根据文献[1]，直接用Pearson r来计算置信区间，由于它不是正态分布，因此计算起来比较复杂，需要进行以下步骤：
1.将r转换为z '，
1.计算z'的置信区间。z'的抽样分布近似正态分布，标准误差为1/sqrt（n-3）。
1.将置信区间转换回r。
以下是一些示例代码：

def r_to_z(r):
    return math.log((1 + r) / (1 - r)) / 2.0

def z_to_r(z):
    e = math.exp(2 * z)
    return((e - 1) / (e + 1))

def r_confidence_interval(r, alpha, n):
    z = r_to_z(r)
    se = 1.0 / math.sqrt(n - 3)
    z_crit = stats.norm.ppf(1 - alpha/2)  # 2-tailed z critical value

    lo = z - z_crit * se
    hi = z + z_crit * se

    # Return a sequence
    return (z_to_r(lo), z_to_r(hi))

参考编号：

1. http://onlinestatbook.com/2/estimation/correlation_ci.html

使用rpy2和心理测试库（您需要安装R，并首先在R中运行install.packages（“心理测试”））

from rpy2.robjects.packages import importr
psychometric=importr('psychometric')
psychometric.CIr(r=.9, n = 100, level = .95)

其中，0.9是相关性，n是样本量，0.95是置信水平

这里有一个解决方案，它使用自举法来计算置信区间，而不是Fisher变换（它假设二元正态性等），借用this answer：

import numpy as np

def pearsonr_ci(x, y, ci=95, n_boots=10000):
    x = np.asarray(x)
    y = np.asarray(y)

   # (n_boots, n_observations) paired arrays
    rand_ixs = np.random.randint(0, x.shape[0], size=(n_boots, x.shape[0]))
    x_boots = x[rand_ixs]
    y_boots = y[rand_ixs]

    # differences from mean
    x_mdiffs = x_boots - x_boots.mean(axis=1)[:, None]
    y_mdiffs = y_boots - y_boots.mean(axis=1)[:, None]

    # sums of squares
    x_ss = np.einsum('ij, ij -> i', x_mdiffs, x_mdiffs)
    y_ss = np.einsum('ij, ij -> i', y_mdiffs, y_mdiffs)

    # pearson correlations
    r_boots = np.einsum('ij, ij -> i', x_mdiffs, y_mdiffs) / np.sqrt(x_ss * y_ss)

    # upper and lower bounds for confidence interval
    ci_low = np.percentile(r_boots, (100 - ci) / 2)
    ci_high = np.percentile(r_boots, (ci + 100) / 2)
    return ci_low, ci_high

bennylp给出的答案大部分是正确的，但在计算第三个函数中的临界值时有一个小误差。
它应该改为：

def r_confidence_interval(r, alpha, n):
    z = r_to_z(r)
    se = 1.0 / math.sqrt(n - 3)
    z_crit = stats.norm.ppf((1 + alpha)/2)  # 2-tailed z critical value

    lo = z - z_crit * se
    hi = z + z_crit * se

    # Return a sequence
    return (z_to_r(lo), z_to_r(hi))

这里有另一个帖子供参考：Scipy - two tail ppf function for a z value?

我知道上面已经建议了自举，下面提出了它的另一种变体，它可能更适合一些其他的设置。

#1对数据进行抽样（成对的X和Y，也可以添加其他权重），拟合原始模型，记录r2，附加它。然后从记录的所有R2的分布中提取置信区间。
#2此外，可以拟合抽样数据，并使用抽样数据模型预测非抽样X （也可以提供连续范围来扩展预测，而不是使用原始X），以获得Y帽的置信区间。

因此，在示例代码中：

import numpy as np
from scipy.optimize import curve_fit
import pandas as pd
from sklearn.metrics import r2_score

x = np.array([your numbers here])
y = np.array([your numbers here])

### define list for R2 values

r2s = []

### define dataframe to append your bootstrapped fits for Y hat ranges

ci_df = pd.DataFrame({'x': x})

### define how many samples you want

how_many_straps = 5000

### define your fit function/s

def func_exponential(x,a,b):
    return np.exp(b) * np.exp(a * x)

### fit original, using log because fitting exponential

polyfit_original = np.polyfit(x
                              ,np.log(y)
                              ,1
                              ,# w= could supply weight for observations here)
                              )

for i in range(how_many_straps+1):

    ### zip into tuples attaching X to Y, can combine more variables as well
    zipped_for_boot = pd.Series(tuple(zip(x,y)))

    ### sample zipped X & Y pairs above with replacement
    zipped_resampled = zipped_for_boot.sample(frac=1, 
                                              replace=True)

    ### creater your sampled X & Y 
    boot_x = []
    boot_y = []

    for sample in zipped_resampled:
        boot_x.append(sample[0])
        boot_y.append(sample[1])

    ### predict sampled using original fit
    y_hat_boot_via_original_fit = func_exponential(np.asarray(boot_x),
                                                   polyfit_original[0], 
                                                   polyfit_original[1])       

    ### calculate r2 and append
    r2s.append(r2_score(boot_y,  y_hat_boot_via_original_fit))

    ### fit sampled
    polyfit_boot = np.polyfit(boot_x
                              ,np.log(boot_y)
                              ,1
                              ,# w= could supply weight for observations here)
                              )

    ### predict original via sampled fit or on a range of min(x) to Z
    y_hat_original_via_sampled_fit = func_exponential(x,
                                                      polyfit_boot[0], 
                                                      polyfit_boot[1])     

    ### insert y hat into dataframe for calculating y hat confidence intervals
    ci_df["trial_" + str(i)] = y_hat_original_via_sampled_fit

### R2 conf interval

low = round(pd.Series(r2s).quantile([0.025, 0.975]).tolist()[0],3)
up = round(pd.Series(r2s).quantile([0.025, 0.975]).tolist()[1],3)
F"r2 confidence interval = {low} - {up}"