title: Open specific excel file/sheet to get a specific content cell location
body: <p>I have a list of excel files and their corresponding sheet number. I need python to go to those sheets and find out the cell location for a particular content. Can someone point out the error in my code? Thanks in advance</p>
  
  <pre><code>import xlrd
  value = 'Avg.'
  filename = ('C:/002 Av SW W of 065 St SW 2011-Jul-05.xls', 'C:/003 Avenue SW West of 058 Street SW 2012-Jun-23.xls')
  sheetnumber = ('505840', '505608')
  dictionary = dict(zip(filename, sheetnumber))
  for item in dictionary:
      book = xlrd.open_workbook(item)
      sheet = book.sheet_by_name(dictionary[key])
      for row in range(sheet.nrows):
          for column in range(sheet.ncols):
              if sheet.cell(row,column).value == value:
                  print row, column
  </code></pre>