title: Is there any method to calculate confident score of a random Forest Regression (Continuous output) prediction in python?
body: <p>I'm currently predicting the waiting time of a person at a location.</p>
  
  <p>Test Input:</p>
  
  <pre><code>Location_Name  Arrival_Time
  NSP            10:21:2019 05:10:01 PM
  </code></pre>
  
  <p>Output:</p>
  
  <pre><code>Wait_Time: 1 hr 32 min 27 sec
  </code></pre>
  
  <p>What I wish now, is to tell how confident(%) I am that this predicted output is correct.</p>
  
  <p>In order to achieve it, I have tried Confident Intervals and score concept on Random Forest regressor but not able to achieve it. </p>
  
  <p>predict_proba() function is not callable for regressor. So, I'm stuck with it.</p>
  
  <pre><code>#Python problem
  
  print('------- Fitting the model ---------')
  r_leg = RandomForestRegressor(max_depth=3, random_state=32,n_estimators=200,oob_score=True)
  
  r_leg.fit(X_train,Y_train)
  
  #define function to calculate rmse and print score
  def rmse(x,y): return math.sqrt(((x-y)**2).mean())
  
  def print_score(m):
    res = [rmse(m.predict(X_train), Y_train.to_numpy()),m.score(X_train, Y_train.to_numpy())]
  #        if hasattr(m, 'oob_score_'): res.append(m.oob_score_)
    print(res)
  
  print_score(r_leg)
  
  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, m-h, m+h
  
  mean,LB,UB = mean_confidence_interval(Y_train.DayDiff)
  </code></pre>
  
  <p>The output should return the confident percentage like,</p>
  
  <pre><code>"I'm 80% sure that the person will wait at NSP for 1 hr 32 min 27 sec"
  </code></pre>