Related
I've spent hours on trying to do what I thought was a simple task, which is to add labels onto an XY plot while using seaborn.
Here's my code
import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline
df_iris=sns.load_dataset("iris")
sns.lmplot('sepal_length', # Horizontal axis
'sepal_width', # Vertical axis
data=df_iris, # Data source
fit_reg=False, # Don't fix a regression line
size = 8,
aspect =2 ) # size and dimension
plt.title('Example Plot')
# Set x-axis label
plt.xlabel('Sepal Length')
# Set y-axis label
plt.ylabel('Sepal Width')
I would like to add to each dot on the plot the text in "species" column.
I've seen many examples using matplotlib but not using seaborn.
Any ideas? Thank you.
One way you can do this is as follows:
import seaborn as sns
import matplotlib.pyplot as plt
import pandas as pd
%matplotlib inline
df_iris=sns.load_dataset("iris")
ax = sns.lmplot('sepal_length', # Horizontal axis
'sepal_width', # Vertical axis
data=df_iris, # Data source
fit_reg=False, # Don't fix a regression line
size = 10,
aspect =2 ) # size and dimension
plt.title('Example Plot')
# Set x-axis label
plt.xlabel('Sepal Length')
# Set y-axis label
plt.ylabel('Sepal Width')
def label_point(x, y, val, ax):
a = pd.concat({'x': x, 'y': y, 'val': val}, axis=1)
for i, point in a.iterrows():
ax.text(point['x']+.02, point['y'], str(point['val']))
label_point(df_iris.sepal_length, df_iris.sepal_width, df_iris.species, plt.gca())
Here's a more up-to-date answer that doesn't suffer from the string issue described in the comments.
import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline
df_iris=sns.load_dataset("iris")
plt.figure(figsize=(20,10))
p1 = sns.scatterplot(x='sepal_length', # Horizontal axis
y='sepal_width', # Vertical axis
data=df_iris, # Data source
size = 8,
legend=False)
for line in range(0,df_iris.shape[0]):
p1.text(df_iris.sepal_length[line]+0.01, df_iris.sepal_width[line],
df_iris.species[line], horizontalalignment='left',
size='medium', color='black', weight='semibold')
plt.title('Example Plot')
# Set x-axis label
plt.xlabel('Sepal Length')
# Set y-axis label
plt.ylabel('Sepal Width')
Thanks to the 2 other answers, here is a function scatter_text that makes it possible to reuse these plots several times.
import seaborn as sns
import matplotlib.pyplot as plt
def scatter_text(x, y, text_column, data, title, xlabel, ylabel):
"""Scatter plot with country codes on the x y coordinates
Based on this answer: https://stackoverflow.com/a/54789170/2641825"""
# Create the scatter plot
p1 = sns.scatterplot(x, y, data=data, size = 8, legend=False)
# Add text besides each point
for line in range(0,data.shape[0]):
p1.text(data[x][line]+0.01, data[y][line],
data[text_column][line], horizontalalignment='left',
size='medium', color='black', weight='semibold')
# Set title and axis labels
plt.title(title)
plt.xlabel(xlabel)
plt.ylabel(ylabel)
return p1
Use the function as follows:
df_iris=sns.load_dataset("iris")
plt.figure(figsize=(20,10))
scatter_text('sepal_length', 'sepal_width', 'species',
data = df_iris,
title = 'Iris sepals',
xlabel = 'Sepal Length (cm)',
ylabel = 'Sepal Width (cm)')
See also this answer on how to have a function that returns a plot:
https://stackoverflow.com/a/43926055/2641825
Below is a solution that does not iterate over rows in the data frame using the dreaded for loop.
There are many issues regarding iterating over a data frame.
The answer is don't iterate! See this link.
The solution below relies on a function (plotlabel) within the petalplot function, which is called by df.apply.
Now, I know readers will comment on the fact that I use scatter and not lmplot, but that is a bit besides the point.
