绘制 95% 置信区间误差条 python pandas dataframes

Question

I want to show 95% confidence interval with Python pandas, matpolib... But I stucked, because for usual .std() I would do smth like this:

import pandas as pd
import numpy as np

import matplotlib

matplotlib.use('Agg')

import matplotlib.pyplot as plt
import math

data = pd.read_table('output.txt',sep=r'\,', engine='python')
Ox = data.groupby(['Ox'])['Ox'].mean()
Oy = data.groupby(['Ox'])['Oy'].mean()
std = data.groupby(['Ox'])['Oy'].std()

plt.plot(Ox, Oy , label = 'STA = '+ str(x))
plt.errorbar(Ox, Oy, std, label = 'errorbar', linewidth=2)

plt.legend(loc='best', prop={'size':9.2})

plt.savefig('plot.pdf')
plt.close()

But I haven't found something in pandas methods which can help me. Does anybody know?

Answer 1

Using 2 * std to estimate the 95 % interval

In a normal distribution, the interval [μ - 2σ, μ + 2σ] covers 95.5 %, so you can use 2 * std to estimate the 95 % interval:

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

df = pd.DataFrame()
df['category'] = np.random.choice(np.arange(10), 1000, replace=True)
df['number'] = np.random.normal(df['category'], 1)

mean = df.groupby('category')['number'].mean()
std = df.groupby('category')['number'].std()

plt.errorbar(mean.index, mean, xerr=0.5, yerr=2*std, linestyle='')
plt.show()

Result:

Using percentiles

If your distribution is skewed, it is better to use asymmetrical errorbars and get your 95% interval from the percentiles.

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from scipy.stats import skewnorm

df = pd.DataFrame()
df['category'] = np.random.choice(np.arange(10), 1000, replace=True)
df['number'] = skewnorm.rvs(5, df['category'], 1)

mean = df.groupby('category')['number'].mean()
p025 = df.groupby('category')['number'].quantile(0.025)
p975 = df.groupby('category')['number'].quantile(0.975)

plt.errorbar(
    mean.index,
    mean,
    xerr=0.5,
    yerr=[mean - p025, p975 - mean],
    linestyle='',
)
plt.show()

Result:

Answer 2

For a normal distribution ~95% of the values lie within a window of 4 standard deviations around the mean, or in other words, 95% of the values are within plus/minus 2 standard deviations from the mean. See, eg 68–95–99.7-rule .

plt.errorbar 's yerr argument specifies the length of the single sided errorbar. Thus taking

plt.errorbar(x,y,yerr=2*std)

where std is the standard deviation shows the errorbars of the 95% confidence interval.

Answer 3

To obtain the 95% confidence interval you'll need to define a function.

Tweaking: Compute a confidence interval from sample data

def mean_confidence_interval(data, confidence=0.95):
    a = 1.0 * np.array(data)
    n = len(a)
    m, se = np.mean(a), scipy.stats.sem(a)
    h = se * scipy.stats.t.ppf((1 + confidence) / 2., n-1)
    return m

def bound_confidence_interval(data, confidence=0.95):
    a = 1.0 * np.array(data)
    n = len(a)
    m, se = np.mean(a), scipy.stats.sem(a)
    h = se * scipy.stats.t.ppf((1 + confidence) / 2., n-1)
    return h

Then define:

mean = df.groupby(by='yourquery').agg(mean_confidence_interval)
bound = df.groupby(by='yourquery').agg(bound_confidence_interval)

Finally plot with a library of your choice: eg plotly

import plotly.graph_objects as go

   fig = go.Figure(data=go.Scatter(
        x=mean[yourquery],
        y=mean[yourquery2],
        error_y=dict(
            type='data', # value of error bar given in data coordinates
            array=bound[yourquery2],
            visible=True)
    ))
    fig.show()