英文:
Use seaborn object interface to plot overlapping density plots, added inside a for loop, each having its own color/label shown in a legend
问题
import seaborn.objects as so
num_samples = 2000
normal_distr = st.norm(1,1)
sp = so.Plot()
for s_n in range(10,num_samples,400):
sample_normal = normal_distr.rvs(s_n)
sp = sp.add(so.Line(), so.KDE(), x=sample_normal, color=s_n, label=s_n)
sp.show()
英文:
Using seaborn python library, I am trying to make several density plots overlapping each other in the same figure and I want to color/label each of the lines. Using seaborn objects interface I am able to make the density plots within a for loop. But I cannot add color/label to each density plot.
I understand that there are other ways e.g., I create a dataframe with all the data and corresponding labels first and then pass it to seaborn plot(). But I was just wondering if below code (using seaborn objects interface) could work with some modifications. Please advise.
- Plot using Seaborn objects
Code:
Here I am setting color=s_n which is the number of samples that I drew from the normal distribution. I want to label each density plot with the number of samples (please also the see the desired plot towards the end of post)
import scipy.stats as st
import seaborn.objects as so
num_samples = 2000
normal_distr = st.norm(1,1)
sp = so.Plot()
for s_n in range(10,num_samples,400):
sample_normal = normal_distr.rvs(s_n)
sp = sp.add(so.Line(),so.KDE(),x=sample_normal,color=s_n)
sp.show()
The plots looks like this and it does not color/label each density line separately.
- Desired Plot
If I directly use seaborn kdeplot, I can get the desired plot (below). But I was just wondering if I can use seaborn objects instead of direct kdeplot
Code using kdeplot:
import scipy.stats as st
import seaborn as sns
import matplotlib.pyplot as plt
num_samples = 2000
normal_distr = st.norm(1,1)
for s_n in range(10,num_samples,400):
sample_normal = normal_distr.rvs(s_n)
sns.kdeplot(x=sample_normal, label=s_n)
plt.legend()
The (desired) plot:
答案1
得分: 1
我猜这里的诀窍是准备好你的数据框,这样你就可以放弃循环并使用color参数,因为它是设计用来使用的:
import scipy.stats as st
import seaborn.objects as so
import pandas as pd
num_samples = 2000
normal_distr = st.norm(1, 1)
df = pd.concat([
pd.DataFrame(
{'sn': str(s_n),
'values': normal_distr.rvs(s_n)}
)
for s_n in range(10, num_samples, 400)
])
这会看起来像这样:
sn values
0 10 0.976926
1 10 -0.501831
2 10 1.748071
3 10 0.968493
4 10 0.593531
... ... ...
1605 1610 0.311484
1606 1610 1.332424
1607 1610 1.531519
1608 1610 1.240953
1609 1610 -0.793144
然后打印可以在一行内完成:
so.Plot(df, x='values').add(so.Line(), so.KDE(common_norm=False), color='sn').show()
英文:
I guess the trick here would be to prepare your df so that you can forgo the loop and use the color kwarg as it's meant to be used:
import scipy.stats as st
import seaborn.objects as so
import pandas as pd
num_samples = 2000
normal_distr = st.norm(1,1)
df = pd.concat([
pd.DataFrame(
{'sn': str(s_n),
'values': normal_distr.rvs(s_n)}
)
for s_n in range(10,num_samples,400)
])
This would look like this:
sn values
0 10 0.976926
1 10 -0.501831
2 10 1.748071
3 10 0.968493
4 10 0.593531
... ... ...
1605 1610 0.311484
1606 1610 1.332424
1607 1610 1.531519
1608 1610 1.240953
1609 1610 -0.793144
Then printing can be done in a single line:
so.Plot(df, x='values').add(so.Line(), so.KDE(common_norm=False), color='sn').show()
Output:
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