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline
df_iris=sns.load_dataset("iris")
def petalplot(df):
def plotlabel(xvar, yvar, label):
ax.text(xvar+0.002, yvar, label)
fig = plt.figure(figsize=(30,10))
ax = sns.scatterplot(x = 'sepal_length', y = 'sepal_width', data=df)
# The magic starts here:
df.apply(lambda x: plotlabel(x['sepal_length'], x['sepal_width'], x['species']), axis=1)
plt.title('Example Plot')
plt.xlabel('Sepal Length')
plt.ylabel('Sepal Width')
petalplot(df_iris)
Same idea with Scott Boston's answer, however with Seaborn v0.12+, you can leverage seaborn.FacetGrid.apply to add labels on plots and set up your figure in one go:
import seaborn as sns
import pandas as pd
%matplotlib inline
sns.set_theme()
df_iris = sns.load_dataset("iris")
(
sns.lmplot(
data=df_iris,
x="sepal_length",
y="sepal_width",
fit_reg=False,
height=8,
aspect=2
)
.apply(lambda grid: [
grid.ax.text(r["sepal_length"]+.02, r["sepal_width"], r["species"])
for r in df_iris.to_dict(orient="records")
])
.set(title="Example Plot")
.set_axis_labels("Sepal Length", "Sepal Width")
)
Or, if you don't need to use lmplot, also from v0.12, you can use the seaborn.objects interface. This way we don't need to manually iterate over the Iris dataframe nor refer to df_iris or column names sepal_... multiple times.
import seaborn.objects as so
(
so.Plot(df_iris, x="sepal_length", y="sepal_width", text="species")
.add(so.Dot())
.add(so.Text(halign="left"))
.label(title="Example plot", x="Sepal Length", y="Sepal Width")
.layout(size=(20, 10))
)
This produces the below figure:
Use the powerful declarative API to avoid loops (seaborn>=0.12).
Specifically, put x,y, and annotations into a pandas data frame and call plotting.
Here is an example from my own research work.
import seaborn.objects as so
import pandas as pd
df = pd.DataFrame(..,columns=['phase','P(X=1)','text'])
fig,ax = plt.subplots()
p = so.Plot(df,x='phase',y='P(X=1)',text='text').add(so.Dot(marker='+')).add(so.Text(halign='left'))
p.on(ax).show()
I am trying to make a stacked histogram using matplotlib by looping through the categories in the dataframe and assigning the bar color based on a dictionary.
I get this error on the ax1.hist() call. How should I fix it?
AttributeError: 'numpy.ndarray' object has no attribute 'hist'
Reproducible Example
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.gridspec as gridspec
%matplotlib inline
plt.style.use('seaborn-whitegrid')
y = [1,5,9,2,4,2,5,6,1]
cat = ['A','B','B','B','A','B','B','B','B']
df = pd.DataFrame(list(zip(y,cat)), columns =['y', 'cat'])
fig, axes = plt.subplots(3,3, figsize=(5,5), constrained_layout=True)
fig.suptitle('Histograms')
ax1 = axes[0]
mycolorsdict = {'A':'magenta', 'B':'blue'}
for key, batch in df.groupby(['cat']):
ax1.hist(batch.y, label=key, color=mycolorsdict[key],
density=False, cumulative=False, edgecolor='black',
orientation='horizontal', stacked=True)
Updated effort, still not working
This is close, but it is not stacking (should see stacks at y=5); I think maybe because of the loop?
mycolorsdict = {'A':'magenta', 'B':'blue'}
for ii, ax in enumerate(axes.flat):
for key, batch in df.groupby(['cat']):
ax.hist(batch.y,
label=key, color=mycolorsdict[key],density=False, edgecolor='black',
cumulative=False, orientation='horizontal', stacked=True)
To draw on a specific subplot, two indices are needed (row, column), so axes[0,0] for the first subplot. The error message comes from using ax1 = axes[0] instead of ax1 = axes[0,0].
Now, to create a stacked histogram via ax.hist(), all the y-data need to be provided at the same time. The code below shows how this can be done starting from the result of groupby. Also note, that when your values are discrete, it is important to explicitly set the bin boundaries making sure that the values fall precisely between these boundaries. Setting the boundaries at the halves is one way.
