title: Replacing punctuation in a data frame based on punctuation list
body: <p>Using Canopy and Pandas, I have data frame a which is defined by:</p>
  
  <pre><code>a=pd.read_csv('text.txt')
  
  df=pd.DataFrame(a)
  
  df.columns=["test"]
  </code></pre>
  
  <p>test.txt is a single column file that contains a list of string that contains text, numerical and punctuation. </p>
  
  <p>Assuming df looks like:</p>
  
  <hr>
  
  <blockquote>
    <p>test</p>
    
    <p>%hgh&amp;12</p>
    
    <p>abc123!!!</p>
    
    <p>porkyfries</p>
  </blockquote>
  
  <hr>
  
  <p>I want my results to be:</p>
  
  <hr>
  
  <blockquote>
    <p>test</p>
    
    <p>hgh12</p>
    
    <p>abc123</p>
    
    <p>porkyfries</p>
  </blockquote>
  
  <hr>
  
  <p>Effort so far:</p>
  
  <pre><code>from string import punctuation /-- import punctuation list from python itself
  
  a=pd.read_csv('text.txt')
  
  df=pd.DataFrame(a)
  
  df.columns=["test"] /-- define the dataframe
  
  
  for p in list(punctuation):
  
       ...:     df2=df.med.str.replace(p,'')
  
       ...:     df2=pd.DataFrame(df2);
  
       ...:     df2
  </code></pre>
  
  <p>The command above basically just returns me with the same data set.
   Appreciate any leads.</p>
  
  <p>Edit: Reason why I am using Pandas is because data is huge, spanning to bout 1M rows, and future usage of the coding will be applied to list that go up to 30M rows.
  Long story short, I need to clean data in a very efficient manner for big data sets.</p>