Matplotlib create text with colors corresponding to series
I am trying to create a plot with different colors for different series. The question arose when I tried to add the data in the figure as a text box.
The code I used is as follows:
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
import scipy.stats as stats
df = pd.DataFrame('x': [21000, 16900, 18200, 32000, 35000, 7500], 'y':[3000, 2100, 1500, 3000, 2500, 2000], 'z':['a', 'b', 'c', 'd', 'e', 'f'])
fig, ax = plt.subplots(figsize=(8,6))
text_list =
color_list =
for i, row in df.iterrows():
mu, sigma, group = row['x'], row['y'], row['z']
x = np.linspace(mu - 4*sigma, mu + 4*sigma, 100)
sns.lineplot(x, stats.norm.pdf(x, mu, sigma), ax=ax)
color = ax.get_lines()[-1].get_c()
ax = plt.gca()
ax.text(mu*1.05, max(stats.norm.pdf(x, mu, sigma)), group, fontsize=16, color=color) #only retrieve RGB so blank text is not too light
text = r'0: 1 $pm$ 2'.format(group, mu, sigma)
text_list.append(text)
color_list.append(color)
plt.gcf().text(0.68, 0.6, 'n'.join(text_list), bbox=dict(facecolor='white', edgecolor='black', pad=10.0, alpha=1), fontsize=14)
fig.show()
Which produces the following graph:
The texts within the bbox are all black. Ideally, each line in the text box should have a color identical to the corresponding series in the plot.
I was able to save two lists of texts and colors in the text_box_content and color_list. I also tried to add plt.gcf().text()
within the for loop with dynamically-updated text locations, but the bounding boxes are created for each row instead of an overall-bounding box for all text.
It would be nicer if there is something conceptually similar to
plt.gcf().text(zip(text_list, color_list))
so each line can have its own color?
python matplotlib visualization seaborn
add a comment |
I am trying to create a plot with different colors for different series. The question arose when I tried to add the data in the figure as a text box.
The code I used is as follows:
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
import scipy.stats as stats
df = pd.DataFrame('x': [21000, 16900, 18200, 32000, 35000, 7500], 'y':[3000, 2100, 1500, 3000, 2500, 2000], 'z':['a', 'b', 'c', 'd', 'e', 'f'])
fig, ax = plt.subplots(figsize=(8,6))
text_list =
color_list =
for i, row in df.iterrows():
mu, sigma, group = row['x'], row['y'], row['z']
x = np.linspace(mu - 4*sigma, mu + 4*sigma, 100)
sns.lineplot(x, stats.norm.pdf(x, mu, sigma), ax=ax)
color = ax.get_lines()[-1].get_c()
ax = plt.gca()
ax.text(mu*1.05, max(stats.norm.pdf(x, mu, sigma)), group, fontsize=16, color=color) #only retrieve RGB so blank text is not too light
text = r'0: 1 $pm$ 2'.format(group, mu, sigma)
text_list.append(text)
color_list.append(color)
plt.gcf().text(0.68, 0.6, 'n'.join(text_list), bbox=dict(facecolor='white', edgecolor='black', pad=10.0, alpha=1), fontsize=14)
fig.show()
Which produces the following graph:
The texts within the bbox are all black. Ideally, each line in the text box should have a color identical to the corresponding series in the plot.
I was able to save two lists of texts and colors in the text_box_content and color_list. I also tried to add plt.gcf().text()
within the for loop with dynamically-updated text locations, but the bounding boxes are created for each row instead of an overall-bounding box for all text.
It would be nicer if there is something conceptually similar to
plt.gcf().text(zip(text_list, color_list))
so each line can have its own color?
python matplotlib visualization seaborn
add a comment |
I am trying to create a plot with different colors for different series. The question arose when I tried to add the data in the figure as a text box.
The code I used is as follows:
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
import scipy.stats as stats
df = pd.DataFrame('x': [21000, 16900, 18200, 32000, 35000, 7500], 'y':[3000, 2100, 1500, 3000, 2500, 2000], 'z':['a', 'b', 'c', 'd', 'e', 'f'])
fig, ax = plt.subplots(figsize=(8,6))
text_list =
color_list =
for i, row in df.iterrows():
mu, sigma, group = row['x'], row['y'], row['z']
x = np.linspace(mu - 4*sigma, mu + 4*sigma, 100)
sns.lineplot(x, stats.norm.pdf(x, mu, sigma), ax=ax)
color = ax.get_lines()[-1].get_c()
ax = plt.gca()
ax.text(mu*1.05, max(stats.norm.pdf(x, mu, sigma)), group, fontsize=16, color=color) #only retrieve RGB so blank text is not too light
text = r'0: 1 $pm$ 2'.format(group, mu, sigma)
text_list.append(text)
color_list.append(color)
plt.gcf().text(0.68, 0.6, 'n'.join(text_list), bbox=dict(facecolor='white', edgecolor='black', pad=10.0, alpha=1), fontsize=14)
fig.show()
Which produces the following graph:
The texts within the bbox are all black. Ideally, each line in the text box should have a color identical to the corresponding series in the plot.