Things can be simplified a lot using seaborn's histplot(). Here is a breakdown of the parameters used:
data=df the dataframe
y='y' gives the dataframe column for histogram. Use x= (instead of y=) for a vertical histogram.
hue='cat' gives the dataframe column to create mulitple groups
palette=mycolorsdict; the palette defines the coloring; there are many ways to assign a palette, one of which is a dictionary on the hue values
discrete=True: when working with discrete data, seaborn sets the appropriate bin boundaries
multiple='stack' creates a stacked histogram, depending on the hue categories
alpha=1: default seaborn sets an alpha of 0.75; optionally this can be changed
ax=axes[0, 1]: draw on the 2nd subplot of the 1st row
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
plt.style.use('seaborn-whitegrid')
y = [1, 5, 9, 2, 4, 2, 5, 6, 1]
cat = ['A', 'B', 'B', 'B', 'A', 'B', 'B', 'B', 'B']
df = pd.DataFrame({'y':y, 'cat':cat})
fig, axes = plt.subplots(3, 3, figsize=(20, 10), constrained_layout=True)
fig.suptitle('Histograms')
mycolorsdict = {'A': 'magenta', 'B': 'blue'}
groups = df.groupby(['cat'])
axes[0, 0].hist([batch.y for _, batch in groups],
label=[key for key, _ in groups], color=[mycolorsdict[key] for key, _ in groups], density=False,
edgecolor='black',
cumulative=False, orientation='horizontal', stacked=True, bins=np.arange(0.5, 10))
axes[0, 0].legend()
sns.histplot(data=df, y='y', hue='cat', palette=mycolorsdict, discrete=True, multiple='stack', alpha=1, ax=axes[0, 1])
plt.show()
I am plotting a certain categorical value over the map of a city. The line of code I use to plot is the following:
fig = plt.figure(figsize=(12, 12))
ax = plt.gca()
urban_data.plot(column="category", cmap="viridis", ax=ax, categorical=True, /
k=4, legend=True, linewidth=0.5, /
legend_kwds={'fontsize':'19', 'loc':'lower left'})
where urban data is a geopandas dataframe, and I am using matplotlib as plotting library. The argument legend_kwds allows me to control minor things on the legend, like the position or the font size, but I cannot decide major things like, for example, the order of the entries in the legend box. In fact my categories are ranked, let's say 1-2-3-4, but I always get them displayed in a different order.
Is it possible to have more control over the legend? For example by calling it outside the gdf.plot() function? And, if so, how do I match the colors in the legend with those in the map, which are discrete values (that I don't know exactly) of a viridis colormap?
EDIT: here is a verifiable example. Unfortunately shapefiles need other files to work, and here a geometry (an area, not a point) column is needed, so I have to ask you to download this shpfile of the US. Everything you need is within this folder. Here's the code to reproduce the issue. The plot in output is bad because I did not care about the coordinates system here, but the important thing is the legend.
import geopandas as gpd
import numpy as np
import matplotlib.pyplot as plt
gdf=gpd.read_file('.../USA_adm1.shp')
clusters=np.random.randint(0,4, size=52)
gdf['cluster']=clusters
clusdict={1: 'lower-middle', 2: 'upper-middle', 3: 'upper', 0: 'lower'}
gdf['cluster']=gdf['cluster'].map(clusdict)
fig = plt.figure(figsize=(12, 12))
ax = plt.gca()
gdf.plot(column='cluster',cmap='viridis', categorical=True, legend=True, ax=ax)
The bad news is that categories in legends produced by geopandas are sorted and this is hardcoded (see source-code here).
One solution is hence to have the categorical column such that if it is sorted, it would correspond to the desired order. Using integers seems fine for that. Then one can replace the names in the legend, once it is produced in the correct order.
