title: How to find the next Max Value in an array
body: <p>I want to find the next max value after getting the first Max value in the same array for the following:</p>
  
  <pre><code>             Open    High     Low   Close    Volume
  Date                                                
  2017-07-03  370.24  371.35  351.50  352.62   6305401
  2017-07-05  347.20  347.24  326.33  327.09  17046701
  2017-07-06  317.26  320.79  306.30  308.83  19324495
  2017-07-07  313.50  317.00  307.38  313.22  14176915
  2017-07-10  312.90  317.94  303.13  316.05  13820889
  2017-07-11  316.00  327.28  314.30  327.22  11559402
  2017-07-12  330.40  333.10  324.50  329.52  10346127
  2017-07-13  330.11  331.60  319.97  323.41   8594466
  2017-07-14  323.19  328.42  321.22  327.78   5625211
  2017-07-17  325.54  327.10  313.45  319.57   9876912
  2017-07-18  317.50  329.13  315.66  328.24   6373720
  2017-07-19  328.23  331.65  323.22  325.26   6357014
  2017-07-20  326.90  330.22  324.20  329.92   5166188
  2017-07-21  329.46  331.26  325.80  328.40   4901606
  
  import datetime as dt
  from datetime import timedelta as td
  import matplotlib.pyplot as plt
  from matplotlib import style
  import pandas as pd
  import pandas_datareader.data as web
  import numpy as np
  
  start = dt.datetime(2017, 7, 1)
  # retrieving data from google
  df = web.DataReader('TSLA', 'google', start)
  Dates = df.index
  Highs = df['High'] # Getting only the values from the 'High' Column.
  Highest_high = np.amax(Highs)  # returns the Highest value
  &gt; 371.35 #Output
  
  Highests_index = Highs.argmax()  # returns the index of Highest value
  &gt; 2017-07-03 00:00:00 #Output
  Highest_high_2 = ? # Highest High excluding the previous high discovered.
  Highest_index_2 = ? # shall have index of the above Highest_high_2.
  </code></pre>
  
  <p>How to find <code>Highest_high_2</code> and <code>Highest_index_2</code>?</p>
  
  <p>I used numpy's argmax() to find the index of of largest value and amax to find the largest value alone.</p>
  
  <p>Any help will be appreciated.</p>