I was able to save two lists of texts and colors in the text_box_content and color_list. I also tried to add plt.gcf().text()
within the for loop with dynamically-updated text locations, but the bounding boxes are created for each row instead of an overall-bounding box for all text.
It would be nicer if there is something conceptually similar to
plt.gcf().text(zip(text_list, color_list))
so each line can have its own color?
python matplotlib visualization seaborn
I am trying to create a plot with different colors for different series. The question arose when I tried to add the data in the figure as a text box.
The code I used is as follows:
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
import scipy.stats as stats
df = pd.DataFrame('x': [21000, 16900, 18200, 32000, 35000, 7500], 'y':[3000, 2100, 1500, 3000, 2500, 2000], 'z':['a', 'b', 'c', 'd', 'e', 'f'])
fig, ax = plt.subplots(figsize=(8,6))
text_list =
color_list =
for i, row in df.iterrows():
mu, sigma, group = row['x'], row['y'], row['z']
x = np.linspace(mu - 4*sigma, mu + 4*sigma, 100)
sns.lineplot(x, stats.norm.pdf(x, mu, sigma), ax=ax)
color = ax.get_lines()[-1].get_c()
ax = plt.gca()
ax.text(mu*1.05, max(stats.norm.pdf(x, mu, sigma)), group, fontsize=16, color=color) #only retrieve RGB so blank text is not too light
text = r'0: 1 $pm$ 2'.format(group, mu, sigma)
text_list.append(text)
color_list.append(color)
plt.gcf().text(0.68, 0.6, 'n'.join(text_list), bbox=dict(facecolor='white', edgecolor='black', pad=10.0, alpha=1), fontsize=14)
fig.show()
Which produces the following graph:
The texts within the bbox are all black. Ideally, each line in the text box should have a color identical to the corresponding series in the plot.
I was able to save two lists of texts and colors in the text_box_content and color_list. I also tried to add plt.gcf().text()
within the for loop with dynamically-updated text locations, but the bounding boxes are created for each row instead of an overall-bounding box for all text.
It would be nicer if there is something conceptually similar to
plt.gcf().text(zip(text_list, color_list))
so each line can have its own color?
python matplotlib visualization seaborn
python matplotlib visualization seaborn
asked Nov 14 '18 at 17:16
Xiaoyu LuXiaoyu Lu
499515
499515
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You could create a legend and colorize each legend item by the color of the line it belongs to. If you don't want to show the line itself in the legend, you could instead show the corresponding letter as a legend handle.
import pandas as pd
import matplotlib.pyplot as plt
from matplotlib.legend_handler import HandlerBase
from matplotlib.text import Text
import seaborn as sns
import numpy as np
import scipy.stats as stats
class TextHandler(HandlerBase):
def create_artists(self, legend,tup ,xdescent, ydescent,
width, height, fontsize,trans):
tx = Text(width/2.,height/2,tup[0], fontsize=fontsize,
ha="center", va="center", color=tup[1], fontweight="bold")
return [tx]
df = pd.DataFrame('x': [21000, 16900, 18200, 32000, 35000, 7500],
'y':[3000, 2100, 1500, 3000, 2500, 2000],
'z':['a', 'b', 'c', 'd', 'e', 'f'])
fig, ax = plt.subplots(figsize=(8,6))
handles =
labels =
for i, row in df.iterrows():
mu, sigma, group = row['x'], row['y'], row['z']
x = np.linspace(mu - 4*sigma, mu + 4*sigma, 100)
sns.lineplot(x, stats.norm.pdf(x, mu, sigma), ax=ax)
color = ax.get_lines()[-1].get_c()
ax.text(mu*1.05, max(stats.norm.pdf(x, mu, sigma)), group, fontsize=16, color=color)
handles.append((":".format(group), color))
labels.append(" $pm$ ".format(mu, sigma))
leg = ax.legend(handles=handles, labels=labels, handler_map=tuple : TextHandler(),
facecolor='white', edgecolor='black', borderpad=0.9, framealpha=1,
fontsize=10, handlelength=0.5)
for h, t in zip(leg.legendHandles, leg.get_texts()):
t.set_color(h.get_color())
plt.show()
This is inspired by and makes partial use of the code from my answer to this question, which also holds an alternative in case you do not want to use a legend.