import geopandas as gpd
import numpy as np
import matplotlib.pyplot as plt
gdf=gpd.read_file('data/USA_adm/USA_adm1.shp')
clusters=np.random.randint(0,4, size=52)
gdf['cluster']=clusters
clusdict={1: 'lower-middle', 2: 'upper-middle', 3: 'upper', 0: 'lower'}
fig = plt.figure(figsize=(12, 12))
ax = plt.gca()
gdf.plot(column='cluster',cmap='viridis', categorical=True, legend=True, ax=ax)
def replace_legend_items(legend, mapping):
for txt in legend.texts:
for k,v in mapping.items():
if txt.get_text() == str(k):
txt.set_text(v)
replace_legend_items(ax.get_legend(), clusdict)
plt.show()
I had to alter the accepted answer (the second line in the function) from #ImportanceOfBeingErnest a bit to get it to work (maybe there have been updates since),
import geopandas as gpd
import numpy as np
import matplotlib.pyplot as plt
gdf=gpd.read_file('data/USA_adm/USA_adm1.shp')
clusters=np.random.randint(0,4, size=52)
gdf['cluster']=clusters
clusdict={1: 'lower-middle', 2: 'upper-middle', 3: 'upper', 0: 'lower'}
fig = plt.figure(figsize=(12, 12))
ax = plt.gca()
gdf.plot(column='cluster',cmap='viridis', categorical=True, legend=True, ax=ax)
def replace_legend_items(legend, mapping):
for txt in legend.get_texts():
for k,v in mapping.items():
if txt.get_text() == str(k):
txt.set_text(v)
replace_legend_items(ax.get_legend(), clusdict)
plt.show()
Assuming that you have 4 legends, you can do the following to set them in whatever order you like. The following code shows how to put them in the following order (using index): 0, 2, 3, 1.
Here ax is the axis object which you have define using ax = plt.gca()
handles,labels = ax.get_legend_handles_labels()
handles = [handles[0], handles[2], handles[3], handles[1]]
labels = [labels[0], labels[2], labels[3], labels[1]]
ax.legend(handles, labels)
Let me give you an example:
Default order
fig, ax = plt.subplots()
x = np.arange(5)
plt.plot(x, x, label=r'$y=x$')
plt.plot(x, 2*x, label=r'$y=2x$')
plt.plot(x, 3*x, label=r'$y=3x$')
plt.plot(x, 4*x, label=r'$y=4x$')
plt.legend(fontsize=16)
Manually changed order
fig, ax = plt.subplots()
x = np.arange(5)
plt.plot(x, x, label=r'$y=x$')
plt.plot(x, 2*x, label=r'$y=2x$')
plt.plot(x, 3*x, label=r'$y=3x$')
plt.plot(x, 4*x, label=r'$y=4x$')
handles,labels = ax.get_legend_handles_labels()
handles = [handles[0], handles[2],handles[3], handles[1]]
labels = [labels[0], labels[2], labels[3], labels[1]]
ax.legend(handles, labels, fontsize=16)
One can also use list comprehension using a pre-specified order list as
order = [0, 2, 3, 1]
handles,labels = ax.get_legend_handles_labels()
handles = [handles[i] for i in order]
labels = [labels[i] for i in order]
ax.legend(handles, labels, fontsize=16)
I've spent hours on trying to do what I thought was a simple task, which is to add labels onto an XY plot while using seaborn.
Here's my code
import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline
df_iris=sns.load_dataset("iris")
sns.lmplot('sepal_length', # Horizontal axis
'sepal_width', # Vertical axis
data=df_iris, # Data source
fit_reg=False, # Don't fix a regression line
size = 8,
aspect =2 ) # size and dimension
plt.title('Example Plot')
# Set x-axis label
plt.xlabel('Sepal Length')
# Set y-axis label
plt.ylabel('Sepal Width')
I would like to add to each dot on the plot the text in "species" column.
I've seen many examples using matplotlib but not using seaborn.
Any ideas? Thank you.
One way you can do this is as follows:
import seaborn as sns
import matplotlib.pyplot as plt
import pandas as pd
%matplotlib inline
df_iris=sns.load_dataset("iris")
ax = sns.lmplot('sepal_length', # Horizontal axis
'sepal_width', # Vertical axis
data=df_iris, # Data source
fit_reg=False, # Don't fix a regression line
size = 10,
aspect =2 ) # size and dimension
plt.title('Example Plot')
# Set x-axis label
plt.xlabel('Sepal Length')
# Set y-axis label
plt.ylabel('Sepal Width')
def label_point(x, y, val, ax):
a = pd.concat({'x': x, 'y': y, 'val': val}, axis=1)
for i, point in a.iterrows():
ax.text(point['x']+.02, point['y'], str(point['val']))
label_point(df_iris.sepal_length, df_iris.sepal_width, df_iris.species, plt.gca())
Here's a more up-to-date answer that doesn't suffer from the string issue described in the comments.
