
</code></pre>

<a href="https://i.stack.imgur.com/KWmhU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KWmhU.png" alt="enter image description here"></a>

You can also highlight the negative regions by choosing a different colour for the lines:
<a href="https://i.stack.imgur.com/aI7oe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aI7oe.png" alt="enter image description here"></a>

My answer is based on a great post from July, 2017. The latter also tackles the case when the first element is NaN or in your case a negative number:
<a href="https://stackoverflow.com/questions/45176584/dotted-lines-instead-of-a-missing-value-in-matplotlib">Dotted lines instead of a missing value in matplotlib</a>

blocks|key|1604121|text|不太优雅，但这里有一些不使用循环的东西，这是我想出来的(基于上面的答案)，可以工作。@KRKirov和@Thomas+Kühn，感谢你们的回答，我真的很感谢他们|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1604122|import+pandas+as+pd
import+matplotlib.pyplot+as+plt

#+Create+a+data+frame+from+the+list
a+=+pd.DataFrame([5,6,-1,-1,+8,3,5,7,3,6,-1,3,8,5])

b=a.copy()
b[2]=b[0].shift(1,axis=0)
b[4]=(b[0]!=-1)+&+(b[2]==-1)
b[5]=b[4].shift(-1,axis=0)
b[0]+=+(b[5]+%7C+b[4])
c=b[0]
d=pd.DataFrame(c)

#+Prepare+a+boolean+mask
mask+=+a+>+0

#+New+data+frame+with+missing+values+filled+with+the+last+element+of+++
#+the+previous+segment.+Choose+'bfill'+to+use+the+first+element+of+
#+the+next+segment.
a_masked+=+a[mask].fillna(method+=+'ffill')

#+Prepare+the+plot
fig,+ax+=+plt.subplots()
line,+=+ax.plot(a_masked,+'b:o',+lw+=+1)
ax.plot(a[mask],+color=line.get_color(),+lw=1.5,+marker+=+'o')
ax.plot(a_masked[d],+color=line.get_color(),+lw=1.5,+marker+=+'o')
plt.show()|code-block|syntax|javascript|1604123|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Not too elegant, but here's something that doesn't use loops which I came up with (based on the above answers) which works. @KRKirov and @Thomas Kühn , thank you for your answers, I really appreciate them

<pre><code>import pandas as pd
import matplotlib.pyplot as plt

# Create a data frame from the list
a = pd.DataFrame([5,6,-1,-1, 8,3,5,7,3,6,-1,3,8,5])

b=a.copy()
b[2]=b[0].shift(1,axis=0)
b[4]=(b[0]!=-1) &amp; (b[2]==-1)
b[5]=b[4].shift(-1,axis=0)
b[0] = (b[5] | b[4])
c=b[0]
d=pd.DataFrame(c)

# Prepare a boolean mask
mask = a &gt; 0

# New data frame with missing values filled with the last element of 
# the previous segment. Choose 'bfill' to use the first element of 
# the next segment.
a_masked = a[mask].fillna(method = 'ffill')

# Prepare the plot
fig, ax = plt.subplots()
line, = ax.plot(a_masked, 'b:o', lw = 1)
ax.plot(a[mask], color=line.get_color(), lw=1.5, marker = 'o')
ax.plot(a_masked[d], color=line.get_color(), lw=1.5, marker = 'o')
plt.show()
</code></pre>

I would like to replace part of my plot where the function dips down to '-1' with a dashed line carrying on from the previous point (see plots below).

Here's some code I've written, along with its output:

<pre><code>import numpy as np

import matplotlib.pyplot as plt
y = [5,6,8,3,5,7,3,6,-1,3,8,5]

plt.plot(np.linspace(1,12,12),y,'r-o')
plt.show()

for i in range(1,len(y)):
 if y[i]!=-1:
 plt.plot(np.linspace(i-1,i,2),y[i-1:i+1],'r-o')
 else:
 y[i]=y[i-1]
 plt.plot(np.linspace(i-1,i,2),y[i-1:i+1],'r--o')
plt.ylim(-1,9)
plt.show()
</code></pre>

Here's the original plot

<a href="https://i.stack.imgur.com/Rlm5j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rlm5j.png" alt="original plot"></a>

Modified plot:

<a href="https://i.stack.imgur.com/pHCAC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pHCAC.png" alt="modified plot"></a>

The code I've written works (it produces the desired output), but it's inefficient and takes a long time when I actually run it on my (much larger) dataset. Is there a smarter way to go about doing this?

Replacing part of a plot with a dotted line

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想将函数下降到'-1‘的部分曲线图替换为从上一点开始的虚线(参见下面的曲线图)。下面是我写的一些代码，以及它的输出：import numpy as npimport matplotlib.pyplot as plty = [5,6,8,3,5,7,3,6,-1,3,8,5]plt.plot(np.linspace(1...

问用虚线替换图的一部分
EN

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问用虚线替换图的一部分EN