This is an elegant and simple solution. And by putting everything in the legend I can also freely adjust the location of the box and everything else very easily. Thanks.
– Xiaoyu Lu
Nov 14 '18 at 20:22
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53305557%2fmatplotlib-create-text-with-colors-corresponding-to-series%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could create a legend and colorize each legend item by the color of the line it belongs to. If you don't want to show the line itself in the legend, you could instead show the corresponding letter as a legend handle.
import pandas as pd
import matplotlib.pyplot as plt
from matplotlib.legend_handler import HandlerBase
from matplotlib.text import Text
import seaborn as sns
import numpy as np
import scipy.stats as stats
class TextHandler(HandlerBase):
def create_artists(self, legend,tup ,xdescent, ydescent,
width, height, fontsize,trans):
tx = Text(width/2.,height/2,tup[0], fontsize=fontsize,
ha="center", va="center", color=tup[1], fontweight="bold")
return [tx]
df = pd.DataFrame('x': [21000, 16900, 18200, 32000, 35000, 7500],
'y':[3000, 2100, 1500, 3000, 2500, 2000],
'z':['a', 'b', 'c', 'd', 'e', 'f'])
fig, ax = plt.subplots(figsize=(8,6))
handles =
labels =
for i, row in df.iterrows():
mu, sigma, group = row['x'], row['y'], row['z']
x = np.linspace(mu - 4*sigma, mu + 4*sigma, 100)
sns.lineplot(x, stats.norm.pdf(x, mu, sigma), ax=ax)
color = ax.get_lines()[-1].get_c()
ax.text(mu*1.05, max(stats.norm.pdf(x, mu, sigma)), group, fontsize=16, color=color)
handles.append((":".format(group), color))
labels.append(" $pm$ ".format(mu, sigma))
leg = ax.legend(handles=handles, labels=labels, handler_map=tuple : TextHandler(),
facecolor='white', edgecolor='black', borderpad=0.9, framealpha=1,
fontsize=10, handlelength=0.5)
for h, t in zip(leg.legendHandles, leg.get_texts()):
t.set_color(h.get_color())
plt.show()
This is inspired by and makes partial use of the code from my answer to this question, which also holds an alternative in case you do not want to use a legend.
This is an elegant and simple solution. And by putting everything in the legend I can also freely adjust the location of the box and everything else very easily. Thanks.
– Xiaoyu Lu
Nov 14 '18 at 20:22
add a comment |
You could create a legend and colorize each legend item by the color of the line it belongs to. If you don't want to show the line itself in the legend, you could instead show the corresponding letter as a legend handle.
import pandas as pd
import matplotlib.pyplot as plt
from matplotlib.legend_handler import HandlerBase
from matplotlib.text import Text
import seaborn as sns
import numpy as np
import scipy.stats as stats
class TextHandler(HandlerBase):
def create_artists(self, legend,tup ,xdescent, ydescent,
width, height, fontsize,trans):
tx = Text(width/2.,height/2,tup[0], fontsize=fontsize,
ha="center", va="center", color=tup[1], fontweight="bold")
return [tx]
df = pd.DataFrame('x': [21000, 16900, 18200, 32000, 35000, 7500],
'y':[3000, 2100, 1500, 3000, 2500, 2000],
'z':['a', 'b', 'c', 'd', 'e', 'f'])
fig, ax = plt.subplots(figsize=(8,6))
handles =
labels =
for i, row in df.iterrows():
mu, sigma, group = row['x'], row['y'], row['z']
x = np.linspace(mu - 4*sigma, mu + 4*sigma, 100)
sns.lineplot(x, stats.norm.pdf(x, mu, sigma), ax=ax)
color = ax.get_lines()[-1].get_c()
ax.text(mu*1.05, max(stats.norm.pdf(x, mu, sigma)), group, fontsize=16, color=color)
handles.append((":".format(group), color))
labels.append(" $pm$ ".format(mu, sigma))
leg = ax.legend(handles=handles, labels=labels, handler_map=tuple : TextHandler(),
facecolor='white', edgecolor='black', borderpad=0.9, framealpha=1,
fontsize=10, handlelength=0.5)
for h, t in zip(leg.legendHandles, leg.get_texts()):
t.set_color(h.get_color())
plt.show()
This is inspired by and makes partial use of the code from my answer to this question, which also holds an alternative in case you do not want to use a legend.
This is an elegant and simple solution. And by putting everything in the legend I can also freely adjust the location of the box and everything else very easily. Thanks.