import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline
df_iris=sns.load_dataset("iris")
plt.figure(figsize=(20,10))
p1 = sns.scatterplot(x='sepal_length', # Horizontal axis
y='sepal_width', # Vertical axis
data=df_iris, # Data source
size = 8,
legend=False)
for line in range(0,df_iris.shape[0]):
p1.text(df_iris.sepal_length[line]+0.01, df_iris.sepal_width[line],
df_iris.species[line], horizontalalignment='left',
size='medium', color='black', weight='semibold')
plt.title('Example Plot')
# Set x-axis label
plt.xlabel('Sepal Length')
# Set y-axis label
plt.ylabel('Sepal Width')
Thanks to the 2 other answers, here is a function scatter_text that makes it possible to reuse these plots several times.
import seaborn as sns
import matplotlib.pyplot as plt
def scatter_text(x, y, text_column, data, title, xlabel, ylabel):
"""Scatter plot with country codes on the x y coordinates
Based on this answer: https://stackoverflow.com/a/54789170/2641825"""
# Create the scatter plot
p1 = sns.scatterplot(x, y, data=data, size = 8, legend=False)
# Add text besides each point
for line in range(0,data.shape[0]):
p1.text(data[x][line]+0.01, data[y][line],
data[text_column][line], horizontalalignment='left',
size='medium', color='black', weight='semibold')
# Set title and axis labels
plt.title(title)
plt.xlabel(xlabel)
plt.ylabel(ylabel)
return p1
Use the function as follows:
df_iris=sns.load_dataset("iris")
plt.figure(figsize=(20,10))
scatter_text('sepal_length', 'sepal_width', 'species',
data = df_iris,
title = 'Iris sepals',
xlabel = 'Sepal Length (cm)',
ylabel = 'Sepal Width (cm)')
See also this answer on how to have a function that returns a plot:
https://stackoverflow.com/a/43926055/2641825
Below is a solution that does not iterate over rows in the data frame using the dreaded for loop.
There are many issues regarding iterating over a data frame.
The answer is don't iterate! See this link.
The solution below relies on a function (plotlabel) within the petalplot function, which is called by df.apply.
Now, I know readers will comment on the fact that I use scatter and not lmplot, but that is a bit besides the point.
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline
df_iris=sns.load_dataset("iris")
def petalplot(df):
def plotlabel(xvar, yvar, label):
ax.text(xvar+0.002, yvar, label)
fig = plt.figure(figsize=(30,10))
ax = sns.scatterplot(x = 'sepal_length', y = 'sepal_width', data=df)
# The magic starts here:
df.apply(lambda x: plotlabel(x['sepal_length'], x['sepal_width'], x['species']), axis=1)
plt.title('Example Plot')
plt.xlabel('Sepal Length')
plt.ylabel('Sepal Width')
petalplot(df_iris)
Same idea with Scott Boston's answer, however with Seaborn v0.12+, you can leverage seaborn.FacetGrid.apply to add labels on plots and set up your figure in one go:
import seaborn as sns
import pandas as pd
%matplotlib inline
sns.set_theme()
df_iris = sns.load_dataset("iris")
(
sns.lmplot(
data=df_iris,
x="sepal_length",
y="sepal_width",
fit_reg=False,
height=8,
aspect=2
)
.apply(lambda grid: [
grid.ax.text(r["sepal_length"]+.02, r["sepal_width"], r["species"])
for r in df_iris.to_dict(orient="records")
])
.set(title="Example Plot")
.set_axis_labels("Sepal Length", "Sepal Width")
)
Or, if you don't need to use lmplot, also from v0.12, you can use the seaborn.objects interface. This way we don't need to manually iterate over the Iris dataframe nor refer to df_iris or column names sepal_... multiple times.
import seaborn.objects as so
(
so.Plot(df_iris, x="sepal_length", y="sepal_width", text="species")
.add(so.Dot())
.add(so.Text(halign="left"))
.label(title="Example plot", x="Sepal Length", y="Sepal Width")
.layout(size=(20, 10))
)
This produces the below figure:
Use the powerful declarative API to avoid loops (seaborn>=0.12).
Specifically, put x,y, and annotations into a pandas data frame and call plotting.
Here is an example from my own research work.
import seaborn.objects as so
import pandas as pd
df = pd.DataFrame(..,columns=['phase','P(X=1)','text'])
fig,ax = plt.subplots()
p = so.Plot(df,x='phase',y='P(X=1)',text='text').add(so.Dot(marker='+')).add(so.Text(halign='left'))
p.on(ax).show()
Hi I wanted to draw a histogram with a boxplot appearing the top of the histogram showing the Q1,Q2 and Q3 as well as the outliers. Example phone is below. (I am using Python and Pandas)
I have checked several examples using matplotlib.pyplot but hardly came out with a good example. And I also wanted to have the histogram curve appearing like in the image below.
I also tried seaborn and it provided me the shape line along with the histogram but didnt find a way to incorporate with boxpot above it.
can anyone help me with this to have this on matplotlib.pyplot or using pyplot
import numpy as np
import seaborn as sns
import matplotlib.pyplot as plt
sns.set(style="ticks")
x = np.random.randn(100)
f, (ax_box, ax_hist) = plt.subplots(2, sharex=True,
gridspec_kw={"height_ratios": (.15, .85)})
sns.boxplot(x, ax=ax_box)
sns.distplot(x, ax=ax_hist)
ax_box.set(yticks=[])
sns.despine(ax=ax_hist)
sns.despine(ax=ax_box, left=True)
From seaborn v0.11.2, sns.distplot is deprecated. Use sns.histplot for axes-level plots instead.
np.random.seed(2022)
x = np.random.randn(100)
f, (ax_box, ax_hist) = plt.subplots(2, sharex=True, gridspec_kw={"height_ratios": (.15, .85)})
sns.boxplot(x=x, ax=ax_box)
sns.histplot(x=x, bins=12, kde=True, stat='density', ax=ax_hist)
ax_box.set(yticks=[])
sns.despine(ax=ax_hist)
sns.despine(ax=ax_box, left=True)
Solution using only matplotlib, just because:
# start the plot: 2 rows, because we want the boxplot on the first row
# and the hist on the second
fig, ax = plt.subplots(
2, figsize=(7, 5), sharex=True,
gridspec_kw={"height_ratios": (.3, .7)} # the boxplot gets 30% of the vertical space
)
# the boxplot
ax[0].boxplot(data, vert=False)
# removing borders
ax[0].spines['top'].set_visible(False)
ax[0].spines['right'].set_visible(False)
ax[0].spines['left'].set_visible(False)
# the histogram
ax[1].hist(data)
# and we are good to go
plt.show()
Expanding on the answer from #mwaskom, I made a little adaptable function.
import seaborn as sns
def histogram_boxplot(data, xlabel = None, title = None, font_scale=2, figsize=(9,8), bins = None):
""" Boxplot and histogram combined
data: 1-d data array
xlabel: xlabel
title: title
font_scale: the scale of the font (default 2)
figsize: size of fig (default (9,8))
bins: number of bins (default None / auto)
example use: histogram_boxplot(np.random.rand(100), bins = 20, title="Fancy plot")
"""
sns.set(font_scale=font_scale)
f2, (ax_box2, ax_hist2) = plt.subplots(2, sharex=True, gridspec_kw={"height_ratios": (.15, .85)}, figsize=figsize)
sns.boxplot(data, ax=ax_box2)
sns.distplot(data, ax=ax_hist2, bins=bins) if bins else sns.distplot(data, ax=ax_hist2)
if xlabel: ax_hist2.set(xlabel=xlabel)
if title: ax_box2.set(title=title)
plt.show()
histogram_boxplot(np.random.randn(100), bins = 20, title="Fancy plot", xlabel="Some values")
Image
def histogram_boxplot(feature, figsize=(15,10), bins=None):
f,(ax_box,ax_hist)=plt.subplots(nrows=2,sharex=True, gridspec_kw={'height_ratios':(.25,.75)},figsize=figsize)
sns.distplot(feature,kde=False,ax=ax_hist,bins=bins)
sns.boxplot(feature,ax=ax_box, color='Red')
ax_hist.axvline(np.mean(feature),color='g',linestyle='-')
ax_hist.axvline(np.median(feature),color='y',linestyle='--')