– Xiaoyu Lu
Nov 14 '18 at 20:22
add a comment |
You could create a legend and colorize each legend item by the color of the line it belongs to. If you don't want to show the line itself in the legend, you could instead show the corresponding letter as a legend handle.
import pandas as pd
import matplotlib.pyplot as plt
from matplotlib.legend_handler import HandlerBase
from matplotlib.text import Text
import seaborn as sns
import numpy as np
import scipy.stats as stats
class TextHandler(HandlerBase):
def create_artists(self, legend,tup ,xdescent, ydescent,
width, height, fontsize,trans):
tx = Text(width/2.,height/2,tup[0], fontsize=fontsize,
ha="center", va="center", color=tup[1], fontweight="bold")
return [tx]
df = pd.DataFrame('x': [21000, 16900, 18200, 32000, 35000, 7500],
'y':[3000, 2100, 1500, 3000, 2500, 2000],
'z':['a', 'b', 'c', 'd', 'e', 'f'])
fig, ax = plt.subplots(figsize=(8,6))
handles =
labels =
for i, row in df.iterrows():
mu, sigma, group = row['x'], row['y'], row['z']
x = np.linspace(mu - 4*sigma, mu + 4*sigma, 100)
sns.lineplot(x, stats.norm.pdf(x, mu, sigma), ax=ax)
color = ax.get_lines()[-1].get_c()
ax.text(mu*1.05, max(stats.norm.pdf(x, mu, sigma)), group, fontsize=16, color=color)
handles.append((":".format(group), color))
labels.append(" $pm$ ".format(mu, sigma))
leg = ax.legend(handles=handles, labels=labels, handler_map=tuple : TextHandler(),
facecolor='white', edgecolor='black', borderpad=0.9, framealpha=1,
fontsize=10, handlelength=0.5)
for h, t in zip(leg.legendHandles, leg.get_texts()):
t.set_color(h.get_color())
plt.show()
This is inspired by and makes partial use of the code from my answer to this question, which also holds an alternative in case you do not want to use a legend.
You could create a legend and colorize each legend item by the color of the line it belongs to. If you don't want to show the line itself in the legend, you could instead show the corresponding letter as a legend handle.
import pandas as pd
import matplotlib.pyplot as plt
from matplotlib.legend_handler import HandlerBase
from matplotlib.text import Text
import seaborn as sns
import numpy as np
import scipy.stats as stats
class TextHandler(HandlerBase):
def create_artists(self, legend,tup ,xdescent, ydescent,
width, height, fontsize,trans):
tx = Text(width/2.,height/2,tup[0], fontsize=fontsize,
ha="center", va="center", color=tup[1], fontweight="bold")
return [tx]
df = pd.DataFrame('x': [21000, 16900, 18200, 32000, 35000, 7500],
'y':[3000, 2100, 1500, 3000, 2500, 2000],
'z':['a', 'b', 'c', 'd', 'e', 'f'])
fig, ax = plt.subplots(figsize=(8,6))
handles =
labels =
for i, row in df.iterrows():
mu, sigma, group = row['x'], row['y'], row['z']
x = np.linspace(mu - 4*sigma, mu + 4*sigma, 100)
sns.lineplot(x, stats.norm.pdf(x, mu, sigma), ax=ax)
color = ax.get_lines()[-1].get_c()
ax.text(mu*1.05, max(stats.norm.pdf(x, mu, sigma)), group, fontsize=16, color=color)
handles.append((":".format(group), color))
labels.append(" $pm$ ".format(mu, sigma))
leg = ax.legend(handles=handles, labels=labels, handler_map=tuple : TextHandler(),
facecolor='white', edgecolor='black', borderpad=0.9, framealpha=1,
fontsize=10, handlelength=0.5)
for h, t in zip(leg.legendHandles, leg.get_texts()):
t.set_color(h.get_color())
plt.show()
This is inspired by and makes partial use of the code from my answer to this question, which also holds an alternative in case you do not want to use a legend.
answered Nov 14 '18 at 17:46
ImportanceOfBeingErnestImportanceOfBeingErnest
138k13157235
138k13157235
This is an elegant and simple solution. And by putting everything in the legend I can also freely adjust the location of the box and everything else very easily. Thanks.
– Xiaoyu Lu
Nov 14 '18 at 20:22
add a comment |
This is an elegant and simple solution. And by putting everything in the legend I can also freely adjust the location of the box and everything else very easily. Thanks.
– Xiaoyu Lu
Nov 14 '18 at 20:22
This is an elegant and simple solution. And by putting everything in the legend I can also freely adjust the location of the box and everything else very easily. Thanks.
– Xiaoyu Lu
Nov 14 '18 at 20:22
This is an elegant and simple solution. And by putting everything in the legend I can also freely adjust the location of the box and everything else very easily. Thanks.
– Xiaoyu Lu
Nov 14 '18 at 20:22
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53305557%2fmatplotlib-create-text-with-colors-corresponding-to-series%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown